❉❡❛r r❡❛❞❡rs ✱ ❚❤✐s ❞♦❝✉♠❡♥t ✇✐❧❧ ❤❡❧♣ ②♦✉ ❢♦r ②♦✉r ♣r❡♣❛r❛t✐♦♥ ♦❢ ■▼❖ ■♥t❡r♥❛t✐♦♥❛❧ ▼❛t❤❡♠❛t✐❝❛❧ ❖❧②♠♣✐❛❞ ✱ ◆❖ ◆❛t✐♦♥❛❧ ❖❧②♠♣✐❛❞✳ ■t ❝♦♥t❛✐♥s ✶✻✾ ❢✉♥❝t✐♦♥❛❧ ❡q✉❛t✐♦♥s ✇✐t❤ t❤❡ s♦❧✉t✐♦♥s ♦❢ P❛tr✐❝❦ ✧♣❝♦✧ ✳ ▼❛♥② t❤❛♥❦s t♦ P❛tr✐❝❦ ❢♦r ✐ts s♦❧✉t✐♦♥s ♦♥ ▼❛t❤❧✐♥❦s ✱ ✐t ✇✐❧❧ ❤❡❧♣ st✉❞❡♥ts ❢♦r ■▼❖✳ ▼♦✉❜✐♥♦♦❧✳ : x = ❛♥❞ ✶✳ ❉❡t❡r♠✐♥❡ ❛❧❧ ❢✉♥❝t✐♦♥s f : R → R s✉❝❤ t❤❛t t❤❡ s❡t { f (x) x x ∈ R} ✐s ✜♥✐t❡✱ ❛♥❞ ❢♦r ❛❧❧ x ∈ R✱ f (x − − f (x)) = f (x) − x − solution ▲❡t P (x) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (x − − f (x)) = f (x) − x − ▲❡t a ∈ R ❛♥❞ b = f (a) P (a) =⇒ f (a − b − 1) = b − a − P (a − b − 1) =⇒ f (2(a − b) − 1) = 2(b − a) − ❆♥❞ ✇❡ ❣❡t ❡❛s✐❧② f (2n (a − b) − 1) = 2n (b − a) − ∀n ∈ N n ■t✬s t❤❡♥ ✐♠♠❡❞✐❛t❡ t♦ s❡❡ t❤❛t t❤❡ s❡t { f (x) x : x = ❛♥❞ x = (a − b) − ∀n ∈ N} ✐s ✜♥✐t❡ ✐✛ b = a ⇐⇒ f (a) = a ❍❡♥❝❡ t❤❡ ✉♥✐q✉❡ s♦❧✉t✐♦♥ f (x) = x ∀x ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥ ✷✳ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s f : R → R s✉❝❤ t❤❛t ❢♦r ❛❧❧ x, y ∈ R✱ f (f (y + f (x))) = f (x + y) + f (x) + y solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (f (y + f (x))) = f (x + y) + f (x) + y P (x, f (y)) =⇒ f (f (f (x) + f (y))) = f (x + f (y)) + f (x) + f (y) P (y, f (x)) =⇒ f (f (f (x) + f (y))) = f (y + f (x)) + f (x) + f (y) ❙✉❜tr❛❝t✐♥❣✱ ✇❡ ❣❡t f (x + f (y)) = f (y + f (x)) ❙♦ f (f (x + f (y))) = f (f (y + f (x))) ❙♦ ✭✉s✐♥❣ P (x, y) ❛♥❞ P (y, x)✮ ✿ f (x + y) + f (y) + x = f (x + y) + f (x) + y ❙♦ f (x) − x = f (y) − y ❛♥❞ s♦ f (x) = x + a✱ ✇❤✐❝❤ ✐s ♥❡✈❡r ❛ s♦❧✉t✐♦♥✳ f (f (y + f (x))) = f (x + y) + f (x) + y ✸✳ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s f : R+ → R+ s✉❝❤ t❤❛t f (1 + xf (y)) = yf (x + y) ❢♦r ❛❧❧ x, y ∈ R+ solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (1 + xf (y)) = yf (x + y) ✶✮ f (x) ✐s ❛ s✉r❥❡❝t✐✈❡ ❢✉♥❝t✐♦♥ ❂❂❂❂ P ( f (2) = f( f (2) x ) , f (2) x ) =⇒ f (2) f (2) f ( f (2) + ) x x f( ) x ✶ ❆♥❞ s♦ x = f (something) ◗✳❊✳❉✳ ✷✮ f (x) ✐s ❛♥ ✐♥❥❡❝t✐✈❡ ❢✉♥❝t✐♦♥ ❂❂❂ ▲❡t a > b > s✉❝❤ t❤❛t f (a) = f (b) ▲❡t T = b − a > ❈♦♠♣❛r✐♥❣ P (x, a) ❛♥❞ P (x, b)✱ ✇❡ ❣❡t af (x + a) = bf (x + b) ❛♥❞ s♦ f (x) = ab f (x + T ) ∀x > a ❆♥❞ s♦ f (x) = b n a f (x + nT ) ∀x > a, n ∈ N ▲❡t t❤❡♥ y s✉❝❤ t❤❛t f (y) > ✭s✉❝❤ y ❡①✐sts s✐♥❝❡ f (x) ✐s ❛ s✉r❥❡❝t✐♦♥✱ ❛❝❝♦r❞✐♥❣ t♦ ✶✮ ❛❜♦✈❡✮ ▲❡t n ❣r❡❛t ❡♥♦✉❣❤ t♦ ❤❛✈❡ y + nT − > yf (y)+(nT −1)f (y) −1 −1 ) = yf ( y+nT P ( y+nT f (y)−1 , y) =⇒ f (1 + f (y)−1 f (y)−1 + y) ✇❤✐❝❤ ♠❛② ❜❡ ✇r✐tt❡♥ ✿ −1 −1 f ( yf (y)+nT + nT ) = yf ( yf (y)+nT ) f (y)−1 f (y)−1 a n b −1 + nT ) = ❛♥❞ s✐♥❝❡ f ( yf (y)+nT f (y)−1 ∀n✱ ✇❤✐❝❤ ✐s ✐♠♣♦ss✐❜❧❡ ◗✳❊✳❉✳ −1 f ( yf (y)+nT )✱ ✇❡ ❣❡t y = f (y)−1 a n b ✸✮ f (1) = ❂❂❂ P (1, 1) =⇒ f (1 + f (1)) = f (2) ❛♥❞ s♦✱ s✐♥❝❡ f (x) ✐s ✐♥❥❡❝t✐✈❡✱ f (1) = ◗✳❊✳❉✳ ✹✮ ❚❤❡ ♦♥❧② s♦❧✉t✐♦♥ ✐s f (x) = x1 ❂❂❂ P (1, x) =⇒ f (1+f (x)) = xf (1+x) ❛♥❞ s♦ f (1 + x) = x1 f (1 + f (x)) P ( f (x1 ) , x1 ) =⇒ f (1 + x) = x1 f ( f (x1 ) + x1 ) x x ❆♥❞ s♦ ✭❝♦♠♣❛r✐♥❣ t❤❡s❡ t✇♦ ❧✐♥❡s✮ ✿ f (1 + f (x)) = f ( f (x1 ) x1 ) x ❆♥❞ s♦ ✭✉s✐♥❣ ✐♥❥❡❝t✐✈✐t②✮ ✿ + f (x) = x f( x ) ❚❤✐s ✐♠♣❧✐❡s ✭❝❤❛♥❣✐♥❣ x → x1 ✮ ✿ f (x) = ❆♥❞ s♦ f (x) = x x f (x)+1− x + x ❛♥❞ s♦ f ( x1 ) = x f (x)+1− x x f( x )+1−x +1−x ❲❤✐❝❤ ❣✐✈❡s x2 f (x)2 − 2xf (x) + = ❆♥❞ s♦ f (x) = ✱ ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥ x ✹✳ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s f : R → R s❛t✐s❢②✐♥❣ t❤❡ ❡q✉❛❧✐t② f (y) + f (x + f (y)) = y + f (f (x) + f (f (y))) solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (y) + f (x + f (y)) = y + f (f (x) + f (f (y))) P (f (x), 0) =⇒ f (0) + f (f (x) + f (0)) = f (f (f (x)) + f (f (0))) P (f (0), x) =⇒ f (x) + f (f (x) + f (0)) = x + f (f (f (x)) + f (f (0))) ❙✉❜tr❛❝t✐♥❣✱ ✇❡ ❣❡t f (x) = x + f (0) P❧✉❣❣✐♥❣ ❜❛❝❦ f (x) = x + a ✐♥ ♦r✐❣✐♥❛❧ ❡q✉❛t✐♦♥✱ ✇❡ ❣❡t a = ❛♥❞ t❤❡ ✉♥✐q✉❡ s♦❧✉t✐♦♥ f (x) = x∀x ✷ ✺✳ ❋✐♥❞ ❛❧❧ ♥♦♥✲❝♦♥st❛♥t r❡❛❧ ♣♦❧②♥♦♠✐❛❧s f (x) s✉❝❤ t❤❛t ❢♦r ❛♥② r❡❛❧ x t❤❡ ❢♦❧❧♦✇✐♥❣ ❡q✉❛❧✐t② ❤♦❧❞s✿ f (sinx + cosx) = f (sinx) + f (cosx) solution ■❢ f (x) ✐s ♥♦♥ ❝♦♥st❛♥t✱ ❧❡t n > ✐ts ❞❡❣r❡❡ ❛♥❞ ❲❧♦❣ ❝♦♥s✐❞❡r f (x) ✐s ♠♦♥✐❝✳ ❯s✐♥❣ ❤❛❧❢✲t❛♥❣❡♥t✱ t❤❡ ❡q✉❛t✐♦♥ ♠❛② ❜❡ ✇r✐tt❡♥ f +f 1−x2 1+x2 1+2x−x2 1+x2 =f 2x 1+x2 ∀x ▼✉❧t✐♣❧②✐♥❣ ❜② (1 + x2 )n ✱ ❛♥❞ s❡tt✐♥❣ t❤❡♥ x = i✱ ✇❡ ❣❡t (2 + 2i)n = (2i)n + 2n ❛♥❞ s♦ n = ✭❧♦♦❦ ❛t ♠♦❞✉❧✉s✮✳ ❍❡♥❝❡ t❤❡ s♦❧✉t✐♦♥s✿ f (x) = ax ∀a ∈ R∗ ✻✳ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s f : N → Z s✉❝❤ t❤❛t ❢♦r ❛❧❧ x, y ∈ N ❤♦❧❞s f (x+|f (y)|) = x + f (y) solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (x + |f (y)|) = x + f (y) ■❢ |f (a)| < a ❢♦r s♦♠❡ a ∈ N✱ t❤❡♥ P (a − |f (a)|, a) =⇒ |f (a)| = a ❛♥❞ s♦ ❝♦♥tr❛❞✐❝t✐♦♥✳ ❙♦ |f (x)| ≥ x ∀x ∈ N ■❢ f (a) < ❢♦r s♦♠❡ a ∈ N✱ t❤❡♥ P (−f (a), a) =⇒ f (−2f (a)) = ❛♥❞ s♦ ❝♦♥tr❛❞✐❝t✐♦♥ ✇✐t❤ f (x) ≥ x ∀x ∈ N ❙♦ f (x) ≥ ∀x ∈ N ❆s ❛ ❝♦♥s❡q✉❡♥❝❡ |f (x)| = f (x) ❛♥❞ t❤❡ ♣r♦❜❧❡♠ ❜❡❝♦♠❡s ✿ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s f : N → N ∪ {0} s✉❝❤ t❤❛t f (x + f (y)) = x + f (y) ∀x, y ∈ N ▲❡t t❤❡♥ m = min(f (N)) ❛♥❞ ✇❡ ❣❡t f (x) = x ∀x > m ❬❍❡♥❝❡ t❤❡ s♦❧✉t✐♦♥s ▲❡t a ∈ N f (x) = x ∀x ≥ a f (x) ❝❛♥ t❛❦❡ ❛♥② ✈❛❧✉❡ ✐♥ [a − 1, +∞) ❢♦r x ∈ [1, a − 1] ✼✳ ❉❡t❡r♠✐♥❡ ❛❧❧ ♣❛✐rs ♦❢ ❢✉♥❝t✐♦♥s f, g : Q → Q s❛t✐s❢②✐♥❣ t❤❡ ❢♦❧❧♦✇✐♥❣ ❡q✉❛❧✐t② f (x + g(y)) = g(x) + 2y + f (y), ❢♦r ❛❧❧ x, y ∈ Q solution ■❢ f (x) ✐s ❛ s♦❧✉t✐♦♥✱ t❤❡♥ s♦ ✐s f (x) + c✳ ❙♦ ❲❧♦❣ ❝♦♥s✐❞❡r t❤❛t f (0) = ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (x + g(y)) = g(x) + 2y + f (y) P (−g(0), 0) =⇒ g(−g(0)) = P (−g(0), −g(0)) =⇒ g(0) = P (x, 0) =⇒ f (x) = g(x) ✸ ❙♦ ✇❡ ❛r❡ ❧♦♦❦✐♥❣ ❢♦r f (x) s✉❝❤ t❤❛t f (0) = ❛♥❞ f (x + f (y)) = f (x) + 2y + f (y) ▲❡t Q(x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (x + f (y)) = f (x) + 2y + f (y) Q(x − f (x), x) =⇒ f (x − f (x)) = −2x ❛♥❞ s♦ f (x) ✐s s✉r❥❡❝t✐✈❡ Q(x, y) =⇒ f (x + f (y)) = f (x) + 2y + f (y) Q(0, y) =⇒ f (f (y)) = 2y + f (y) ❙✉❜tr❛❝t✐♥❣✱ ✇❡ ❣❡t f (x + f (y)) = f (x) + f (f (y)) ❛♥❞✱ s✐♥❝❡ s✉r❥❡❝t✐✈❡ ✿ f (x + y) = f (x) + f (y) ❙✐♥❝❡ f (x) ✐s ❢r♦♠ Q → Q✱ t❤✐s ✐♠♠❡❞✐❛t❡❧② ❣✐✈❡s f (x) = ax ❛♥❞✱ ♣❧✉❣❣✐♥❣ t❤✐s ✐♥ Q(x, y) ✿ a2 − a − = ❍❡♥❝❡ t❤❡ t✇♦ s♦❧✉t✐♦♥s ✿ f (x) = 2x + c ❛♥❞ g(x) = 2x ∀x ❛♥❞ ❢♦r ❛♥② r❡❛❧ c✱ ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥ ✽✳ ●✐✈❡♥ t✇♦ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs a ❛♥❞ b✱ s✉♣♣♦s❡ t❤❛t ❛ ♠❛♣♣✐♥❣ f : R+ → R+ s❛t✐s✜❡s t❤❡ ❢✉♥❝t✐♦♥❛❧ ❡q✉❛t✐♦♥ f (f (x)) + af (x) = b(a + b)x Pr♦✈❡ t❤❛t t❤❡r❡ ❡①✐sts ❛ ✉♥✐q✉❡ s♦❧✉t✐♦♥ ♦❢ t❤✐s ❡q✉❛t✐♦♥✳ solution a+2b > ❛♥❞ ✇❡ ❣❡t t❤r✉ s✐♠♣❧❡ ✐♥❞✉❝t✐♦♥ ✿ f [n] (x) = ((a+b)x+f (x))bn +(bx−f (x))(−a−b)n a+2b ■❢✱ ❢♦r s♦♠❡ x✱ f (x) − bx = 0✱ ✇❡ ❣❡t t❤❛t✱ ❢♦r s♦♠❡ n ❣r❡❛t ❡♥♦✉❣❤✱ f [n] (x) < 0✱ ✇❤✐❝❤ ✐s ✐♠♣♦ss✐❜❧❡✳ ❍❡♥❝❡ t❤❡ ✉♥✐q✉❡ s♦❧✉t✐♦♥ ✿ f (x) = bx ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥ ✾✳ ❋✐♥❞ ❛❧❧ ♥♦♥✲❝♦♥st❛♥t ❢✉♥❝t✐♦♥s f : Z → N s❛t✐s❢②✐♥❣ ❛❧❧ ♦❢ t❤❡ ❢♦❧❧♦✇✐♥❣ ❝♦♥❞✐t✐♦♥s✿ ❛✮f (x − y) + f (y − z) + f (z − x) = 3(f (x) + f (y) + f (z)) − 15 f (x + y + z) ❜✮ k=1 f (k) ≤ 1995 solution ❙❡tt✐♥❣ x = y = z = ✐♥ t❤❡ ❡q✉❛t✐♦♥✱ ✇❡ ❣❡t f (0) = ∈ / N ❛♥❞ s♦ ♥♦ s♦❧✉t✐♦♥ ❙✐♥❝❡ ❖P ✐s ❛ ❜r❛♥❞ ♥❡✇ ✉s❡r ♦♥ t❤✐s ❢♦r✉♠✱ ■✬❧❧ ❝♦♥s✐❞❡r t❤❛t ❤❡ ✐❣♥♦r❡❞ t❤❛t ✇❡ ✉s❡ ❤❡r❡ t❤❡ ♥♦t❛t✐♦♥ N ❢♦r ♣♦s✐t✐✈❡ ✐♥t❡❣❡rs ❛♥❞ t❤❛t ❤❡ ♠❡❛♥t N0 ✱ s❡t ♦❢ ❛❧❧ ♥♦♥ ♥❡❣❛t✐✈❡ ✐♥t❡❣❡rs✳ ■❢ s♦ ✿ ▲❡t P (x, y, z) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (x − y) + f (y − z) + f (z − x) = 3(f (x) + f (y) + f (z)) − f (x + y + z) P (0, 0, 0) =⇒ f (0) = P (x, 0, 0) =⇒ f (−x) = f (x) P (x, −x, 0) =⇒ f (2x) = 4f (x) P (x+1, −1, −x−1) =⇒ f (x+2) = 2f (x+1)−f (x)+2f (1) ❚❤✐s r❡❝✉rr❡♥❝❡ ❞❡✜♥✐t✐♦♥ ✭♣❧✉s f (0) = 0✮ ✐s q✉✐t❡ ❝❧❛ss✐❝❛❧ ❛♥❞ ❤❛s s✐♠♣❧❡ ❣❡♥❡r❛❧ s♦❧✉t✐♦♥ f (x) = ax2 f (x) ∈ N0 ∀x ∈ Z =⇒ a ≥ f (x) ♥♦♥ ❝♦♥st❛♥t =⇒ a > 15 a k=1 k = 1240a ≤ 1995 =⇒ a ≤ 15 k=1 f (k) = ❬✉❪❬❜❪❍❡♥❝❡ t❤❡ ✉♥✐q✉❡ s♦❧✉t✐♦♥ ♦❢ t❤❡ ♠♦❞✐✜❡❞ ♣r♦❜❧❡♠❬✴❜❪❬✴✉❪ ✿ f (x) = x2 ∀x✱ ✹ ✶✵✳ ✳❉❡t❡r♠✐♥❡ ❛❧❧ t❤❡ ❢✉♥❝t✐♦♥s f : R → R s✉❝❤ t❤❛t✿ f (x + yf (x)) + f (xf (y) − y) = f (x) − f (y) + 2xy ❍❡r❡ ✐s ❛ r❛t❤❡r ❤❡❛✈② solution ✿ ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (x+yf (x))+f (xf (y)−y) = f (x)−f (y)+2xy ✶✮ f (x) ✐s ❛♥ ♦❞❞ ❢✉♥❝t✐♦♥ ❛♥❞ f (x) = ⇐⇒ x = ❂❂ P (0, 0) =⇒ f (0) = P (0, x) =⇒ f (−x) = −f (x) ❙✉♣♣♦s❡ f (a) = 0✳ ❚❤❡♥ P (a, a) =⇒ = 2a2 =⇒ a = ❛♥❞ s♦ f (x) = ⇐⇒ x = ◗✳❊✳❉ ✷✮ f (x) ✐s ❛❞❞✐t✐✈❡ ❂❂❂ ▲❡t t❤❡♥ x = s✉❝❤ t❤❛t f (x) = ✿ P (x, fx+y (x) ) x+y x+y x+y ) − ) = f (x) − f ( ) + 2x f (xf ( fx+y (x) f (x) f (x) f (x) x+y P ( fx+y (x) , −x) =⇒ −f (xf ( f (x) ) − x+y f (x) ) =⇒ f (2x + y) + x+y − f (y) = f ( fx+y (x) ) + f (x) − 2x f (x) ❆❞❞✐♥❣ t❤❡s❡ t✇♦ ❧✐♥❡s✱ ✇❡ ❣❡t ✿ f (2x+y) = 2f (x)+f (y) ✇❤✐❝❤ ✐s ♦❜✈✐♦✉s❧② st✐❧❧ tr✉❡ ❢♦r x = ❛♥❞ s♦ ✿ ◆❡✇ ❛ss❡rt✐♦♥ Q(x, y) ✿ f (2x + y) = 2f (x) + f (y) ∀x, y Q(x, 0) =⇒ f (2x) = 2f (x) ❛♥❞ s♦ Q(x, y) ❜❡❝♦♠❡s f (2x + y) = f (2x) + f (y) ❛♥❞ s♦ f (x + y) = f (x) + f (y) ❛♥❞ f (x) ✐s ❛❞❞✐t✐✈❡✳ ◗✳❊✳❉✳ ✸✮ f (x) s♦❧✉t✐♦♥ ✐♠♣❧✐❡s −f (x) s♦❧✉t✐♦♥ ❛♥❞ s♦ ✇❧♦❣ ❝♦♥s✐❞❡r ❢r♦♠ ♥♦✇ f (1) ≥ ❂❂❂❂ P (y, x) =⇒ f (y + xf (y)) + f (yf (x) − x) = f (y) − f (x) + 2xy =⇒ −f (−y + x(−f (y))) − f (y(−f (x)) + x) = −f (x) − (−f (y)) + 2xy ◗✳❊✳❉ ✹✮ f (x) ✐s ❜✐❥❡❝t✐✈❡ ❛♥❞ f (1) = ❂❂❂❂ ❯s✐♥❣ ❛❞❞✐t✐✈❡ ♣r♦♣❡rt②✱ t❤❡ ♦r✐❣✐♥❛❧ ❛ss❡rt✐♦♥ ❜❡❝♦♠❡s R(x, y) ✿ f (xf (y))+ f (yf (x)) = 2xy R(x, 21 ) =⇒ f (xf ( 21 ) + f (x) ) = x ❛♥❞ f (x) ✐s s✉r❥❡❝t✐✈❡✳ ❙♦ ∃a s✉❝❤ t❤❛t f (a) = ❚❤❡♥ R(a, a) =⇒ a2 = ❛♥❞ s♦ a = ✭r❡♠❡♠❜❡r t❤❛t ✐♥ ✸✮ ✇❡ ❝❤♦♦s❡❞ f (1) ≥ 0✮ ✺✮ f (x) = x ❂❂❂❂ R(x, 1) =⇒ f (x)+f (f (x)) = 2x ❛♥❞ s♦ f (x) ✐s ✐♥❥❡❝t✐✈❡✱ ❛♥❞ s♦ ❜✐❥❡❝t✐✈❡✳ R(xf (x), 1) =⇒ f (xf (x))+f (f (xf (x))) = 2xf (x) R(x, x) =⇒ f (xf (x)) = x2 ❛♥❞ s♦ f (x2 ) = f (f (xf (x))) ❈♦♠❜✐♥✐♥❣ t❤❡s❡ t✇♦ ❧✐♥❡s✱ ✇❡ ❣❡t f (x2 ) + x2 = 2xf (x) ❙♦ f ((x+y)2 )+(x+y)2 = 2(x+y)f (x+y) ❛♥❞ s♦ f (xy)+xy = xf (y)+yf (x) ✺ ❙♦ ✇❡ ❤❛✈❡ t❤❡ ♣r♦♣❡rt✐❡s ✿ R(x, y) ✿ f (xf (y)) + f (yf (x)) = 2xy A(x, y) ✿ f (xy) = xf (y) + yf (x) − xy B(x) ✿ f (f (x)) = 2x − f (x) ❙♦ ✿ ✭❛✮ ✿ R(x, x) =⇒ f (xf (x)) = x2 ✭❜✮ ✿ A(x, f (x)) =⇒ f (xf (x)) = xf (f (x)) + f (x)2 − xf (x) ✭❝✮ ✿ B(x) =⇒ f (f (x)) = 2x − f (x) ❆♥❞ s♦ ✲✭❛✮✰✭❜✮✰①✭❝✮ ✿ = x2 + f (x)2 − 2xf (x) = (f (x) − x)2 ◗✳❊✳❉✳ ✻✮ s②♥t❤❡s✐s ♦❢ s♦❧✉t✐♦♥s ❂❂❂❂ ❯s✐♥❣ ✸✮ ❛♥❞ ✺✮✱ ✇❡ ❣❡t t✇♦ s♦❧✉t✐♦♥s ✭✐t✬s ❡❛s② t♦ ❝❤❡❝❦ ❜❛❝❦ t❤❛t t❤❡s❡ t✇♦ ❢✉♥❝t✐♦♥s ✐♥❞❡❡❞ ❛r❡ s♦❧✉t✐♦♥s✮ ✿ f (x) = x ∀x f (x) = −x ∀x❬✴q✉♦t❡❪ ✶✶✳ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s f ❞❡✜♥❡❞ ♦♥ r❡❛❧ ♥✉♠❜❡rs ❛♥❞ t❛❦✐♥❣ r❡❛❧ ✈❛❧✉❡s s✉❝❤ t❤❛t f (x)2 + 2yf (x) + f (y) = f (y + f (x)) ❢♦r ❛❧❧ r❡❛❧ ♥✉♠❜❡rs x, y ❬ solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (x)2 + 2yf (x) + f (y) = f (y + f (x)) f (x) = ∀x ✐s ❛ s♦❧✉t✐♦♥✳ ❙♦ ✇❡✬❧❧ ❧♦♦❦ ❢r♦♠ ♥♦✇ ❢♦r ♥♦♥ ❛❧❧✲③❡r♦ s♦❧✉t✐♦♥s✳ (a) ▲❡t f (a) = ✿ P (a, u−f 2f (a) ) =⇒ u = f (s♦♠❡t❤✐♥❣) − f (s♦♠❡t❤✐♥❣ ❡❧s❡) ❛♥❞ s♦ ❛♥② r❡❛❧ ♠❛② ❜❡ ✇r✐tt❡♥ ❛s ❛ ❞✐✛❡r❡♥❝❡ f (v) − f (w) P (w, −f (w)) =⇒ −f (w)2 + f (−f (w)) = f (0) P (v, −f (w)) =⇒ f (v)2 − 2f (v)f (w) + f (−f (w)) = f (f (v) − f (w)) ❙✉❜tr❛❝t✐♥❣ t❤❡ ✜rst ❢r♦♠ t❤❡ s❡❝♦♥❞ ✐♠♣❧✐❡s f (v)2 − 2f (v)f (w) + f (w)2 = f (f (v) − f (w)) − f (0) ❛♥❞ s♦ f (f (v) − f (w)) = (f (v) − f (w))2 + f (0) ❆♥❞ s♦ f (x) = x2 + f (0) ∀x ∈ R ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥✳ ❍❡♥❝❡ t❤❡ t✇♦ s♦❧✉t✐♦♥s ✿ f (x) = ∀x f (x) = x2 + a ∀x ✶✷✳ Pr♦✈❡ t❤❛t f (x + y + xy) = f (x) + f (y) + f (xy) ✐s ❡q✉✐✈❛❧❡♥t t♦ f (x + y) = f (x) + f (y)✳ solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (x + y + xy) = f (x) + f (y) + f (xy) ✶✮ f (x + y) = f (x) + f (y) =⇒ P (x, y) ❂❂❂❂❂❂❂ ❚r✐✈✐❛❧✳ ✷✮ P (x, y) =⇒ f (x + y) = f (x) + f (y) ∀x, y ❂❂❂❂❂❂ P (x, 0) =⇒ f (0) = P (x, −1) =⇒ f (−x) = −f (x) ✷✳✶✮ ♥❡✇ ❛ss❡rt✐♦♥ R(x, y) ✿ f (x + y) = f (x) + f (y) ∀x, y s✉❝❤ t❤❛t x + y = −2 ✖✖✖✖✖✖✖✕ x−y x+y ▲❡t x, y s✉❝❤ t❤❛t x + y = −2 ✿ P ( x+y , x+y−2 ) =⇒ f (x) = f ( ) + 2 x−y x −y ) + f ( x+y−2 ) f ( x+y−2 2 y−x x+y x−y x −y P ( x+y , x+y−2 ) =⇒ f (y) = f ( ) − f ( x+y−2 ) − f ( x+y−2 ) ✻ ❆❞❞✐♥❣ t❤❡s❡ t✇♦ ❧✐♥❡s ❣✐✈❡s ♥❡✇ ❛ss❡rt✐♦♥ Q(x, y) ✿ f (x)+f (y) = 2f ( x+y ) ∀x, y s✉❝❤ t❤❛t x + y = −2 Q(x + y, 0) =⇒ f (x + y) = 2f ( x+y ) ❛♥❞ s♦ f (x + y) = f (x) + f (y) ◗✳❊✳❉✳ ✷✳✷✮ f (x + y) = f (x) + f (y) ∀x, y s✉❝❤ t❤❛t x + y = −2 ■❢ x = −2✱ t❤❡♥ y = ❛♥❞ f (x + y) = f (x) + f (y) ■❢ x = −2✱ t❤❡♥ (x + 2) + (−2) = −2 ❛♥❞ t❤❡♥ R(x + 2, −2) =⇒ f (x) = f (x + 2) + f (−2) ❛♥❞ s♦ f (x) + f (−2 − x) = f (−2) ❛♥❞ s♦ f (x) + f (y) = f (x + y) ◗✳❊✳❉✳ ✶✸✳ ✜♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s f : R −→ R s✉❝❤ t❤❛t f (f (x) + y) = 2x + f (f (y) − x) ❢♦r ❛❧❧ x, y r❡❛❧s solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (f (x) + y) = 2x + f (f (y) − x) , −f ( f (0)−x )) =⇒ x = f (f (−f ( f (0)−x )) − P ( f (0)−x 2 ✐s s✉r❥❡❝t✐✈❡✳ f (0)−x ) ❛♥❞ s♦ f (x) ❙♦ ✿ ∃u s✉❝❤ t❤❛t f (u) = ∃v s✉❝❤ t❤❛t f (v) = x + u ❆♥❞ t❤❡♥ P (u, v) =⇒ f (x) = x − u ✇❤✐❝❤ ✐♥❞❡❡❞ ✱✐s ❛ s♦❧✉t✐♦♥ ❍❡♥❝❡ t❤❡ ❛♥s✇❡r ✿ f (x) = x + c ✶✹✳ ✜♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s f : R −→ R s✉❝❤ t❤❛t f (x2 + f (y)) = y + f (x)2 ❢♦r ❛❧❧ x, y r❡❛❧s solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (x2 + f (y)) = y + f (x)2 P (0, y) =⇒ f (f (y)) = y + f (0)2 ❛♥❞ t❤❡♥ ✿ P (x, f (y − f (0)2 )) =⇒ f (x2 + y) = f (y − f (0)2 ) + f (x)2 ❙❡tt✐♥❣ x = ✐♥ t❤✐s ❧❛st ❡q✉❛❧✐t②✱ ✇❡ ❣❡t f (y) = f (y − f (0)2 ) + f (0)2 ❛♥❞ s♦ f (x2 + y) = f (y) + f (x)2 − f (0)2 ❙❡tt✐♥❣ y = ✐♥ t❤✐s ❧❛st ❡q✉❛❧✐t②✱ ✇❡ ❣❡t f (x2 ) = f (0) + f (x)2 − f (0)2 ❛♥❞ s♦ f (x2 + y) = f (y) + f (x2 ) − f (0) ▲❡t t❤❡♥ g(x) = f (x) − f (0)✳ ❲❡ ❣♦t g(x + y) = g(x) + g(y) ∀x ≥ 0, ∀y ■t✬s ✐♠♠❡❞✐❛t❡ t♦ ❡st❛❜❧✐s❤ g(0) = ❛♥❞ g(−x) = −g(x) ❛♥❞ s♦ g(x + y) = g(x) + g(y) ∀x, y P (x, 0) =⇒ f (x2 +f (0)) = f (x)2 =⇒ f (x2 +f (0))−f (0) = f (x)2 −f (0) ❛♥❞ s♦ g(x) ≥ −f (0) ∀x ≥ f (0) ❙♦ g(x) ✐s ❛ s♦❧✉t✐♦♥ ♦❢ ❈❛✉❝❤② ❡q✉❛t✐♦♥ ✇✐t❤ ❛ ❧♦✇❡r ❜♦✉♥❞ ♦♥ s♦♠❡ ♥♦♥ ❡♠♣t② ♦♣❡♥ ✐♥t❡r✈❛❧✳ ❙♦ g(x) = ax ❛♥❞ f (x) = ax + b P❧✉❣❣✐♥❣ t❤✐s ❜❛❝❦ ✐♥ ♦r✐❣✐♥❛❧ ❡q✉❛t✐♦♥✱ ✇❡ ❣❡t a = ❛♥❞ b = ❛♥❞ t❤❡ ✉♥✐q✉❡ s♦❧✉t✐♦♥ f (x) = x ✼ ✶✺✳ ❋✐♥❞ ❛❧❧ a ∈ R ❢♦r ✇❤✐❝❤ t❤❡r❡ ❡①✐sts ❛ ♥♦♥✲❝♦♥st❛♥t ❢✉♥❝t✐♦♥ f : (0, 1] → R s✉❝❤ t❤❛t a + f (x + y − xy) + f (x)f (y) ≤ f (x) + f (y) ❢♦r ❛❧❧ x, y ∈ (0, 1] solution ▲❡t g(x) ❢r♦♠ [0, 1) → R s✉❝❤ t❤❛t g(x) = f (1 − x) − a + f (x + y − xy) + f (x)f (y) ≤ f (x) + f (y) ⇐⇒ g((1 − x)(1 − y)) + g(1 − x)g(1 − y) ≤ −a ⇐⇒ g(xy) + g(x)g(y) ≤ −a ∀x, y ∈ [0, 1) ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ g(xy) + g(x)g(y) ≤ −a P (0, 0) =⇒ g(0) + g(0)2 ≤ −a ⇐⇒ a ≤ ■❢ a < 4 ✿ ▲❡t ✉s ❝♦♥s✐❞❡r g(x) = − 21 − (g(0) + 12 )2 ❛♥❞ s♦ a ≤ ∀x ∈ (0, 1) ❛♥❞ g(0) = − 21 − − a = − 12 ✭s♦ t❤❛t g(x) ✐s ♥♦t ❝♦♥st❛♥t✮ ✿ ■❢ x = y = ✿ g(xy) + g(x)g(y) = −a ≤ −a ■❢ x = ❛♥❞ y = ✿ g(xy) + g(x)g(y) = − 14 − − a < − 14 < −a ■❢ x, y = ✿ g(xy) + g(x)g(y) = − 14 < −a 1 ■❢ a = 41 ✿ P (0, 0) √=⇒√ g(0) + g(0)2 ≤ −√ ❛♥❞ s♦ g(0) = − P (x, 0) 1 =⇒ g(x) ≥ − P ( x, x) =⇒ g(x) + g( x)2 ≤ − =⇒ g(x) ≤ − 14 ▲❡t t❤❡♥ t❤❡ s❡q✉❡♥❝❡ un ❞❡✜♥❡❞ ❛s ✿ u0 = − 41 un+1 = − 14 − a2n ■t✬s ❡❛s② t♦ s❤♦✇ ✇✐t❤ ✐♥❞✉❝t✐♦♥ t❤❛t − 21 ≤ g(x) ≤ an < ∀x ∈ [0, 1) ■t✬s t❤❡♥ ❡❛s② t♦ s❤♦✇ t❤❛t an ✐s ❛ ❞❡❝r❡❛s✐♥❣ s❡q✉❡♥❝❡ ✇❤♦s❡ ❧✐♠✐t ✐s − 21 ❆♥❞ s♦ t❤❡ ✉♥✐q✉❡ s♦❧✉t✐♦♥ ❢♦r a = 14 ✐s g(x) = − 21 ✇❤✐❝❤ ✐s ♥♦t ❛ s♦❧✉t✐♦♥ ✭s✐♥❝❡ ❝♦♥st❛♥t✮✳ ❍❡♥❝❡ t❤❡ ❛♥s✇❡r ✿ a ∈ (−∞, ) ✶✻✳ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s f : Q → C s❛t✐s❢②✐♥❣ ✭✐✮ ❋♦r ❛♥② x1 , x2 , , x2010 ∈ Q✱ f (x1 +x2 + .+x2010 ) = f (x1 )f (x2 ) f (x2010 )✳ ✭✐✐✮ f (2010)f (x) = f (2010)f (x) ❢♦r ❛❧❧ x ∈ Q✳ ❬ solution ▲❡t a = f (0) ❯s✐♥❣ x1 = x2 = = xp = x ❛♥❞ xp+1 = = x2010 = 0✱ ✭✐✮ =⇒ f (px) = a2010−p f (x)p ∀x ∈ Q, ∀0 ≤ p ≤ 2010 ∈ Z ❙❡tt✐♥❣ x = ✐♥ t❤❡ ❛❜♦✈❡ ❡q✉❛t✐♦♥✱ ✇❡ ❣❡t a = a2010 ❛♥❞ s♦ ✿ ❊✐t❤❡r a = ❛♥❞ s♦ f (x) = ∀x✱ ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥✳ ❊✐t❤❡r a2009 = ❛♥❞ ✇❡ ❣❡t f (px) = a1−p f (x)p ▲❡t t❤❡♥ g(x) = f (x) ❛♥❞ ✇❡ ❣♦t g(px) = g(x)p ∀0 ≤ p ≤ 2010 ∈ Z ❆ a s✐♠♣❧❡ ✐♥❞✉❝t✐♦♥ ✉s✐♥❣ ✭✐✮ s❤♦✇s t❤❛t g(px) = g(x)p ∀p ∈ N ∪ {0} ❆♥❞ ✐t✬s t❤❡♥ ✐♠♠❡❞✐❛t❡ t♦ ❣❡t g( xp ) = g(x) p ❛♥❞ s♦ g(x) = cx ∀x ∈ Q ❙♦ f (x) = a · cx ✭✐✐✮ ✐♠♣❧✐❡s t❤❡♥ c = c ❛♥❞ s♦ c ∈ R ✽ ❍❡♥❝❡ t❤❡ s♦❧✉t✐♦♥s ✿ f (x) = ∀x 2kπ f (x) = ei 2009 cx ✇✐t❤ k ∈ Z ❛♥❞ c ∈ R ✭❛❝❝♦r❞✐♥❣ t♦ ♠❡✱ ❜❡tt❡r t♦ s❛② c ∈ R+ ✮ ✶✼✳ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s f : R → R✱ s❛t✐s❢②✐♥❣✿ f (x) = maxy∈R (2xy − f (y)) ❢♦r ❛❧❧ x ∈ R✳ solution ✶✮ f (x) ≥ x2 ∀x ❂❂❂❂ f (x) ≥ 2xy − f (y) ∀x, y ✳ ❈❤♦♦s✐♥❣ y = x✱ ✇❡ ❣❡t f (x) ≥ x2 ◗✳❊✳❉ ✷✮ f (x) ≤ x2 ∀x ❂❂❂❂ ▲❡t x ∈ R ❙✐♥❝❡ f (x) = maxy∈R (2xy − f (y))✱ ∃ ❛ s❡q✉❡♥❝❡ yn s✉❝❤ t❤❛t limn→+∞ (2xyn − f (yn )) = f (x) ❙♦ limn→+∞ (f (yn ) − yn2 + (x − yn )2 ) = x2 − f (x) ❆♥❞ s✐♥❝❡ ✇❡ ❦♥♦✇ t❤❛t f (yn ) − yn2 ≥ 0✱ t❤❡♥ LHS ≥ ❛♥❞ s♦ RHS ≥ ◗✳❊✳❉ ❙♦ f (x) = x2 ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥ ✶✽✳ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s f : R → R s❛t✐s❢②✐♥❣ f (f (x) + y) = f (x2 − y) + 4f (x)y ❢♦r ❛❧❧ x, y ∈ R✳ ❬ solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (f (x) + y) = f (x2 − y) + 4f (x)y (x) P (x, x −f ) =⇒ f (x)(f (x) − x2 ) = ❛♥❞ s♦ ✿ ∀x✱ ❡✐t❤❡r f (x) = 0✱ ❡✐t❤❡r f (x) = x2 f (x) = ∀x ✐s ❛ s♦❧✉t✐♦♥ f (x) = x2 ∀x ✐s ❛❧s♦ ❛ s♦❧✉t✐♦♥✳ ❙✉♣♣♦s❡ ♥♦✇ t❤❛t ∃a = s✉❝❤ t❤❛t f (a) = ❚❤❡♥ ✐❢ ∃b = s✉❝❤ t❤❛t f (b) = ✿ f (b) = b2 ❛♥❞ P (a, b) =⇒ b2 = f (a2 − b) ❛♥❞ s♦ b2 = (a2 − b)2 2 ❛♥❞ s♦ b = a2 ❙♦ t❤❡r❡ ✐s ❛ ✉♥✐q✉❡ s✉❝❤ b ✭❡q✉❛❧ t♦ a2 ✮ ❇✉t t❤❡♥ t❤❡r❡ ❛t ❛t ♠♦st t✇♦ s✉❝❤ a ✭a ❛♥❞ −a✮ ❆♥❞ ✐t ✐s ✐s ✐♠♣♦ss✐❜❧❡ t♦ ❤❛✈❡ ❛t ♠♦st ♦♥❡ x = s✉❝❤ t❤❛t f (x) = x2 ❛♥❞ ❛t ♠♦st t✇♦ x = s✉❝❤ t❤❛t f (x) = ❙♦ ✇❡ ❤❛✈❡ ♦♥❧② t✇♦ s♦❧✉t✐♦♥s ✿ f (x) = ∀x f (x) = x2 ∀x ✶✾✳ ❋✐♥❞ ❛❧❧ ❝♦♥t✐♥♦✉s ❢✉♥❝t✐♦♥s R → R s✉❝❤ t❤❛t ✿ f (x + f (y + f (z))) = f (x) + f (f (y)) + f (f (f (z))) solution ▲❡t P (x, y, z) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (x + f (y + f (z))) = f (x) + f (f (y)) + f (f (f (z))) ❙✉❜tr❛❝t✐♥❣ P (0, y − f (z), z) ❢r♦♠ P (x, y − f (z), z)✱ ✇❡ ❣❡t f (x + f (y)) = f (x) + f (f (y)) − f (0) ▲❡t g(x) = f (x) − f (0) ❛♥❞ A = f (R) ✾ ❲❡ ❣♦t g(x + y) = g(x) + g(y) ∀x ∈ R✱ ∀y ∈ A ❆♥❞ ❛❧s♦ g(x − y) = g(x) − g(y) ∀x ∈ R✱ ∀y ∈ A g(x + y1 + y2 ) = g(x + y1 ) + g(y2 ) = g(x) + g(y1 ) + g(y2 ) = g(x) + g(y1 + y2 ) ∀x ∈ R✱ ∀y1 , y2 ∈ A g(x + y1 − y2 ) = g(x + y1 ) − g(y2 ) = g(x) + g(y1 ) − g(y2 ) = g(x) + g(y1 − y2 ) ∀x ∈ R✱ ∀y1 , y2 ∈ A ❆♥❞✱ ✇✐t❤ s✐♠♣❧❡ ✐♥❞✉❝t✐♦♥✱ g(x + y) = g(x) + g(y) ∀x✱ ∀y ✜♥✐t❡ s✉♠s ❛♥❞ ❞✐✛❡r❡♥❝❡s ♦❢ ❡❧❡♠❡♥ts ♦❢ ❆ ■❢ ❝❛r❞✐♥❛❧ ♦❢ ❆ ✐s 1✱ ✇❡ ❣❡t f (x) = c ❛♥❞ s♦ f (x) = ■❢ ❝❛r❞✐♥❛❧ ♦❢ ❆ ✐s ♥♦t ❛♥❞ s✐♥❝❡ f (x) ✐s ❝♦♥t✐♥✉♦✉s✱ ∃u < v s✉❝❤ t❤❛t [u, v] ⊆ A ❛♥❞ ❛♥② r❡❛❧ ♠❛② ❜❡ r❡♣r❡s❡♥t❡❞ ❛s ✜♥✐t❡ s✉♠s ❛♥❞ ❞✐✛❡r❡♥❝❡s ♦❢ ❡❧❡♠❡♥ts ♦❢ [u, v] ❙♦ g(x + y) = g(x) + g(y) ∀x, y ❛♥❞ s♦✱ s✐♥❝❡ ❝♦♥t✐♥✉♦✉s✱ g(x) = ax ❛♥❞ f (x) = ax + b P❧✉❣❣✐♥❣ t❤✐s ✐♥ ♦r✐❣✐♥❛❧ ❡q✉❛t✐♦♥✱ ✇❡ ❣❡t b(a + 2) = ❍❡♥❝❡ t❤❡ s♦❧✉t✐♦♥s ✿ f (x) = ax f (x) = b − 2x ✷✵✳ ▲❡t a ❜❡ ❛ r❡❛❧ ♥✉♠❜❡r ❛♥❞ ❧❡t f : R → R ❜❡ ❛ ❢✉♥❝t✐♦♥ s❛t✐s❢②✐♥❣✿ f (0) = 21 ❛♥❞ f (x + y) = f (x)f (a − y) + f (y)f (a − x)✱ ∀x, y ∈ R✳ Pr♦✈❡ t❤❛t f ✐s ❝♦♥st❛♥t✳ solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (x + y) = f (x)f (a − y) + f (y)f (a − x) P (0, 0) =⇒ f (a) = 21 P (x, 0) =⇒ f (x) = f (a − x) ❛♥❞ s♦ P (x, y) ♠❛② ❛❧s♦ ❜❡ ✇r✐tt❡♥ Q(x, y) ✿ f (x + y) = 2f (x)f (y) Q(a, −x) =⇒ f (a − x) = f (−x) ❛♥❞ s♦ f (x) = f (−x) ❚❤❡♥✱ ❝♦♠♣❛r✐♥❣ Q(x, y) ❛♥❞ Q(x, −y)✱ ✇❡ ❣❡t f (x + y) = f (x − y) ❛♥❞ u−v ❝❤♦♦s✐♥❣ x = u+v ❛♥❞ y = ✱ ✇❡ ❣❡t f (u) = f (v) ✷✶✳ ❋✐♥❞ ❛❧❧ ❝♦♥t✐♥✉♦✉s ❢✉♥❝t✐♦♥s f : R → R s✉❝❤ t❤❛t f (x)3 = − x · x2 + 7x · f (x) + 16 · f (x)2 , ∀x ∈ R 12 solution ❚❤✐s ❡q✉❛t✐♦♥ ♠❛② ❜❡ ✇r✐tt❡♥ (f (x)+ x2 )2 (f (x)+ x3 ) = ❛♥❞ s♦ ✹ s♦❧✉t✐♦♥s ✿ ❙✶ ✿ f (x) = − x2 ∀x ❙✷ ✿ f (x) = − x3 ∀x ❙✸ ✿ f (x) = − x2 ∀x < ❛♥❞ f (x) = − x3 ∀x ≥ ❙✹ ✿ f (x) = − x2 ∀x > ❛♥❞ f (x) = − x3 ∀x ≤ ✶✵ ❆♥❞ s♦ P (x + y) = P (x) + P (y) + a ∀x, y ≥ ❛♥❞ ❢♦r s♦♠❡ a ∈ R ❆♥❞ s♦ P (x + y) = P (x) + P (y) + a ∀x, y ❛♥❞ ❢♦r s♦♠❡ a ∈ R ❆♥❞ s♦ P (x) − a ✐s ❛ ❝♦♥t✐♥✉♦✉s s♦❧✉t✐♦♥ ♦❢ ❈❛✉❝❤②✬s ❡q✉❛t✐♦♥✳ ❙♦ P (x) = cx + a ❢♦r s♦♠❡ a, b ❛♥❞✱ ♣❧✉❣❣✐♥❣ ✐♥ ♦r✐❣✐♥❛❧ ❡q✉❛t✐♦♥✱ ✇❡ ❣❡t t❤❡ s♦❧✉t✐♦♥s ✿ P (x) = ∀x P (x) = 2− 2009 ∀x P (x) = x ∀x ✶✸✾✳ ●✐✈❡♥ f (x) = ax3 + bx2 + cx + d ✱ s✉❝❤ t❤❛t f (0) = 1✱ f (1) = 2✱ f (2) = 4✱ f (3) = 8✳ ❋✐♥❞ t❤❡ ✈❛❧✉❡ ♦❢ f (4) solution f (0) = ⇐⇒ d = ✭❡✶✮ ✿ f (1) = ⇐⇒ a + b + c = ✭❡✷✮ ✿ f (2) = ⇐⇒ 8a + 4b + 2c = ✭❡✸✮ ✿ f (3) = ⇐⇒ 27a + 9b + 3c = ✭❡✷✮✲✷✭❡✶✮ ✿ 6a + 2b = ✭❡✸✮✲✸✭❡✶✮ ✿ 12a + 3b = ❚❤✐s ❣✐✈❡s a = b=0 ❆♥❞ s♦ c = ❛♥❞ f (x) = x3 +5x+6 6 ❛♥❞ ❛♥❞ f (4) = 15 ✶✹✵✳ ❉♦❡s ❚❤❡r❡ ❊①✐st ❆ ❋✉♥❝t✐♦♥ f :N →N ∀n ≥ f (f (n − 1)) = f (n + 1) − f (n) solution f (n + 1) − f (n) ≥ ∀n ≥ ❛♥❞ s♦ f (n) ≥ f (2) + n − ≥ n − ∀n ≥ ❙♦ ✿ ∀n ≥ ✿ f (n−1) ≥ n−2 ≥ ❛♥❞ s♦ f (f (n−1)) ≥ f (n−1)−1 ≥ n−3 ❛♥❞ s♦ f (n + 1) − f (n) ≥ n − ❆❞❞✐♥❣ t❤❡s❡ ❧✐♥❡s ❢♦r n = 5, 6, 7✱ ✇❡ ❣❡t f (8) − f (5) ≥ ❛♥❞ s♦ f (8) ≥ 10✳ ▲❡t t❤❡♥ a = f (8) ≥ 10 ❆❞❞✐♥❣ t❤❡♥ t❤❡ ❧✐♥❡s f (f (n − 1)) = f (n + 1) − f (n) ❢♦r n = → a − 1✱ a−2 ✇❡ ❣❡t f (a) − f (2) = k=1 f (f (k)) ❆♥❞✱ s✐♥❝❡ a ≥ 10✱ ✇❡ ❝❛♥ ✇r✐t❡ f (a) − f (2) = f (f (8)) + ❛♥❞ s♦✱ s✐♥❝❡ f (8) = a✱ t❤✐s ❜❡❝♦♠❡s ✿ −f (2) = ✐♠♣♦ss✐❜❧❡ s✐♥❝❡ LHS < ✇❤✐❧❡ RHS > ❆♥❞ s♦ ♥♦ s♦❧✉t✐♦♥ ✻✾ a−2 k=1,k=8 a−2 k=1,k=8 f (f (k)) f (f (k))✱ ❝❧❡❛r❧② ✶✹✶✳ ■❢ f : R → R s❛t✐s✜❡s f (x + y) = f (x) · f (y)❀ t❤❡♥ ✐s ✐t ♥❡❝❡ss❛r② ❢♦r f (x) t♦ ❜❡ ♦❢ t❤❡ ❢♦r♠ ax ❢♦r s♦♠❡ a ∈ R ❄ solution ◆♦✳ ❋✐rst ♥♦t✐❝❡ t❤❛t f (u) = =⇒ f (x) = f (x − u)f (u) = ∀x ❛♥❞ s♦ ❛ s♦❧✉t✐♦♥ f (x) = ∀w ❚❤❡♥✱ ✐❢ f (x) = ∀x✱ ✇❡ ❣❡t f (x) = f ( x2 )2 > ❛♥❞ s♦ ✇❡ ❝❛♥ ✇r✐t❡ g(x) = ln f (x) ❛♥❞ ✇❡ ❤❛✈❡ t❤❡ ❡q✉❛t✐♦♥ g(x+y) = g(x)+g(y) ❙♦ g(x) ✐s ❛♥② s♦❧✉t✐♦♥ ♦❢ ❈❛✉❝❤②✬s ❡q✉❛t✐♦♥ ❛♥❞ ✇❡ ❤❛✈❡ t❤❡ ❣❡♥❡r❛❧ s♦❧✉t✐♦♥s ✿ f (x) = ∀x f (x) = eg(x) ✇❤❡r❡ g(x) ✐s ❛♥② s♦❧✉t✐♦♥ ♦❢ ❈❛✉❝❤②✬s ❡q✉❛t✐♦♥✳ ❬✉❪■❢ ②♦✉ ❛❞❞ s♦♠❡ ❝♦♥str❛✐♥ts ❬✴✉❪✭❝♦♥t✐♥✉✐t②✱ ♦r ln f (x) ✉♣♣❡r ❜♦✉♥❞❡❞ ♦r ❧♦✇❡r ❜♦✉♥❞❡❞ ♦♥ s♦♠❡ ✐♥t❡r✈❛❧✱ t❤❡♥ ✇❡ ❣❡t g(x) = cx ❛♥❞ s♦ f (x) = ax ❢♦r s♦♠❡ a > 0❬ ✶✹✷✳ ●✐✈❡ ❛❧❧ ❢✉♥❝t✐♦♥s f : R+ → R+ s✉❝❤ t❤❛t (x + y)f (f (x)y) = x2 (f (f (x) + f (y)) ❢♦r ❛❧❧ x, y ♣♦s✐t✐✈❡ r❡❛❧✳ solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ (x + y)f (f (x)y) = x2 f (f (x) + f (y)) ■❢ f (a) = f (b)✱ t❤❡♥✱ ❝♦♠♣❛r✐♥❣ P (a, y) ❛♥❞ P (b, y)✱ ✇❡ ❣❡t ❛♥❞ s♦ a = b ❛♥❞ f (x) ✐s ✐♥❥❡❝t✐✈❡✳ √ √ a+y a2 = b+y b2 √ P ( 1+2 , 1) =⇒ f (f ( 1+2 )) = f (f ( 1+2 ) + f (1)) √ √ ❆♥❞ s♦✱ s✐♥❝❡ ✐♥❥❡❝t✐✈❡ ✿ f ( 1+2 ) = f ( 1+2 ) + f (1) ❛♥❞ f (1) = 0✱ ✐♠♣♦s✲ s✐❜❧❡✳ ❙♦ ♥♦ s♦❧✉t✐♦♥ t♦ t❤✐s ❡q✉❛t✐♦♥✳ ✶✹✸✳ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s f : R → R s✉❝❤ t❤❛t f (x + y) = max(f (x), y) + min(x, f (y)) solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (x + y) = max(f (x), y) + min(x, f (y)) ✭❛✮ ✿ P (x, 0) =⇒ f (x) = max(f (x), 0) + min(x, f (0)) ✭❜✮ ✿ P (0, x) =⇒ f (x) = min(0, f (x)) + max(f (0), x) ❯s✐♥❣ t❤❡ ❢❛❝t t❤❛t max(u, v) + min(u, v) = u + v ✱ t❤❡ s✉♠ ✭❛✮✰✭❜✮ ✐♠♣❧✐❡s f (x) = x + f (0) ❚❤❡♥ P (0, f (0)) =⇒ f (0) = min(0, 2f (0)) ❛♥❞ s♦ f (0) = ❍❡♥❝❡ t❤❡ ✉♥✐q✉❡ s♦❧✉t✐♦♥ ✿ f (x) = x ∀x✱ ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥ ✶✹✹✳ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s f : R → R s✉❝❤ t❤❛t ❢♦r ❡✈❡r② x, y ∈ R f (x + f (y)) = f (x − f (y)) + 4xf (y) solution ✼✵ ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (x + f (y)) = f (x − f (y)) + 4xf (y) f (x) = ∀x ✐s ❛ s♦❧✉t✐♦♥ ❛♥❞ ❧❡t ✉s ❢r♦♠ ♥♦✇ ❧♦♦❦ ❢♦r ♥♦♥ ❛❧❧③❡r♦ s♦❧✉t✐♦♥s✳ ▲❡t u s✉❝❤ t❤❛t f (u) = ▲❡t A = {2f (x) ∀x ∈ R} P ( 8fx(u) , u) =⇒ x = 2f ( 8fx(u) + f (u)) − 2f ( 8fx(u) − f (u)) ❙♦ ❛♥② x ∈ R ♠❛② ❜❡ ✇r✐tt❡♥ ❛s x = a − b ✇❤❡r❡ a, b ∈ A ▲❡t t❤❡♥ g(x) = f (x) − x2 ▲❡t a = 2f (y) ∈ A P (x + f (y), y) =⇒ f (x + a) = f (x) + 2ax + a2 ❛♥❞ s♦ g(x + a) = g(x) ∀x ∈ R✱ ∀a ∈ A ❙♦ g(x − b) = g(x) ∀x ∈ R✱ ∀b ∈ A ❙♦ g(x + a − b) = g(x − b) = g(x) ∀x ∈ R✱ ∀a, b ∈ A ❆♥❞ s✐♥❝❡ ✇❡ ❛❧r❡❛❞② ♣r♦✈❡❞ t❤❛t ❛♥② r❡❛❧ ♠❛② ❜❡ ✇r✐tt❡♥ ❛s a − b ✇✐t❤ a, b ∈ A✱ ✇❡ ❣❡t g(x + y) = g(x) ∀x, y ∈ R ❛♥❞ s♦ g(x) = c ❍❡♥❝❡ t❤❡ t✇♦ s♦❧✉t✐♦♥s ✿ f (x) = ∀x f (x) = x2 + c ∀x ❛♥❞ ❢♦r ❛♥② c ∈ R✱ ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥ ✶✹✺✳ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s f : Z+ → Z+ s✉❝❤ t❤❛t f (a) + b ❞✐✈✐❞❡s (f (b) + a)2 ❢♦r ❛❧❧ a, b ♣♦s✐t✐✈❡ ✐♥t❡❣❡rs✳ solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (x) + y|(f (y) + x)2 ▲❡t x > ❛♥❞ p > f (x) ♣r✐♠❡✳ P (p − f (x), x) =⇒ f (p − f (x)) + x|p2 ❛♥❞ s♦ f (p − f (x)) ∈ {p − x, p2 − x} ▲❡t Ax = {p ♣r✐♠❡ ✐♥t❡❣❡rs > f (x) s✉❝❤ t❤❛t f (p − f (x)) = p2 − x} ❋♦r p ∈ Ax ✿ P (p − f (x), y) =⇒ p2 − x + y|(f (y) + p − f (x))2 ❆♥❞ s♦ ✭s✉❜tr❛❝t✐♥❣ LHS ❢r♦♠ RHS ✮ ✿ p2 +y−x|x−y+(f (y)−f (x))(2p+ f (y) − f (x)) ❇✉t✱ ❢♦r p ❣r❡❛t ❡♥♦✉❣❤✱ |LHS| > |RHS| ❛♥❞ RHS ❝❛♥t ❜❡ ③❡r♦ ❢♦r ❛♥② y ❛♥❞ ❛♥② p ❛♥❞ s♦ ✐♠♣♦ss✐❜✐❧✐t② ❙♦ Ax ✐s ✉♣♣❡r ❜♦✉♥❞❡❞ ❛♥❞ ∃Nx s✉❝❤ t❤❛t ∀p > Nx f (p − f (x)) = p − x ❚❤❡♥✱ ❋♦r p > Nx ✿ P (p − f (x), y) =⇒ p − x + y|(f (y) + p − f (x))2 ❆♥❞ s♦ ✭s✉❜tr❛❝t✐♥❣ LHS ❢r♦♠ RHS ✮ ✿ p + y − x|(f (y) − f (x) − y + x)(2p + f (y) − f (x) + y − x) ❆♥❞ ✭s✉❜tr❛❝t✐♥❣ 2(f (y) − f (x) − y + x)LHS ❢r♦♠ RHS ❂ ✿ p + y − x|(f (y) − f (x) − y + x)2 ❇✉t✱ ❢♦r p ❣r❡❛t ❡♥♦✉❣❤✱ |LHS| > |RHS| ❛♥❞ s♦ RHS ♠✉st ❜❡ ③❡r♦ ❢♦r ❛♥② y ❛♥❞ s♦ f (y) − y = f (x) − x ❙♦ f (x) = x + a ∀x ❛♥❞ ❢♦r ❛♥② a ∈ Z≥0 ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥ ✶✹✻✳ ∀x, y ∈ Z+ f (f (x) + f (y)) = x + y ✜♥❞ ❛❧❧ f : Z+ → Z+ ✳ solution ■❢ f (x1 ) = f (x2 )✱ ✇❡ ❣❡t x1 = x2 ❛♥❞ t❤❡ ❢✉♥❝t✐♦♥ ✐s ✐♥❥❡❝t✐✈❡ ✼✶ ❚❤❡♥ f (f (x + 1) + f (1)) = x + = f (f (x) + f (2)) ❛♥❞✱ s✐♥❝❡ ✐♥❥❡❝t✐✈❡✱ f (x + 1) = f (x) + f (2) − f (1) ❙♦ f (x) = (f (2) − f (1))x + 2f (1) − f (2) = ax + b ❢♦r s♦♠❡ ✐♥t❡❣❡rs a, b ❜❡❝❛✉s❡ ❲r✐t❡ ✿ f (2) = f (1) + f (2) − f (1) f (3) = f (2) + f (2) − f (1) f (4) = f (3) + f (2) − f (1) ✳✳✳ f (x) = f (x − 1) + f (2) − f (1) ❆♥❞ ❛❞❞ ❛❧❧ t❤❡s❡ ❧✐♥❡s✳ P❧✉❣❣✐♥❣ t❤✐s ❜❛❝❦ ✐♥ ♦r✐❣✐♥❛❧ ❡q✉❛t✐♦♥✱ ✇❡ ❣❡t a = ±1 ❛♥❞ b = ❛♥❞✱ s✐♥❝❡ ✐♥ Z+ ✿ ❆ ✉♥✐q✉❡ s♦❧✉t✐♦♥ f (x) = x ∀x ✶✹✼✳ f (x) = f (x + y)f (x − y) ✜♥❞ ❛❧❧ f : R → R ❢✉♥❝t✐♦♥s solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (x)2 = f (x + y)f (x − y) ✐❢ f (u) = ❢♦r s♦♠❡ u✱ t❤❡♥ P (x, u − x) =⇒ f (x) = ❛♥❞ ✇❡ ❣❡t t❤❡ ❛❧❧③❡r♦ s♦❧✉t✐♦♥✳ ❙♦ ❧❡t ✉s ❝♦♥s✐❞❡r ❢r♦♠ ♥♦✇ t❤❛t f (x) = ∀x P ( x2 , x2 ) =⇒ f (x) f (0) = f( x 2) f (0)2 ❛♥❞ s♦ f (x) f (0) > ∀x ▲❡t t❤❡♥ g(x) = ln ff (x) (0) ✿ ✇❡ ❣❡t t❤❡ ♥❡✇ ❛ss❡rt✐♦♥ Q(x, y) ✿ 2g(x) = g(x + y) + g(x − y) ✇✐t❤ g(0) = Q(x, x) =⇒ 2g(x) = g(2x) ❛♥❞ s♦ t❤❡ ❡q✉❛t✐♦♥ ✐s g(2x) = g(x + y) + g(x − y) ❆♥❞ s♦ g((x+y)+(x−y)) = g(x+y)+g(x−y) ❛♥❞ s♦ g(x+y) = g(x)+g(y) ❆♥❞ s♦ g(x) ✐s ❛♥② s♦❧✉t✐♦♥ ♦❢ ❈❛✉❝❤② ❡q✉❛t✐♦♥✳ ❬✉❪❬❜❪❍❡♥❝❡ t❤❡ s♦❧✉t✐♦♥s ❬✴❜❪❬✴✉❪✿ f (x) = a · eh(x) ∀x ❛♥❞ ❢♦r ❛♥② r❡❛❧ a ❛♥❞ ❛♥② h(x) s♦❧✉t✐♦♥ ♦❢ ❈❛✉❝❤② ❡q✉❛t✐♦♥✱ ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥ ◆♦t✐❝❡ t❤❛t a = ❣✐✈❡s t❤❡ ❛❧❧③❡r♦ s♦❧✉t✐♦♥ ✶✹✽✳ ∀x ∈ Q+ ✜♥❞ ❛❧❧ f ❢✉♥❝t✐♦♥s f : Q+ → Q+ a)f (x + 1) = f (x) + b)f (x2 ) = f (x)2 solution ❋r♦♠ ❛✮ ✇❡ ❣❡t f (x + n) = f (x) + n ❋r♦♠ ❜✮ ✇❡ ❣❡t f (( pq + q)2 ) = f ( pq + q)2 ❆♥❞ s♦ f ( pq2 + 2p + q ) = (f ( pq ) + q)2 = f ( pq )2 + 2qf ( pq ) + q ✼✷ ❇✉t LHS = f ( pq2 ) + 2p + q = f ( pq )2 + 2p + q ❛♥❞ s♦ p = qf ( pq ) ❛♥❞s♦ f ( pq ) = pq ❍❡♥❝❡ t❤❡ s♦❧✉t✐♦♥ ✿ f (x) = x ∀x ∈ Q+ ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥✳ ✶✹✾✳ ∀x, y ∈ Z+ ✶✳ f (2) = ✷✳f (mn) = f (m)f (n) ✸✳f (n + 1) ≥ f (n) ✜♥❞ ❛❧❧ f : Z+ → Z+ ❢✉♥❝t✐♦♥s✳ solution ❯s✐♥❣ m = n = ✐♥ ✷✱ ✇❡ ❣❡t f (1) = ▲❡t u > ∈ N✳ ▲❡t a, b, c, d ∈ N s✉❝❤ t❤❛t a b ≥ ln u ln ≥ c d ❚❤✐s ✐♠♣❧✐❡s 2a ≥ ub ❛♥❞ ud ≥ 2c ❛♥❞ ❛♥❞ s♦✱ ✉s✐♥❣ ✸ ✿ f (2a ) ≥ f (ub ) ❛♥❞ f (ud ) ≥ f (2c ) ❛♥❞ s♦✱ ✉s✐♥❣ ✶ ❛♥❞ ✷ ✿ 2a ≥ f (u)b ❛♥❞ f (u)d ≥ 2c ❛♥❞ s♦ ✿ a b ≥ ln(f (u)) ln(2) ≥ c d ln u ln ≥ c d ❙♦ a b ❙♦ ln(f (u)) ln ≥ = ✐♠♣❧✐❡s a b ≥ ln(f (u)) ln(2) ≥ c d ln u ln ❙♦ f (u) = u ❍❡♥❝❡ t❤❡ r❡s✉❧t ✿ f (n) = n ∀n ∈ N✱ ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥✳ ✶✺✵✳ ❉❡t❡r♠✐♥❡ ❛❧❧ ❢✉♥❝t✐♦♥s f ❢r♦♠ t❤❡ ♥♦♥♥❡❣❛t✐✈❡ ✐♥t❡❣❡rs t♦ t❤❡ ♥♦♥♥❡❣❛t✐✈❡ ✐♥t❡❣❡rs s✉❝❤ t❤❛t f (1) = ❛♥❞✱ ❢♦r ❛❧❧ x ❛♥❞ y ✐♥ t❤❡ ♥♦♥♥❡❣❛t✐✈❡ ✐♥t❡❣❡rs✿ f (x)2 + f (y)2 = f (x2 + y )✳ solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (x)2 + f (y)2 = f (x2 + y ) ✶✮ f (x) = x ∀ ✐♥t❡❣❡r x ∈ [0, 9] ❂❂❂❂ P (0, 0) =⇒ f (0) = P (1, 0) =⇒ f (1) = P (1, 1) =⇒ f (2) = P (2, 0) =⇒ f (4) = P (2, 1) =⇒ f (5) = P (5, 0) =⇒ f (25) = 25 P (5, 5) =⇒ f (50) = 50 P (3, 4) =⇒ f (3) = P (7, 1) =⇒ f (7) = P (2, 2) =⇒ f (8) = P (3, 0) =⇒ f (9) = P (9, 2) =⇒ f (85) = 85 P (6, 7) =⇒ f (6) = ◗✳❊✳❉✳ ✷✮ f (x) = x ∀x ❂❂❂ ▲❡t x ≥ P (2x+1, x−2) =⇒ f (2x+1)2 +f (x−2)2 = f (5x2 + 5) P (2x − 1, x + 2) =⇒ f (2x − 1)2 + f (x + 2)2 = f (5x2 + 5) ❛♥❞ s♦ f (2x + 1)2 = f (2x − 1)2 + f (x + 2)2 − f (x − 2)2 P (2x + 2, x − 4) =⇒ f (2x + 2)2 + f (x − 4)2 = f (5x2 + 20) P (2 − 2, x + 4) =⇒ f (2x − 2)2 + f (x + 4)2 = f (5x2 + 20) ❆♥❞ s♦ f (2x + 2)2 = f (2x − 2)2 + f (x + 4)2 − f (x − 4)2 ❆♥❞ s♦ ❦♥♦✇❧❡❞❣❡ ♦❢ f (n) ✉♣ t♦ 2x ≥ ❣✐✈❡s ✉♥✐q✉❡ ❦♥♦✇❧❡❞❣❡ ♦❢ f (2x+1) ❛♥❞ f (2x + 2) ❆♥❞ s✐♥❝❡ f (x) ✐s q✉✐t❡ ❞❡✜♥❡❞ ✉♣ t♦ f (9)✱ t❤❡r❡ ✐s ❛t ♠♦st ♦♥❡ s♦❧✉t✐♦♥ f (x) ❆♥❞ s✐♥❝❡ f (x) = x ∀x ✐s ♦❜✈✐♦✉s❧② ❛ s♦❧✉t✐♦♥✱ t❤✐s ✐s t❤❡ ✉♥✐q✉❡ ♦♥❡✳ ✼✸ ✶✺✶✳ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s f : R → R s✉❝❤ t❤❛t f (x2 + y + f (y)) = 2y + (f (x))2 ❢♦r ❡✈❡r② x, y ∈ R✳ solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (x2 + y + f (y)) = 2y + f (x)2 ✶✮ f (x) = ⇐⇒ x = ❂❂❂ P (0, − 21 f (0)2 ) =⇒ f (s♦♠❡t❤✐♥❣) = ❛♥❞ s♦ ∃u s✉❝❤ t❤❛t f (u) = ▲❡t u s✉❝❤ t❤❛t f (u) = 0✱ t❤❡♥✱ ❝♦♠♣❛r✐♥❣ P (u, 0) ❛♥❞ P (−u, 0)✱ ✇❡ ❣❡t t❤❛t f (u) = f (−u) = ❛♥❞ s♦ ✿ P (0, u) =⇒ = 2u + f (0)2 P (0, −u) =⇒ = −2u + f (0)2 ❆♥❞ s♦ u = ◗✳❊✳❉✳ ✷✮ f (x) ✐s ✐♥❥❡❝t✐✈❡ ❂❂❂ P (0, − 12 f (x)2 ) =⇒ f (x2 − 21 f (x)2 +f (− 21 f (x)2 )) = ❆♥❞ s♦✱ ✉s✐♥❣ ✶✮ ❛❜♦✈❡ ✿ x2 − 21 f (x)2 + f (− 21 f (x)2 ) = ❚❤❡♥ f (x1 ) = f (x2 ) ✐♠♣❧✐❡s |x1 | = |x2 | ❈♦♠♣❛r✐♥❣ P (x, y) ❛♥❞ P (−x, y)✱ ✇❡ ❣❡t f (−x) = ±f (x) ▲❡t t❤❡♥ t s✉❝❤ t❤❛t f (−t) = f (t) P (0, t) =⇒ f (t + f (t)) = 2t ❛♥❞ s♦ P (t + f (t), 0) =⇒ f ((t + f (t))2 ) = 4t2 P (0, −t) =⇒ f (−t + f (t)) = −2t ❛♥❞ s♦ P (−t + f (t), 0) =⇒ f ((−t + f (t))2 ) = 4t2 ❙♦ f ((t + f (t))2 ) = f ((−t + f (t))2 ) ❛♥❞ s♦ ✭s❡❡ s♦♠❡ ❧✐♥❡s ❛❜♦✈❡✮ |(t + f (t))2 | = |(−t + f (t))2 | ❲❤✐❝❤ ✐♠♣❧✐❡s tf (t) = ❛♥❞ s♦ t = ✭✉s✐♥❣ ✶✮ ❛❜♦✈❡✮ ❙♦ f (−x) = −f (x) ∀x ❆♥❞ t❤❡♥ ✧f (x1 ) = f (x2 ) ✐♠♣❧✐❡s |x1 | = |x2 |✧ ❜❡❝♦♠❡s ✧f (x1 ) = f (x2 ) ✐♠♣❧✐❡s x1 = x2 ✧ ✭✉s✐♥❣ ❛❣❛✐♥ ✶✮ ❛❜♦✈❡✮ ◗✳❊✳❉✳ ✸✮ x + f (x) ✐s s✉r❥❡❝t✐✈❡ ❂❂❂ P (0, 21 f (x)) =⇒ f ( 12 f (x) + f ( 12 f (x))) = f (x) ❆♥❞ s♦✱ s✐♥❝❡ ✐♥❥❡❝t✐✈❡✱ 12 f (x) + f ( 12 f (x)) = x ◗✳❊✳❉✳ ✹✮ f (x) = x ∀x ❂❂❂ P (x, 0) =⇒ f (x2 ) = f (x)2 P (0, y) =⇒ f (y + f (y)) = 2y ❙♦ P (x, y) ❜❡❝♦♠❡s f (x2 + y + f (y)) = f (x2 ) + f (y + f (y)) ❆♥❞ s✐♥❝❡ x + f (x) ✐s s✉r❥❡❝t✐✈❡✱ t❤✐s ❜❡❝♦♠❡s f (x + y) = f (x) + f (y) ∀x ≥ 0✱ ∀y ❙✐♥❝❡ f (−x) = −f (x)✱ t❤✐s ✐♠♣❧✐❡s f (x + y) = f (x) + f (y) ∀x, y ❆♥❞ s✐♥❝❡ f (x2 ) = f (x)2 ✱ ✇❡ ❣❡t t❤❛t f (x) ≥ ∀x ≥ ❛♥❞ s♦ f (x + y) = f (x) + f (y) ✐♠♣❧✐❡s t❤❛t f (x) ✐s ♥♦♥ ❞❡❝r❡❛s✐♥❣✳ ❙♦✱ ❛s ❛ ♠♦♥♦t♦♥♦✉s s♦❧✉t✐♦♥ ♦❢ ❈❛✉❝❤②✬s ❡q✉❛t✐♦♥✱ f (x) = ax ∀x P❧✉❣❣✐♥❣ t❤✐s ❜❛❝❦ ✐♥ ♦r✐❣✐♥❛❧ ❡q✉❛t✐♦♥✱ ✇❡ ❣❡t a = ❆♥❞ s♦ t❤❡ ✉♥✐q✉❡ s♦❧✉t✐♦♥ f (x) = x ∀x ✶✺✷✳ ❉❡t❡r♠✐♥❡ ❛❧❧ s✉❝❤ ❢✉♥t✐♦♥s f, g, h ❢r♦♠ R+ t♦ ✐ts❡❧❢✱ t❤❛t f (g(h(x)) + y) + h(z + f (y)) = g(y) + h(y + f (z)) + x✳ solution ■ s✉♣♣♦s❡❞ t❤❛t t❤❡ ❞♦♠❛✐♥ ♦❢ ❢✉♥❝t✐♦♥❛❧ ❡q✉❛t✐♦♥ ✐s t❤❡ s❛♠❡ t❤❛♥ ❞♦♠❛✐♥ ♦❢ ❢✉♥❝t✐♦♥s ✭❜❡tt❡r t♦ ✐♥❞✐❝❛t❡ ❜♦t❤ ❞♦♠❛✐♥s✮✳ ✼✹ ▲❡t P (x, y, z) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (g(h(x)) + y) + h(z + f (y)) = g(y) + h(y + f (z)) + x P (x, y, y) =⇒ f (g(h(x)) + y) = g(y) + x ❛♥❞ s♦ h(x) ✐s ✐♥❥❡❝t✐✈❡ ❙✉❜tr❛❝t✐♥❣ P (x, y, y) ❢r♦♠ P (x, y, z)✱ ✇❡ ❣❡t h(z + f (y)) = h(y + f (z)) ❛♥❞ s♦✱ s✐♥❝❡ h(x) ✐s ✐♥❥❡❝t✐✈❡ ✿ z + f (y) = y + f (z) ❛♥❞ s♦ f (x) = x + a ❢♦r s♦♠❡ a ≥ P❧✉❣❣✐♥❣ t❤✐s ✐♥ P (1, x, x)✱ ✇❡ ❣❡t g(h(1)) + x + a = g(x) + ❛♥❞ s♦ g(x) = x + b ❢♦r s♦♠❡ b ≥ P❧✉❣❣✐♥❣ f (x) = x + a ❛♥❞ g(x) = x + b ✐♥ ♦r✐❣✐♥❛❧ ❡q✉❛t✐♦♥✱ ✇❡ ❣❡t h(x) = x − a ❛♥❞ s♦ a = ❍❡♥❝❡ t❤❡ s♦❧✉t✐♦♥s ✿ (f, g, h) = (x, x + b, x) ❢♦r ❛♥② r❡❛❧ b ≥ ✶✺✸✳ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s f : R → R s✉❝❤ t❤❛t ❢♦r ❛❧❧ x, y ✐♥ R✱ xf (x + xy) = xf (x) + f (x2 ).f (y) solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ xf (x + xy) = xf (x) + f (x2 )f (y) P (0, 0) =⇒ f (0) = ■❢ f (1) = 0✱ t❤❡♥ P (1, x − 1) =⇒ f (x) = ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥ ▲❡t ✉s ❢r♦♠ ♥♦✇ ❝♦♥s✐❞❡r t❤❛t f (1) = a = n −1 ■❢ a = 1✱ P (1, x) =⇒ f (x+1) = af (x)+a ❛♥❞ ✇❡ ❡❛s✐❧② ❣❡t f (n) = a aa−1 ∀n ∈ N P❧✉❣❣✐♥❣ t❤✐s ❡①♣r❡ss✐♦♥ ✐♥ P (m, n)✱ ✇❡ s❡❡ t❤❛t t❤✐s ✐s ♥♦t ❛ s♦❧✉t✐♦♥ ✭r❛t❤❡r ✉❣❧②✱ ■ t❤✐♥❦✮✳ ❙♦ a = ❛♥❞ P (1, x) =⇒ f (x + 1) = f (x) + ❛♥❞ s♦ f (n) = n ❛♥❞ f (x + n) = f (x) + n P (x, −1) =⇒ f (x2 ) = xf (x) P❧✉❣❣✐♥❣ t❤✐s ✐♥ P (x, y)✱ ✇❡ ❣❡t xf (x(y + 1)) = xf (x)(f (y) + 1) = xf (x)f (y + 1) ❆♥❞ s♦ f (xy) = f (x)f (y) P (x, y) ❜❡❝♦♠❡s t❤❡♥ xf (x)f (y+1) = xf (x)+f (x)2 f (y) ⇐⇒ xf (x)(f (y)+ 1) = xf (x) + f (x)2 f (y) ❆♥❞ s♦✱ s❡tt✐♥❣ y = ✿ f (x)(f (x) − x) = ❛♥❞ s♦ ∀x✱ ❡✐t❤❡r f (x) = 0✱ ❡✐t❤❡r f (x) = x ❇✉t✱ ✐❢ ❢♦r s♦♠❡ x = 0✱ ✇❡ ❤❛✈❡ f (x) = 0✱ t❤❡♥ f (x + 1) = f (x) + ✐♠♣❧✐❡s f (x + 1) = ✇❤✐❝❤ ✐s ✐♠♣♦ss✐❜❧❡ s✐♥❝❡ ❡✐t❤❡r f (x + 1) = x + = 1✱ ❡✐t❤❡r f (x + 1) = = ❙♦ f (x) = x ∀x✱ ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥✳ ❬✉❪❬❜❪❍❡♥❝❡ t❤❡ ❛♥s✇❡r ❬✴❜❪❬✴✉❪✿ f (x) = ∀x f (x) = x ∀x❬ ✶✺✹✳ ❋✐♥❞ ❛❧❧ ♣♦❧②♥♦♠✐❛❧s P (x) ∈ R[X]✱ degP = ✇✐t❤ t❤❡ ♣r♦♣❡rt② t❤❛t P (x2 ) = −P (x)P (−x) solution P (x) ✐s ♦❜✈✐♦✉s❧② ♠♦♥✐❝ ❛♥❞ ♠❛② ❜❡ ✇r✐tt❡♥ x3 + ax2 + bx + c ❛♥❞ t❤❡ ❡q✉❛t✐♦♥ ✐s ✿ x6 + ax4 + bx2 + c = (x3 + bx + ax2 + c)(x3 + bx − (ax2 + c) = = x2 (x2 + b)2 − (ax2 + c)2 ❛♥❞ s♦ ✿ ✼✺ a = 2b − a2 b = b2 − 2ac c = −c2 ❛♥❞ s♦ c = ♦r c = −1 c = ❣✐✈❡s b = b2 ❛♥❞ s♦ b = ♦r b = c = ❛♥❞ b = ❣✐✈❡s a = ♦r a = −1 ❛♥❞ s♦ t✇♦ s♦❧✉t✐♦♥s x3 ❛♥❞ x3 − x2 c = ❛♥❞ b = ❣✐✈❡s a = ♦r a = −2 ❛♥❞ s♦ t✇♦ s♦❧✉t✐♦♥s x3 + x2 + x ❛♥❞ x3 − 2x2 + x c = −1 ✐♠♣❧✐❡s a2 + a − 2b = ❛♥❞ b2 − b + 2a = ❛♥❞ s♦ t✇♦ s♦❧✉t✐♦♥s x3 − ❛♥❞ x3 − 3x2 + 3x − ❬✉❪❬❜❪❍❡♥❝❡ t❤❡ s✐① s♦❧✉t✐♦♥s ❬✴❜❪❬✴✉❪✿ P (x) = x3 P (x) = x2 (x − 1) P (x) = x(x2 + x + 1) P (x) = x(x − 1)2 P (x) = x3 − P (x) = (x − 1)3 ✶✺✺✳ ❋✐♥❞ ❛❧❧ ♠♦♥♦t♦♥✐❝ ❢✉♥❝t✐♦♥s u : R → R ✇❤✐❝❤ ❤❛✈❡ t❤❡ ♣r♦♣❡rt② t❤❛t t❤❡r❡ ❡①✐sts ❛ str✐❝t❧② ♠♦♥♦t♦♥✐❝ ❢✉♥❝t✐♦♥ f : R → R s✉❝❤ t❤❛t f (x + y) = f (x)u(x) + f (y) ❢♦r ❛❧❧ x, y ∈ R✳ solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (x + y) = f (x)u(x) + f (y) ❙✉❜tr❛❝t✐♥❣ P (x, 0) ❢r♦♠ P (x, y)✱ ✇❡ ❣❡t f (x + y) = f (x) + f (y) − f (0) ❛♥❞ s♦✱ s✐♥❝❡ str✐❝t❧② ✐♥❝r❡❛s✐♥❣✱ f (x) = ax + b ✇✐t❤ a > ❆♥❞ s♦ x = (x + ab )u(x) ❙❡tt✐♥❣ x = − ab ✱ ✇❡ ❣❡t b = ❛♥❞ s♦ t❤❡ s♦❧✉t✐♦♥ ✿ u(x) = ∀x = ❛♥❞ u(0) = c ❛♥② r❡❛ ✶✺✻✳ ●✐✈❡ ❛❧❧ ❢✉♥❝t✐♦♥s f : R− > R s✉❝❤ t❤❛t f (y)f (z) + f (x)f (x + y + z) = f (x + y)f (x + z) ❢♦r ❛❧❧ x, y, z r❡❛❧✳ solution ■s ✐t ❛ r❡❛❧ ♦❧②♠♣✐❛❞ ❡①❡r❝✐s❡ ❄ ❲✐t❤ ♥♦ ❢♦r❣♦tt❡♥ ❝♦♥str❛✐♥t ✭❧✐❦❡ ❝♦♥t✐✲ ♥✉✐t②✱ ❢♦r ❡①❛♠♣❧❡✮ ❄ ■♥ ✇❤❛t ❝♦♥t❡st ❞✐❞ ②♦✉ ❣❡t t❤✐s ♣r♦❜❧❡♠ ❄ ■t✬s ❡❛s② t♦ s❤♦✇ t❤❛t t❤❡ ❢✉♥❝t✐♦♥❛❧ ❡q✉❛t✐♦♥ ✐s ❡q✉✐✈❛❧❡♥t t♦ f (x)2 − f (y)2 = f (x + y)f (x − y) ❆♥❞ t❤✐s ❡q✉❛t✐♦♥ ❤❛s ✐♥✜♥✐t❡❧② ♠❛♥② s♦❧✉t✐♦♥s✳ ❋♦r ❡①❛♠♣❧❡ ✿ f (x) = ❛♥② s♦❧✉t✐♦♥ ♦❢ ❛❞❞✐t✐✈❡ ❈❛✉❝❤② ❡q✉❛t✐♦♥ f (x) = a sin(g(x)) ✇❤❡r❡ g(x) ✐s ❛♥② s♦❧✉t✐♦♥ ♦❢ ❛❞❞✐t✐✈❡ ❈❛✉❝❤② ❡q✉❛t✐♦♥ f (x) = a sinh(g(x)) ✇❤❡r❡ g(x) ✐s ❛♥② s♦❧✉t✐♦♥ ♦❢ ❛❞❞✐t✐✈❡ ❈❛✉❝❤② ❡q✉❛t✐♦♥ ❆♥❞ ■ ❞♦♥t ❦♥♦✇ ✐❢ t❤❡s❡ ❛r❡ t❤❡ ♦♥❧② s♦❧✉t✐♦♥s✳ ✶✺✼✳ ❋✐♥❞ ❛❧❧ f : R → R t❤❛t s❛t✐s❢② f (x − f (y) + y) = f (x) − f (y) ❛❧❧ r❡❛❧ ♥✉♠❜❡rs x, y ✳ solution ✼✻ ▲❡t g(x) = f (x) − x ❛♥❞ t❤❡ ❡q✉❛t✐♦♥ ❜❡❝♦♠❡s ❛ss❡rt✐♦♥ P (x, y) ✿ g(x − g(y)) = g(x) − y ❚❤✐s ✐♠♣❧✐❡s t❤❛t g(x) ✐s ❛ ❜✐❥❡❝t✐♦♥✳ ❙♦ ∃u s✉✈❤ t❤❛t g(u) = 0✳ P (x, u) ✐♠♣❧✐❡s t❤❡♥ u = P (g(x), x) =⇒ g(g(x)) = x P (x, g(y)) =⇒ g(x − y) = g(x) − g(y) ❙♦ g(x) ✐s ❛♥② ✐♥✈♦❧✉t✐✈❡ s♦❧✉t✐♦♥ ♦❢ ❈❛✉❝❤②✬s ❡q✉❛t✐♦♥✳ ❆♥❞ ✐t✬s✐♠♠❡❞✐❛t❡ t♦ ✈❡r✐❢② t❤❛t t❤✐s ✐s ✐♥❞❡❡❞ ❛ s♦❧✉t✐♦♥✳ ❬✉❪❬❜❪❍❡♥❝❡ t❤❡ ❛♥s✇❡r ❬✴❜❪❬✴✉❪✿ f (x) = x + g(x) ✇❤❡r❡ g(x) ✐s ❛♥② ✐♥✈♦✲ ❧✉t✐✈❡ s♦❧✉t✐♦♥ ♦❢ ❈❛✉❝❤②✬s ❡q✉❛t✐♦♥ ◆♦t✐❝❡ t❤❛t ✇❡ ❤❛✈❡ ✐♥✜♥✐t❡❧② ♠❛♥② s♦❧✉t✐♦♥s✳ ❚❤❡ ♦♥❧② ❝♦♥t✐♥✉♦✉s s♦❧✉✲ t✐♦♥s ❛r❡ f (x) = ∀x ❛♥❞ f (x) = 2x ∀x ◆♦t✐❝❡ t❤❛t t❤❡ ❣❡♥❡r❛❧ s♦❧✉t✐♦♥ ❢♦r ✧✐♥✈♦❧✉t✐✈❡ s♦❧✉t✐♦♥s ♦❢ ❈❛✉❝❤②✬s ❡q✉❛✲ t✐♦♥✧ ♠❛② ❛❧s♦ ❜❡ ✇r✐tt❡♥ ❛s ✿ ▲❡t A, B t✇♦ s✉♣♣❧❡♠❡♥t❛r② s✉❜✈❡❝t♦rs♣❛❝❡s ♦❢ t❤❡ Q✲✈❡❝t♦rs♣❛❝❡ R ▲❡t a(x) ❛♥❞ b(x) t❤❡ ♣r♦❥❡❝t✐♦♥s ♦❢ x ✐♥ A ❛♥❞ B s♦ t❤❛t x = a(x) + b(x) ✇✐t❤ a(x) ∈ A ❛♥❞ b(x) ∈ B ❚❤❡♥ g(x) = a(x) − b(x) ✶✮ ♣r♦♦❢ t❤❛t ❛♥② s✉❝❤ g(x) ✐s ❛♥ ✐♥✈♦❧✉t✐✈❡ s♦❧✉t✐♦♥ ♦❢ ❈❛✉❝❤②✬s ❡q✉❛t✐♦♥ ❛♥❞ s♦ t❤✐s ✐s ❛ s♦❧✉t✐♦♥ ❂❂❂ a(x) ❛♥❞ b(x) ❛r❡ ❛❞❞✐t✐✈❡ ❛♥❞ s♦ g(x) ✐s s♦❧✉t✐♦♥ ♦❢ ❈❛✉❝❤②✬s ❡q✉❛t✐♦♥✳ a(a(x)) = a(x) ❛♥❞ a(b(x)) = ❛♥❞ a(a(x)−b(x)) = a(x) b(a(x)) = 0) ❛♥❞ b(b(x)) = b(x) ❛♥❞ b(a(x)−b(x)) = −b(x) ❆♥❞ s♦ g(g(x)) = a(x)+b(x) = x ◗✳❊✳❉✳ ✷✮ ♣r♦♦❢ t❤❛t ❛♥② s♦❧✉t✐♦♥ ♠❛② ❜❡ ✇r✐tt❡♥ ✐♥ t❤✐s ❢♦r♠ ❛♥❞ s♦ ✐t✬s ❛ ❣❡♥❡r❛❧ s♦❧✉t✐♦♥ ❂❂❂❂ ▲❡t A = {x s✉❝❤ t❤❛t g(x) = x} ▲❡t B = {x s✉❝❤ t❤❛t g(x) = −x} ❖❜✈✐✲ ♦✉s❧②✱ s✐♥❝❡ g(x) ✐s ❛❞❞✐t✐✈❡✱ A, B ❛r❡ s✉❜✈❡❝t♦rs♣❛❝❡s ♦❢ t❤❡ Q✲✈❡❝t♦rs♣❛❝❡ R A ∩ B = {0} ❙✐♥❝❡ g(g(x)) = x✱ ✇❡ ❣❡t t❤❛t g(x + g(x)) = x + g(x) ❛♥❞ s♦ a(x) = x+g(x) ∈ A ❙✐♥❝❡ g(g(x)) = x✱ ✇❡ ❣❡t t❤❛t g(x − g(x)) = g(x) − x ❛♥❞ s♦ b(x) = x−g(x) ∈B ❆♥❞ s✐♥❝❡ a(x) + b(x) = x✱ ✇❡ ❝♦♥❝❧✉❞❡ t❤❛t A, B ❛r❡ s✉♣♣❧❡♠❡♥t❛r② s✉❜✈❡❝t♦rs♣❛❝❡s✳ ❆♥❞ ✇❡ ❝❧❡❛r❧② ❤❛✈❡ g(x) = a(x) − b(x) ◗✳❊✳❉✳ ✶✺✽✳ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥ f : R → R s❛t✐s❢②✐♥❣ t❤❡ ❝♦♥❞✐t✐♦♥✿ f (y + f (x)) = f (x)f (y) + f (f (x)) + f (y) − xy solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (y + f (x)) = f (x)f (y) + f (f (x)) + f (y) − xy ✼✼ f (x) = −1 ∀x ✐s ♥♦t ❛ s♦❧✉t✐♦♥ ❛♥❞ s♦ ❧❡t v s✉❝❤ t❤❛t f (v) = −1 P (v, 0) =⇒ f (0)(f (v) + 1) = ❛♥❞ s♦ f (0) = f (x) = ∀x ✐s ♥♦t ❛ s♦❧✉t✐♦♥ ❛♥❞ s♦ ❧❡t u s✉❝❤ t❤❛t f (u) = P (x, f (u)) =⇒ f (f (x)+f (u)) = f (x)f (f (u))+f (f (x))+f (f (u))−xf (u) P (u, f (x)) =⇒ f (f (x)+f (u)) = f (u)f (f (x))+f (f (x))+f (f (u))−uf (x) ❙✉❜tr❛❝t✐♥❣✱ ✇❡ ❣❡t f (f (x)) + x = f (x) f (ff(u))+u (u) ❛♥❞ s♦ f (f (x)) = af (x) − x ❢♦r s♦♠❡ a ∈ R ❙♦ ✇❡ ❝❛♥ r❡✇r✐t❡ P (x, y) ❛s ♥❡✇ ❛ss❡rt✐♦♥ Q(x, y) ✿ f (y + f (x)) = f (x)f (y) + af (x) − x + f (y) − xy Q(y, −1) =⇒ f (f (y) − 1) = f (y)(f (−1) + a) + f (−1) = cf (y) + d Q(x, f (y) − 1) =⇒ f (f (x) + f (y) − 1) = f (x)(cf (y) + d) + af (x) − x + cf (y) + d − x(f (y) − 1) ❛♥❞ s♦ ✿ f (f (x) + f (y) − 1) = cf (x)f (y) + (a + d)f (x) + (c − x)f (y) + d ❙✇❛♣♣✐♥❣ x, y ✱ ✇❡ ❣❡t f (f (x) + f (y) − 1) = cf (x)f (y) + (a + d)f (y) + (c − y)f (x) + d ❙✉❜tr❛❝t✐♥❣ ✿ (a + d − c + y)f (x) = (a + d − c + x)f (y) ❙❡tt✐♥❣ y = ✐♥ t❤✐s ❧✐♥❡✱ ✇❡ ❣❡t a + d − c = ❛♥❞ s♦ yf (x) = xf (y) ∀x, y ❙❡tt✐♥❣ y = ✐♥ t❤✐s ❡①♣r❡ss✐♦♥✱ ✇❡ ❣❡t f (x) = xf (1) P❧✉❣❣✐♥❣ ✐♥ ♦r✐❣✐♥❛❧ ❡q✉❛t✐♦♥✱ ✇❡ ❣❡t f (1) = ±1 ❬✉❪❬❜❪❆♥❞ s♦ t❤❡ t✇♦ s♦❧✉t✐♦♥s ❬✴❜❪❬✴✉❪✿ f (x) = x ∀x f (x) = −x ∀x ✶✺✾✳ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s√f : R+ → R+ s✉❝❤ t❤❛t f (xyz) + f (x) + f (y) + f (z) = √ √ f ( xy)f ( yz)f ( zx) ❢♦r ♣♦s✐t✐✈❡ r❡❛❧s x, y, z ❛♥❞ ❛❧s♦ f (x) < f (y) ❢♦r 1≤x ∀x > 1✱ ✇❡ ❣❡t t❤❛t f (x) ≥ ∀x > ▲❡t t❤❡♥ u(x) ≥ s✉❝❤ t❤❛t f (x) = u(x) + u(x) ✭✇❤✐❝❤ ❛❧✇❛②s ❡①✐sts s✐♥❝❡ f (x) ≥ 2✮ ❚❤❡ ❛❜♦✈❡ q✉❛❞r❛t✐❝ ✐♠♣❧✐❡s u(xy) = u(x)u(y) ♦r u(xy) = u(x) u(y) ♦r u(xy) = u(y) u(x) ❯s✐♥❣ t❤❡ ❢❛❝t t❤❛t f (x) ✐s ✐♥❝r❡❛s✐♥❣ ❢♦r x ≥ ❛♥❞ s♦ u(x) ✐s ✐♥❝r❡❛s✐♥❣ t♦♦✱ ✇❡ ❣❡t t❤❛t u(xy) = u(x)u(y) ∀x, y ≥ ✼✽ ❙♦ u(x) = xa ✇✐t❤ a > ∀x ≥ P❧✉❣❣✐♥❣ t❤✐s ❜❛❝❦ ✐♥ ♦r✐❣✐♥❛❧ ❡q✉❛t✐♦♥✱ ✇❡ ❣❡t t❤❛t ❛♥② r❡❛❧ a > ✜ts ❛♥❞ s♦ f (x) = xa + x−a ∀x ≥ P (x, x1 , 1) =⇒ f (x2 ) + f ( x12 ) + = 2f (x)f ( x1 ) ❆♥❞ s♦✱ ✉s✐♥❣ f (x2 ) = f (x)2 − ❢♦r x2 ❛♥❞ x12 ✿ (f (x) − f ( x1 ))2 = ❛♥❞ s♦ f ( x1 ) = f (x) ❙♦ f (x) = xa + x−a ∀x ❛♥❞ ❢♦r ❛♥② r❡❛❧ a = ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥ ✶✻✵✳ ❋✐♥❞ ❛❧❧ ❝♦♥t✐♥✉♦✉s ❢✉♥❝t✐♦♥s f : R → R s♦ t❤❛t ✿ f (f (x)) = f (x) + x, ∀x∈R solution ▲❡t x ∈ R ❛♥❞ t❤❡ s❡q✉❡♥❝❡ a0 = x ❛♥❞ an+1 = f (an ) ❲❡ ❣❡t a0 = x ❛♥❞ a1 = f (x) ❛♥❞ an+2 = an+1 + an ✳ ▲❡t r1 < r2 ❜❡ t❤❡ t✇♦ r❡❛❧ r♦♦ts ♦❢ ❡q✉❛t✐♦♥ x2 − x − = 0✳ ❲❡ ❣❡t (f (x)−r2 x)r1n −(f (x)−r1 x)r2n an = r1 −r2 f (x) ✐s ✐♥❥❡❝t✐✈❡✳ ■t✬s ❡❛s② t♦ s❡❡ t❤❛t f (x) ✐s ♥❡✐t❤❡r ✉♣♣❡r ❜♦✉♥❞❡❞✱ ♥❡✐t❤❡r ❧♦✇❡r ❜♦✉♥❞❡❞ ❛♥❞ s♦ f (x) ✐s ❛ ❜✐❥❡❝t✐♦♥ ❢r♦♠ R → R ❙♦ t❤❡ ❡q✉❛❧✐t② an = (f (x)−r2 x)r1n −(f (x)−r1 x)r2n r1 −r2 ✐s tr✉❡ ❛❧s♦ ❢♦r n < ❙❡tt✐♥❣ x = ✐♥ t❤❡ ❡q✉❛t✐♦♥✱ ✇❡ ❣❡t f (f (0)) = f (0) ❛♥❞ s♦ f (0) = 0✱ s✐♥❝❡ ✐♥❥❡❝t✐✈❡✳ f (x) ✐s ✐♥❥❡❝t✐✈❡ ❛♥❞ ❝♦♥t✐♥✉♦✉s✱ ❛♥❞ s♦ ♠♦♥♦t♦♥♦✉s ❛♥❞ (0) s♦ f (x)−f ❤❛s ❛ ❝♦♥st❛♥t s✐❣♥ ❛♥❞ s♦ aan+1 ❤❛s ❛ ❝♦♥st❛♥t s✐❣♥✳ x−0 n ❙♦ (f (x)−r2 x)r1n+1 −(f (x)−r1 x)r2n+1 (f (x)−r2 x)r1n −(f (x)−r1 x)r2n ❤❛s ❛ ❝♦♥st❛♥t s✐❣♥✳ ■❢ f (x) ✐s ❞❡❝r❡❛s✐♥❣ ❛♥❞ f (x) − r1 x = 0✱ t❤❡♥ t❤❡ ❛❜♦✈❡ q✉❛♥t✐t② ❤❛s ❧✐♠✐t r2 > ✇❤❡♥ n → +∞✱ ✐♥ ❝♦♥tr❛❞✐❝t✐♦♥ ✇✐t❤ t❤❡ ❢❛❝t f (x) ❞❡❝r❡❛s✐♥❣✳ ❙♦ t❤❡ ♦♥❧② ❝♦♥t✐♥✉♦✉s ❞❡❝r❡❛s✐♥❣ s♦❧✉t✐♦♥ ♠❛② ❜❡ f (x) = r1 x ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥✳ ■❢ f (x) ✐s ✐♥❝r❡❛s✐♥❣ ❛♥❞ f (x) − r2 x = 0✱ t❤❡♥ t❤❡ ❛❜♦✈❡ q✉❛♥t✐t② ❤❛s ❧✐♠✐t r1 < ✇❤❡♥ n → −∞✱ ✐♥ ❝♦♥tr❛❞✐❝t✐♦♥ ✇✐t❤ t❤❡ ❢❛❝t f (x) ✐♥❝r❡❛s✐♥❣✳ ❙♦ t❤❡ ♦♥❧② ❝♦♥t✐♥✉♦✉s ✐♥❝r❡❛s✐♥❣ s♦❧✉t✐♦♥ ♠❛② ❜❡ f (x) = r2 x ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥✳ ❍❡♥❝❡ t❤❡ ♦♥❧② s♦❧✉t✐♦♥s ✿ f (x) = √ f (x) = − √ 1+ x 5−1 x ✶✻✶✳ ▲❡t f : N− > N ❜❡ ❛ ❢✉♥❝t✐♦♥ s❛t✐s❢②✐♥❣✿ f (f (n)) = 4n−3 (2n ) = 2n+1 −1✱ ❢♦r ❛❧❧ ♥❛t✉r❛❧ ♥ ❋✐♥❞ f (1993)✱ ❝❛♥ ②♦✉ ✜♥❞ ❡①♣❧✐❝✐❡t❧② t❤❡ ✈❛❧✉❡ f (2007)❄ ✇❤❛t ✈❛❧✉❡s ❝❛♥ f (1997) t❛❦❡❄ solution ■ s✉♣♣♦s❡ t❤❛t t❤✐r❞ ❧✐♥❡ ♠✉st ❜❡ r❡❛❞ f (2n ) = 2n+1 − ✼✾ ▲❡t N0 = N ∪ {0} ▲❡t g(n) ❢r♦♠ N0 → N0 ❞❡✜♥❡❞ ❛s g(n) = f (n + 1) − 1✳ ❚❤❡ ❡q✉❛t✐♦♥ ✐s t❤❡♥ g(g(n)) = 4n ✇❤♦s❡ ❣❡♥❡r❛❧ s♦❧✉t✐♦♥ ✐s ✿ ▲❡t A, B t✇♦ ❡q✉✐♥✉♠❡r♦✉s s❡ts ✇❤♦s❡ ✐♥t❡rs❡❝t✐♦♥ ✐s ❡♠♣t② ❛♥❞ ✇❤♦s❡ ✉♥✐♦♥ ✐s t❤❡ s❡t ♦❢ ❛❧❧ ♥❛t✉r❛❧ ♥✉♠❜❡rs ♥♦t ❞✐✈✐s✐❜❧❡ ❜② 4✳ ▲❡t h(x) ❛♥② ❜✐❥❡❝t✐♦♥ ❢r♦♠ A → B ❛♥❞ h−1 (x) ✐t✬s ✐♥✈❡rs❡ ❢✉♥❝t✐♦♥✳ ❚❤❡♥ g(x) ♠❛② ❜❡ ❞❡✜♥❡❞ ❛s ✿ g(0) = ∀x ∈ A ✿ g(x) = h(x) ∀x ∈ B ✿ g(x) = 4h−1 (x) ∀x ∈ N \ (A ∪ B) ✿ g(x) = 4v4 (x) g(x4−v4 (x) ) ❚❤❡ ❝♦♥str❛✐♥t f (2n ) = 2n+1 − ❜❡❝♦♠❡s g(2n − 1) = 2n+1 − ❛♥❞ s♦ ✇❡ ❥✉st ❤❛✈❡ t♦ ❛❞❞ t♦ t❤❡ ♣r❡✈✐♦✉s ❣❡♥❡r❛❧ s♦❧✉t✐♦♥ t❤❡ ❝♦♥str❛✐♥ts ✿ 2n − ∈ A ∀n ∈ N 2n − ∈ B ∀n > ∈ N h(2n − 1) = 2n+1 − ❚❤❡♥ f (1993) = g(1992) + = 4g(498) + ❛♥❞ s✐♥❝❡ 498 ✐s ♥♦t ❞✐✈✐s✐❜❧❡ ❜② ❛♥❞ ✐s ♥♦t ✐♥ t❤❡ ❢♦r♠ 2n − ♥❡✐t❤❡r 2n+1 − 1✱ ✇❡ ❣❡t ♥❡❛r❧② ♥♦ ❝♦♥str❛✐♥t ❢♦r g(498) ✿ ❲❡ ❝❛♥ ♣✉t 498 ✐♥ A ❛♥❞ t❤❡♥ g(498) ∈ B ♠❛② ❜❡ ❛♥② ✈❛❧✉❡ ♥♦t ❞✐✈✐s✐❜❧❡ ❜② ❛♥❞ ♥♦t ✐♥ t❤❡ ❢♦r♠ 2n+1 − ❲❡ ❝❛♥ ♣✉t 498 ✐♥ B ❛♥❞ t❤❡♥ g(498) = 4u ✇❤❡r❡ u ✐s ❛♥② ♥✉♠❜❡r ♥♦t ❞✐✈✐s✐❜❧❡ ❜② ❛♥❞ ♥♦t ✐♥ t❤❡ ❢♦r♠ 2n − ❆♥❞ t❤❡ s❛♠❡ ❝♦♥❝❧✉s✐♦♥s ♠❛② ❜❡ ♦❜t❛✐♥❡❞ ❢♦r g(2006) ❛♥❞ g(1996) ✶✻✷✳ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥ f : R · R → R s✉❝❤ t❤❛t f (f (x, z), f (z, y)) = f (x, y) + z ❢♦r ❛❧❧ r❡❛❧ ♥✉♠❜❡rs ①✱② ❛♥❞ ③ solution ▲❡t P (x, y, z) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (f (x, z), f (z, y)) = f (x, y) + z ▲❡t s(x) = f (x, x) ✇❤❡r❡ ✧s✧ st❛♥❞s ❢♦r ✧s❛♠❡✧ ▲❡t r(x) = f (0, x) ✇❤❡r❡ ✧r✧ st❛♥❞s ❢♦r ✧r✐❣❤t✧ ▲❡t l(x) = f (x, 0) ①❤❡r❡ ✧❧✧ st❛♥❞s ❢♦r ✧❧❡❢t✧ P (x, x, x) =⇒ s(s(x)) = s(x) + x ❛♥❞ s♦ s(x) ✐s ✐♥❥❡❝t✐✈❡ P (0, 0, 0) =⇒ s(s(0)) = s(0) ❛♥❞ s♦✱ s✐♥❝❡ ✐♥❥❡❝t✐✈❡ ✿ s(0) = ❛♥❞ s♦ r(0) = l(0) = ❛♥❞ f (0, 0) = P (x, 0, 0) =⇒ l(l(x)) = l(x) P (0, x, 0) =⇒ r(r(x)) = r(x) P (x, y, 0) =⇒ f (l(x), r(y)) = f (x, y) ❚❤❡♥✱ l(l(x)) = l(x) =⇒ f (x, y) = f (l(x), y) ❙❛♠❡✱ r(r(y)) = r(y) =⇒ f (x, y) = f (x, r(y)) P (0, 0, x) =⇒ f (r(x), l(x)) = x ❙✉♣♣♦s❡ ∃u, v s✉❝❤ t❤❛t l(u) = d(v) = a✳ ❚❤❡♥ ✿ l(l(u)) = l(u) ❛♥❞ s♦ l(a) = a r(r(v)) = r(v) ❛♥❞ s♦ r(a) = a a = f (r(a), l(a)) ❛♥❞ s♦ f (a, a) = a ❛♥❞ s♦ s(a) = a P (a, a, a) =⇒ s(s(a)) = s(a) + a ❛♥❞ s♦ a = 2a ❛♥❞ a = ❙♦ l(R) ∩ r(R) = {0} ❙✉♣♣♦s❡ ♥♦✇ ∃u, v s✉❝❤ t❤❛t f (u, v) = P (u, v, u) =⇒ f (f (u, u), f (u, v)) = f (u, v)+u =⇒ l(s(u)) = u =⇒ u ∈ l(R) P (u, v, v) =⇒ f (f (u, v), f (v, v)) = f (u, v)+v =⇒ r(s(v)) = v =⇒ v ∈ r(R) l(u) = f (u, 0) = f (u, f (u, v)) = f (f (r(u), l(u)), f (l(u), v)) = f (r(u), v) + l(u) =⇒ f (r(u), v) = =⇒ r(u) ∈ l(R) r(v) = f (0, v) = f (f (u, v), v) = f (f (u, r(y)), f (r(v), l(v))) = f (u, l(v))+r(v) =⇒ f (u, l(v)) = =⇒ l(v) ∈ r(R) ❙♦ r(u) ∈ l(R)∩r(R) ✽✵ ❛♥❞ s♦ r(u) = ❛♥❞ s♦ u = f (r(u), l(u)) = r(l(u)) ❛♥❞ r(u) = r(r(l(u))) = r(l(u)) = u ❛♥❞ s♦ u = ❙❛♠❡ ✿ l(v)) ∈ l(R) ∩ r(R) ❛♥❞ s♦ l(v) = ❛♥❞ s♦ v = f (r(v), l(v)) = l(r(v)) ❛♥❞ s♦ l(v) = l(l(r(v))) = l(r(v)) = v ❛♥❞ s♦ v=0 ❙♦ f (x, y) = ⇐⇒ x = y = ❚❤❡♥ P (x, y, −f (x, y)) =⇒ f (f (x, −f (x, y)), f (−f (x, y), y)) = ❛♥❞ s♦ f (x, −f (x, y)) = f (−f (x, y), y) = ❛♥❞ s♦ x = y = 0✱ ✐♠♣♦ss✐❜❧❡ ❙♦ ♥♦ s♦❧✉t✐♦♥ ❢♦r t❤✐s ❡q✉❛t✐♦♥ ✶✻✸✳ ▲❡t f : R → R s❛t✐s❢② f (4x ) = f (2x ) + x ✳ ❋✐♥❞ ❛❧❧ ❢✉♥❝t✐♦♥s f ✇✐t❤ t❤❡ ❣✐✈❡♥ ♣r♦♣❡rt②✳ solution ❖❜✈✐♦✉s❧② f (x) ❝❛♥ t❛❦❡ ❛♥② ✈❛❧✉❡ ✇❡ ✇❛♥t ❢♦r x ≤ ❋♦r x > 0✱ ❧❡t ✉s ✇r✐t❡ f (x) = g(ln x)+log2 x ❛♥❞ t❤❡ ❡q✉❛t✐♦♥ ❜❡❝♦♠❡s g(2x) = g(x) ✇❤✐❝❤ ✐s ✈❡r② ❝❧❛ss✐❝❛❧ ✇✐t❤ s♦❧✉t✐♦♥ ✿ g(x) = u({log2 x} ❢♦r ❛♥② x > ✇❤❡r❡ u(x) ✐s ❛♥② ❢✉♥❝t✐♦♥ ❞❡✜♥❡❞ ♦✈❡r [0, 1) g(0) = a ✇❤❡r❡ a ✐s ❛♥② r❡❛❧ ✇❡ ✇❛♥t g(x) = v({log2 −x} ❢♦r ❛♥② x < ✇❤❡r❡ v(x) ✐s ❛♥② ❢✉♥❝t✐♦♥ ❞❡✜♥❡❞ ♦✈❡r [0, 1) ❬✉❪❬❜❪❍❡♥❝❡ ❛ ❣❡♥❡r❛❧ s♦❧✉t✐♦♥ ♦❢ r❡q✉✐r❡❞ ❡q✉❛t✐♦♥ ❬✴❜❪❬✴✉❪✿ ∀x ≤ ✿ f (x) ✐s ❛♥② ❢✉♥❝t✐♦♥ ✇❡ ✇❛♥t ∀x ∈ (0, 1) ✿ f (x) = log2 x + v({log2 | ln x|}) ✇❤❡r❡ v(x) ✐s ❛♥② ❢✉♥❝t✐♦♥ ❞❡✜♥❡❞ ♦✈❡r [0, 1) f (1) = a ✇❤❡r❡ a ✐s ❛♥② r❡❛❧ ✇❡ ✇❛♥t ∀x > ✿ f (x) = log2 x + u({log2 ln x}) ✇❤❡r❡ u(x) ✐s ❛♥② ❢✉♥❝t✐♦♥ ❞❡✜♥❡❞ ♦✈❡r [0, 1) ✶✻✹✳ f : R → R f (x + y) + f (y + z) + f (z + x)?f (x + 2y + 3z) ❢♦r ❛♥② r❡❛❧ x, y, z solution ❚❤❡r❡ ❛r❡ ♦❜✈✐♦✉s❧② ✐♥✜♥✐t❡❧② ♠❛♥② s♦❧✉t✐♦♥s ❛♥❞ ■ ✇♦♥❞❡r ❤♦✇ ✇❡ ❝❛♥ ✜♥❞ ❛ ❣❡♥❡r❛❧ ❢♦r♠✉❧❛ ❢♦r t❤❡s❡✳ ❙♦♠❡ ❡①❛♠♣❧❡s ✿ f (x) = f (x) = 2π + arctan(x) f (x) = x2 +2 x2 +1 f (x) = + q(x) ✇✐t❤ q✭①✮❂①✲✢♦♦r❢✉♥❝t✐♦♥✭①✮ ✶✻✺✳ ❋✐♥❞ ❛❧❧ f : Q → Q s❛t✐s❢② ✿ f (f (x) + y) = x + f (y), ∀x, y ∈ Q solution ▲❡t P (x, y) ❜❡ t❤❡ ❛ss❡rt✐♦♥ f (f (x) + y) = x + f (y) P (x, 0) =⇒ f (f (x)) = x + f (0) P (f (x), y) =⇒ f (x + y + f (0)) = f (x) + f (y) ❲r✐t✐♥❣ f (x) = g(x + f (0))✱ t❤✐s ❜❡❝♦♠❡s g((x + f (0)) + (y + f (0))) = g(x + f (0)) + g(y + f (0)) ❙♦ g(x + y) = g(x) + g(y) ❛♥❞ g(x) = g(1)x ∀x ∈ Q ❛♥❞ s♦ f (x) = g(1)(x + f (0)) ✽✶ ❙♦ f (x) = ax + b ❛♥❞✱ ♣❧✉❣❣✐♥❣ t❤✐s ✐♥ ♦r✐❣✐♥❛❧ ❡q✉❛t✐♦♥✱ ✇❡ ❣❡t b = ❛♥❞ a2 = ❍❡♥❝❡ t❤❡ t✇♦ s♦❧✉t✐♦♥s ✿ f (x) = x ∀x f (x) = −x ∀x ✶✻✻✳ ❋✐♥❞ ❛❧❧ f : Q+ → Q+ s❛t✐s❢②✿ f (x + 1) = f (x) + f (x2 ) = f (x) ∀x ∈ Q+ solution ❚❤❡ ✜rst ❡q✉❛t✐♦♥ ✐♠♣❧✐❡s f (x + n) = f (x) + n ∀x ∈ Q+ ✱ ∀n ∈ N 2 ❚❤❡♥ f (( pq + q)2 ) = f ( pq2 + 2p + q ) = f ( pq2 ) + 2p + q = f ( pq ) + 2p + q ❇✉t f (( pq + q)2 ) = f ( pq + q) = (f ( pq ) + q)2 = f ( pq ) + 2qf ( pq ) + q ❆♥❞ s♦ 2p = 2qf ( pq ) ❛♥❞ f ( pq ) = p q ❍❡♥❝❡ t❤❡ ✉♥✐q✉❡ s♦❧✉t✐♦♥ f (x) = x ∀x ∈ Q+ ✱ ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥ ✶✻✼✳ ❋✐♥❞ ❛❧❧ ❝♦♥t✐✉♦✉s ❢✉♥❝t✐♦♥ f : R+− > R+ s❛t✐s❢②✐♥❣✿ f (x+ x1 )+f (y+ y1 ) = f (x + y1 ) + f (y + x1 ) ❢♦r ❡✈❡r② ①✱② ❢r♦♠ ❘✰ solution ❈♦♥s✐❞❡r t❤❡♥ a, b > s✉❝❤ t❤❛t a = b ❛♥❞ ab ≥ ❈♦♥s✐❞❡r t❤❡ s②st❡♠ ✿ x ≥ ab ❛♥❞ y ≥ s②st❡♠ ❛❧✇❛②s ❤❛✈❡ ❛ ✉♥✐q✉❡ r❡❛❧ s♦❧✉t✐♦♥ b a x+ y =a y+ x = b ❚❤s ▲❡t t❤❡♥ u = x + x1 ❛♥❞ v = y + y1 ■t✬s ❡❛s② t♦ s❡❡ t❤❛t ✿ f (a) + f (b) = f (u) + f (v) a + b = u + v |u − v| < |a − b| u = v ❛♥❞ uv ≥ ❆♥❞ s♦ ✇❡ ❝❛♥ ❝r❡❛t❡ ❛ s❡q✉❡♥❝❡ (a, b) → (u, v)✱ r❡♣❡❛t✐♥❣ t❤❡ ♣r♦❝❡ss ■t✬s ❡❛s② t♦ s❡❡ t❤❛t t❤❡ t✇♦ ♥✉♠❜❡rs ❤❛✈❡ t❤❡✐r ❞✐✛❡r❡♥❝❡ t❡♥❞✐♥❣ t♦✇❛r❞s ✵ ❛♥❞ s♦ ❤❛✈❡ t❤❡ s❛♠❡ ❧✐♠✐t a+b ❛♥❞ s♦✱ s✐♥❝❡ ❝♦♥t✐♥✉♦✉s✱ f (a) + f (b) = 2f ( a+b ) ∀a, b > s✉❝❤ t❤❛t a = b ❛♥❞ ab ≥ ❚❤✐s ✐s ❛ ❝❧❛ss✐❝❛❧ ❢✉♥❝t✐♦♥❛❧ ❡q✉❛t✐♦♥ ✇❤✐❝❤ ✐♠♣❧✐❡s ❡❛s✐❧② ✭❝♦♥t✐♥✉✐t② ❛❣❛✐♥✮ f (x) = cx + d ∀x ≥ ❯s✐♥❣ t❤❡♥ t❤❡ ❢✉♥❝t♦♥❛❧ ❡q✉❛t✐♦♥ ✇✐t❤ ❢♦r ❡①❛♠♣❧❡ y ≥ 12 ✱ ✇❡ ❣❡t x + 1 1 x , y + y , x + y ≥ ❛♥❞ s♦ f (y + x ) = c(y + x ) + d ❛♥❞ s♦ f (x) = cx + d ∀x > ❆♥❞ ✐t✬s ❡❛s② t♦ ✉s❡ s✐♠✐❧❛r st❡♣s ❛s ♠❛♥② t✐♠❡s ❛s ✇❡ ✇❛♥t t♦ ❣❡t f (x) = cx + d ∀x > ❆♥❞ t❤✐s ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥ ❛s s♦♦♥ ❛s c ≥ ❛♥❞ d ≥ ♦r c = ❛♥❞ d>0 ❍❡♥❝❡ t❤❡ ❛♥s✇❡r ✿ f (x) = ax + b ∀x > ❛♥❞ ❢♦r ❛♥② ✭a > ❛♥❞ b ≥ 0✮ ♦r ✭a = ❛♥❞ b > 0✮ ✽✷ ✶✻✽✳ ❋✐♥❞ t❤❡ f : R → R s✉❝❤ t❤❛t ❢ ✐s ❛ ❝♦♥t✐♥✉♦✉s ❢✉♥❝t✐♦♥ ❛♥❞ s❛t✐s❢② ✿ f (x + y) = f (x) + f (y) + 2xy, ∀x, y ∈ R olution ▲❡t f (x) = x2 + g(x) ❛♥❞ t❤❡ ❡q✉❛t✐♦♥ ❜❡❝♦♠❡s g(x + y) = g(x) + g(y) ❛♥❞ s♦ g(x) = ax✱ s✐♥❝❡ ❝♦♥t✐♥✉♦✉s ❛♥❞ f (x) = x2 + ax ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥✳ x2 → x2 ✶✻✾✳ ▲❡t ❢ ❜❡ ❛ ❝♦♥t✐✉♦✉s ❛♥❞ ✐♥❥❡❝t✐✈❡ ❢✉♥❝t✐♦♥ R − > R ❀ f (1) = ❀ f (2x − f (x)) = x✳ Pr♦✈❡ t❤❛t f (x) = x✳ solution ❙♦ f (x) ✐s str✐❝t❧② ♠♦♥♦t♦♥♦✉s✳ ■❢ f (x) ✐s ❞❡❝r❡❛s✐♥❣✱ t❤❡♥ 2x − f (x) ✐s ✐♥❝r❡❛s✐♥❣ ❛♥❞ f (2x − f (x)) ✐s ❞❡❝r❡❛s✐♥❣✱ ✇❤✐❝❤ ✐s ✇r♦♥❣✳ ❙♦ f (x) ✐s ✐♥❝r❡❛s✐♥❣✳ ■❢ f (a) > a✱ t❤❡♥ 2a − f (a) < a ❛♥❞ f (2a − f (a)) < f (a) ❛♥❞ s♦ f (a) > a✱ ✐♠♣♦ss✐❜❧❡ ■❢ f (a) < a✱ t❤❡♥ 2a − f (a) > a ❛♥❞ f (2a − f (a)) > f (a) ❛♥❞ s♦ f (a) < a✱ ✐♠♣♦ss✐❜❧❡ ❙♦ f (x) = x ∀x✱ ✇❤✐❝❤ ✐♥❞❡❡❞ ✐s ❛ s♦❧✉t✐♦♥ ✽✸