Number Theory in Mathlink docx

Number Theory in Mathlink Notice: There’re some mistakes in some solutions, so you have to read it carefully 1) system of equation including a prime number Solve the following system of equations for all primes p with integer solutions x,y ! (1) p + 1 = 2x^2 (2) p^2 + 1 = 2y^2 Hint 1) OK, this is a German question. You can look up the solution in the 1997 - 1998 book, http://www.unl.edu/amc/a-activities/a4-for-students/mc97-98-01feb.pdf at page 41. 2) it also appears in Po Leung Kuk invitational math competition(? or something like that, just for junior form, i.e. grade 7-9) 5 years ago i think 2) diofant1 Solve the equation in Z: x^4+6x 2 y 2 +y^4=z 2 Solution (x 2 +y 2 ) 2 + (2xy) 2 = z 2 (given) (x 2 +y 2 ) 2 -(2xy) 2 = (x 2 -y 2 ) 2 denote a=x 2 +y 2 and b=2xy the above two equations imply a^4 - b^4 is a perfect square. we will prove that this is not possible assume there xists a nonempty set S s.t when a \varepsilon S ,then there xist natural numbers b and c satisfying a^4 - b^4 = c^2.by well ordering there is a least element ,suppose that element is a .which implies gcd (a,b,c) = gcd(a,b)=gcd(b,c)=gcd(c,a) = 1 consider cases CASE 1 (c is even) =>a and b are odd. and 4lc 2 gcd(a,b)=1 and gcd(a 2 +b 2 ,a 2 -b 2 )=2 let x= \sqrt [(a 2 +b 2 )/2] and y = \sqrt [(a 2 -b 2 )/2] then x and y are both integers and x^4- y^4=(ab) 2 .but a>b implies x<a. which is a contradiction CASE 2 (c is odd) => wlog ,a is odd and b is even .now use pythagorean substitution and there xist integers r and s such that gcd(r,s)=1 and a 2 =r 2 +s 2 ,c 2 =r 2 -s 2 and b 2 =2rs xactly one of r and s is even.assume r' to be even and s' to be odd. r'/2=x 2 and s'=y 2 then a 2 =4x^4+y^4 again by pythagoras,there xist integer m and s.t gcd (m,n)=1 and a=m 2 +n 2 , 2x 2 =2mn and y 2 =m 2 -n 2 .since x 2 =mn and gcd(m,n)=1 m=p 2 and n=q 2 .so y 2 = p^4-q^4.thus (p,q,y) is a new solution with p= \sqrt m \leq x= \sqrt (r'/2)<r' \leq a,which is a contradiction. thus there do not xist solutions to this equation • Nice soln Galois! This a well known problem: there are no numbers x and y s.t. x^2+y^2=z^2 and x^2-y^2=t^2. This was used before on this forum I think. I don't remember for which problem though Bye! • There are many simmilar problems regarding this one which are done using galois method 1)Prove that x^2+xy+y^2=36^2 has no solutions in Natural numbers \{0} 2)Solve in N* the equation 3x^4+10x^2y^2+3y^2=z^2. cheers! 3) Solve in integers : (1) x y = y 2001 (2) x 2 + 15y 2 = 2 2000 Solution For the first one: If x=1 then y=1. If x=-1 then y=-1. x=0=>y=0=>x^y=0^0 contradiction (same if y=0), so x<>0. Let's take the case |x|>1. It's obvious that y>=-1 because if y<=-2 then the left term isn't an integer while the right term is. If y=-1 then x=-1 (we already have that soln). This leaves us with y>0. y=p1^a1* *pk^ak where pi are primes. x must have the same prime divisors, so x=p1^b1* *pk^bk (ai,bi>=1). The equation is translated to y*bi=2001*ai for all i, or p1^a1* *pk^ak*bi=2001*ai for all i from 1 to k (*). Let's assume one of pi doesn't divide 2001 (let's assume it's p1). This means that p1^a1|a1, which is impossible because for any p1>=2 p1^a1>a1. This means that p1|2001. It's the same for all the other primes, so k=1, 2 or 3, so y=3^a1*23^a2*29^a3 or 3^a1*23^a2 or any other case that might come out of this. Let's consider the case when equations are 3^a1*23^a2*29^a3*bi=3*23*29*ai for all i. This is equivalent to 3^(a1-1)*23^(a2-1)*29^(a3-1)*bi=ai for all i. It's pretty obvious that the left terms are greater than the right part if ai>1 for some i, so all ai must be <=1. This means that all possible values of y are the divisors of 2001: 3, 23, 29, 3*23 etc. Conversely, it's easy to see that all divisors of 2001 are suitable values of y. For example, if y=3*23, then x=(3*23)^29 and we're done; if y=29 then x=29^(3*23) and so on. Is it right, Arne? I might have missed sth. 4) vietnam 2003 a problem from vietnam 2003 . Find the largest positive integer n such that the following equations have integer solutions in x, y_1, y_2, , y_n: (x + 1)^2 + y_1^2 = (x + 2)^2 + y_2^2 = = (x + n)^2 + y_n^2. Solution Lemma: If u^2 + v^2 + 2 = 2w^2 then u, v are even and w is odd. Proof: Clearly u, v either both even, either both odd. If u, v odd then LHS is divided by 4 but not by 8, that is impossible, because if RHS divided by 4 then it divided by 8. If u, v are even then LHS not divided by 4 => w is odd. If n>3, we have y_1^2 + y_3^2 + 2 = 2y_2^2 (why?) y_2^2 + y_4^2 + 2 = 2y_3^2 From lemma we have y_2 odd and even at the same time, contradiction. For n=3 x=-2, y_1=0, y_2=1, y_3=0 fulfil. Namdung 5) Some NT'ists know this, but I couldn't even guess the right 43th DeMO (German Mathematics Olympiad) Problem 431146 Given a Cartesian coordinate system in the plane, a point with coordinates (x; y) is called a grid point if and only if his coordinates x and y are integers. Does there exist a circle with exactly five grid points lying on its periphery? Solution Nice solving for a German problem!!! :) This is a problem that needs some spirit of observation ;] There are two ideas that may bring you to its solution: 1) The equation sin^2(x)+cos^2(x)=1 has only four integer solutions; 2) If a grid point lies on the periphery then its symmetric for the center of the circle also lies on the periphery of the circle. 1)One knows that the parametrical equations of the circle are : x=a+Rcos(x) y=b+Rsin(x), where (a,b) are the coordinates of the center of the circle, and R is its radius. But cos(x) and sin(x) are integer only for four points on the circle, so there exist only 4 grid points on the periphery of the circle. (this may be difficult to understand for R irrational, but when one looks at the circle and at the coordinates, everything becomes clear ) 2) This is a solution only for circles with grid center points : Let M(c,d) be a point on the periphery of the circle, then its symetrical point M' (according to the center of the circle) is also grid, because its coordinates are (2a-c,2b-d). So there can be only an even number of grid points on the periphery of the circle, and because 5 is even there doesn't exist a circle with such a property. This solution came to my mind quickly, but its generality is hard to prove. • The complete solution is quite straightforward once you have the explicite answer: The circle with equation passes through the points (0; -8), (5; 2), (-5; 2), (6; 0) and (-6; 0) and through no other lattice points. About the general theorem, I think there is some information on the Mathworld site. 6) nice find all integer values that ((a + 1) / b) + ((b + 1) / a) can assume and determine all pairs of integers a,b, giving these solutions. Hint Nice, but classical. If I'm not wrong, infinite descending shows that the only integer valus are 3 and 4. And if we plug these values, we will find to Pell equations, which aren't hard to be solved. 7) Seem to be easy. Was about to be a test for 10 grade student. Does it exist integer numbers satisfying integer,we can find at least one number so that Disscard 1) x also has to be an integer, right? 2) Wait I must be misreading this Couldn't you just say that x=1, then find three integers that add up to n!? 3) Not really : you have to find a fixed triple which satisfies this for any . After you've found such a triple you're only allowed to move 4) Hi, some questions: i)Find all triple (a.b.c) in such that:there exits satisfied for any n>= then exits an integer x satisfied: in part i) I only prove c=0 if ab<>0. ii)If we change to a polynomial in Z[x] whose degree >2 then what is the conclusion of the problem? 5) I think I've reduced it to the question: do there exist s.t. is a perfect square for all ? Solution there are no such (a,b,c): just use the condition for n=1,2,3,4 (you don't need more) and use the fast for any polynomial P with integer coefficients. let . wlog we have (else translate x) and (actually, is only possible for -1 and 1 by the fact, thus by sending -x to x we can assume ). then we have some x such that . it follows that and , hence or . thus, we have in the first case , and . these equations lead to , thus , hence , contradiction. in the second case , and , thus and . now take the condition for n=4: the discriminant must be a square, contradiction. • Very nice and good solution.But what about . 8) Table with infinite number of fields Email A table with an infinite number of fields contains numbers.Empty fields are considered to contain zeros.In any square with sides along the grid lines is the absolute value of the sum of the numbers equal or less 1.Regarding these conditions, determine the maximum value of the sum of all elements in a rectangle 5 x 1. The problem is called "Der Burische Hammer" after Wolfgang Burmeister Discard 1) Did you by any chance post it in this subforum because the numbers in the table are integers? (you just mentioned they were "numbers") 2) Good question grobber! The numbers are real numbers! 3) You can have squares formed by many small squares. For example, if you label the small squares in the obvious way, the cells for a square of side . The sum of the four numbers in these four cells has absolute value 4) so you can divide up each square? Then I still don't get why it isn't infinite, like replace 10^9 by whatever in the below example (where we split each square in 9 squares so all border squares fullfil the requirement): Perhaps can you reformulate it and/or make a drawing, cuz my english is too bad to understand this I'm afraid 5) No, man, you cannot divide each square. You have an infinite chess board (without any coloring, though ). This board is made up of small squares of fixed dimension. However, from these small squares you can form larger squares, made up of a certain number of small squares. Is it that hard? 6) For example: . are small squares (you can't find smaller squares on your board). However, the array formed by is another (larger) square. The problem says that, for example, in addition to . 7) then what does he mean by: 'an infinite field' and 'squares along the grid lines'? If the fields is infinite, and you may just group together to larger squares, where are the grid lines then??? Gosh I'll better just quit this problem, i don't get the heck of it 8) Dude, think about the integer lattice. That's naturally divided in small squares of side and it has grid lines of equations of the types and . Let be the number written in the cell numbered . The problem says that (I'm just giving some examples, I can't write all of the relations because there are infinitely many ) . It also says that . It also says that , and so on (for all integers ). How on Earth can it still be unclear? 9) Cyclic equations Turkey, 97 Prove that for each prime , there exists a positive integer n and integers not divisible by p which satisfy Hint Lemma: for a prime there is s.t. is a quadratic residue. Proof: Take s.t. (there is such a residue because ). Then satisfies the condition. We now take and . 10) IMO 1988/6, One of the hardest number theory problems If are integers, then is a perfect square. Well I have seen many solutions of this problem. Most assume to the contrary that q is not a perfect square. Then I see nothing which follows uses this assumption. (In fact I think the 'contradiction' holds even if we assume q to be a perfect square.) Then they choose a and b such that a+b is minimal. Then they proceed further to prove that and . In my opinion the proof goes exactly the same way had we assumed that is a perfect cube, fifth power, etc. Will we not get the same 'contradiction' if we assumed to be a perfect square? Can anyone explain this to me? Disscard 1) This is part of the solution I found in one of Titu's books. Let q be the number as stated in the problem. We want to show that among all pairs of nonnegative integers with the property and , the one with minimal has . But why? Can it be that some other and with not minimal have the property that is not a perfect square? If you know the solution (or rather, understand the solution), can you please explain everything in detail? 2) Your problem is WRONG, because if then can not be a square, because is a negative number !!!! 3) The problem is not wrong. It can be proved that if ab+1 were negative, the resulting fraction can NEVER be an integer. By the way how did you type those mathematical equations? 4) There is a solution to this problem in Arthur Engel's book, involving hyperbolas. As has been said, the task is to show that if a solution exists for , then one exists for the same with , in which case must equal . One shows that if a lattice point exists on the upper branch of the hyperbola , then a corresponding point exists directly below it. Then we flip it across the axis of symmetry and repeat until . Arthur Engel also rates it in his book (Problem Solving Strategies) as the hardest IMO problem until 1995. But I think this should probably be in the problem- solving section. 5) Hmm Are you kidding? This is a very standart problem for a descent method. And, of course, it can't be the hardest IMO problem (maybe easiest ). The same thing I can say about IMO95, pr6 (did Engle mean this problem?). 6) I still don't understand your solution. Can you please explain in detail? If is an integer, then is a perfect square. Can anyone teach me how to use Latex? Engel meant IMO 1996/5, a geometry problem. Thank you. But I saw in some other sites that teach us to use some sort of \begin document blablabla. What's the difference. And why is it that your line remains unchanged even when you use $? 7) [quote="Lek Huo Tang"]I still don't understand your solution. Can you please explain in detail? If $\frac{a^2+b^2}{1+ab} = q$is an integer, then $q$ is a perfect square. Can anyone teach me how to use Latex? Engel meant IMO 1996/5, a geometry problem.[/quote] You just need to use $ $ when you write formulas. 11) Rational number Find all numbers positive integer such that: a) is rational number b) is rational number I haven't got any solution. Discard 1) Quite a few of these have been posted and solved around here. Take the second one, for instance. Let . It's an algebraic integer, because it satisfies . This means that is also an algebraic integer, and so is . Since it's also rational, it means that it's integral, and it's easy from here: being in it can only be , and we solve it for these values. 2) Well, I solved this problem but i can0t remember the method, but grobber failed, because . In fact I think the unique solutions was . Good luck looking for a nice solution for this nice problem. Now if you have carefully read, and it is clear that you did not, you would have seen that grobber did not say that is the set of solutions for , but the values that takes, and now we can easily solve each of the 5 equations 3) giving all the solutions, including the one you specified above. This is the Olympiad section of the site. If needed more details on easy problems such as this one, one can use the High-School forums. PS Next time I will not waste anymore time explaining posts of other users in such detail, as for kids, but I will rather directly delete the additional balast 12) Old Equation Friend Prove that the only solution in rational numbers of the equation: is Discard 1) this is a pretty standard method for this sort of problem (fermat's method of infinite descent) must be divisible by 3, and from there we immediately get that must be divisible by 27, which is to say that . this then implies , which then implies that and so . so if is a solution, so is , and by induction is a solution for all natural k. and this is obviously impossible unless . 2) A very beautiful problem, proposed for IMO by Titu Andreescu is 3) pleuristique: the problem asks for rational solutions. My guess is that the solution for this problem (and harazi's problem) involve the factorization of 4) Are you sure? If we multiply by the product of the denominators, we aren't necessarily multiplying by a perfect cube. I solved it using the factorization. The expression factors into [...]... integers which is a UFD, are not Euclidean For example, Barnes and Swinnerton Dyer in two papers in Acta Math (1952) showed that is not Euclidean Davenport, Hua and Inkeri proved that there are only 22 quadratic Euclidean fields A way of showing that a quadratic field has class number 1 is by the theory of binary forms An ideal in a number ring in a quadratic field is a rank 2 module, generated by say, and... few that you'll succede The reason is that the general style of proof in the local cases is by reducing to equations in and work there (using quadratic reciprocity, perhaps) and then use Hensel's lemma to lift to the -adic case So you gain nothing by reverting the process 5) Andrei, thank you very much for enlarging upon this Can I find somewhere this article which you give reference to? On AMS site... classes, and the theory of reduction gives an effectively computable way of listing the reduced forms, which are canonical representatives of the classes If there is only one reduced form, then the ring is a PID 17) Diophantine equation The problem is actually from the Russian Math Olympiad this year Using very advanced number theory, I managed to prove that the following equations do not have integer solutions... biggest/smallest/one special (eg ) (when necessary modulus) of these numbers and compare them') One easy problem you can do this way is proving that solutions with or being zero has only 3) The use of -adic methods to attack diophantine equations in the integers is called the local-to-global principle For example, if you have a polynomial with integer coefficients, you could try to see if it has solutions... farther until Hence, producing desired always has a solution is a unique factorization domain, and since , but it doesn't divide either one of , it means that is not a prime in decomposable: we can write it as Now, since (their norm is , which, in turn, means that it's , we also have , and the only possibility left is , because both , which is a prime in ) must be primes in 4) Nice, these quadratic... as an example that Hasse's global-local principle is wrong for equations of degree greater than two) 16) (u,v) & prime Let be a prime number, and suppose that is an integer such that By cnsidering for various pairs of show that at least one of the equations has a solution Disscard 1) First, note that Choose for some positive integer from the interval Then Combine so that where and and to get , where... know any other methods bounding between squares/cubes (whichever it may be?) what else is there? Disscard 1) I also have a question: Can someone show me how to use p-adic numbers in attacking Diophantine equations ? Not "forced" solution just to show the solution with p-adic numbers but some more "natural" (I hope you understand what I actually mean ) What I've found on Internet is vastly general equations... p-adics I'll be blissed 2) If you have an equation in integers, assume that it has a 'postive' solution and proof out of this fact that it has a 'smaller' 'positive' solution Here 'positive' and 'smaller' can be of various meaning (for example when you have an equation in , you can define 'positive' as 'not all numbers are zero' or as 'none of the numbers is zero' and 'smaller' as 'take the biggest/smallest/one... 1996 on 14) nice equation unknown Find all such that: a) b) 15) x²+5=y³ something classical Prove that there is no solution for the equation in integers Disscard 1) This problem can be easily proved by using Legendre symbol For arbirtary odd prime such that , must have However, For the cases form prime form , right side contains Contradiction! 2) There is some mistake in your solution, because for there... same 5) I think there is wrong wth you because you say that always solution but the problem say another thing has 6) Maxal and grobber are right! Even if you only know the statement of the problem, on can prove that there are already Let be immediatly for some with , then : is even But this gives or equivalently 7) No, because most numbers fields with class number 1, that is, with ring of integers which . Table with infinite number of fields Email A table with an infinite number of fields contains numbers.Empty fields are considered to contain zeros .In any square with sides along the grid lines is. Number Theory in Mathlink Notice: There’re some mistakes in some solutions, so you have to read it carefully 1) system of equation including a prime number Solve the following system. and Inkeri proved that there are only 22 quadratic Euclidean fields. A way of showing that a quadratic field has class number 1 is by the theory of binary forms. An ideal in a number ring in