I use a delicate argument involving properties of hyperbolas. If anyone has a nicer argument, I would be very interested. Let $H_q: x^2+y^2-qxy+1=0$, a hyperbola symmetric in $x,y$. I want to show that only $H_3$ contains lattice points. Since it is symmetric, we work only on the positive branch with $x \leq y$. Observe that $T: (x,y) \rightarrow (qx-y, x)$ allows us to generate new points on the hyperbola, since $(qx-y)^2+x^2-qx(qx-y)+1=x^2+y^2-qxy+1$. If $x<y$, such a new point must be either further down the branch or on a different branch. The other positive branch (which is the only possibility since $x$ is positive) contains the point $(y,x)$ by symmetry, so if our new point lies on this branch, $y=qx-y$, which is equivalent to $y= \frac{qx}{2}$. Intersecting this line with $H_q$, $x^2(q^2-4)=1$, but this is clearly not a lattice point, so this possibility doesn't concern us. So by repeated application of $T$ I must at some point reach a lattice point at which $x=y$, since otherwise I have an infinitely decreasing sequence of positive integers, so here $(q-2)x^2=1$. Both factors are integers, and $x^2$ is positive, so $q=3$ must be the only hyperbola containing lattice points, which was to be shown. This problem seems hard. I doubt I would have found it if I hadn't read Engel's Number Theory chapter.

i have one vieta-jumping again. \[ \frac {x^{2} + y^{2} + 1}{xy} = k \] -> \[ x^{2} - kxy + y^{2} + 1 = 0 \] we assume that we have a solution, otherwise the problem would be trivial. if $ x = y$ we get $ x,y|1$ so $ \frac {x^{2} + y^{2} + 1}{xy} = 3$ and we have finished. $ k \geq 2$ must hold because of AMGM. so if we assume $ x > y$ we get a new solution $ x_{2} < y$ with $ x_{2} = \frac {y^{2} + 1}{x_{1}}$ (by vieta). it must be a "really" $ <$ for $ y > 1$. So we can jump down by vieta and vieta again and get $ y = 1$ and $ x = 1$. because of the fact that we have already done this, the problem got PWN3D and the quotient is always $ 3$. Naphthalin

Nice. I think that's kind of related to my method, but much better.

thx^^ the method of vieta-jumping works really often. too often if you ask me Naphthalin

I'm a beginner and I'm having some problem understanding your solution. What do you mean, using vieta's jumping up and down until, ..." ?? I couldn't get the last part of A3 either. Anyone's answer would be a great help. thnxs

I'll write Napthalin's solution a bit easier for you. Assume there exists $ x,y,k > 0$ with $ x^{2} - kxy + y^{2} + 1 = 0$ (else there is nothing to prove), and take the pair $ (x,y)$ where $ x + y$ is minimal, wlog $ x\ge y$. If $ x = y$ or $ y = 1$ we easily get $ x = y = 1$ and $ k = 3$. For $ k\not = 3$ (by AM-GM this is $ k > 3$) we have $ y > 1$, so solving for $ x$ we get two positive roots $ x,x' = \frac {ky\pm\sqrt {k^2y^2 - 4y^2 - 4}}{2}$. By the minimality of $ x + y$, we need $ x'\ge x > y$, while de Vieta tells us $ xx' = y^2 + 1$, contradiction.

OK,and more(I think it is very nice) *Let $ p,q\in \mathbb Z^+,p\le q$ Find $ a,b$ satisfying $ \frac{(a-b)^2}{pab-q}=k (k\in \mathbb N)$ *$ \frac{(a-b)^2+m}{pab-q}=k(m\mathbb N^*)$ so K is belong to a finite set They are nice,do you think that

Here is my way to express the same ideas than those proposed by Naphtalin and Peter. Let (m, n) be a solution of equation $ m^2 + n^2 - qmn + 1 = 0 $ with $0 < m \le n $ and $m + n$ minimal. We show that m = n. As (qm - n, m) is another solution, we have, by minimality, $ m + n \le qm - n + m$ so $2n \le qm$ and $2n^2 \le qmn$; thus $ 0 = m^2 + n^2 - qmn + 1 \le m^2 - n^2 + 1 $ which is possible only when m = n.

Using the lemma $ x^2+y^2+z^2=kxyz $ has solutions only for $ k=1 $ or $ k=3 $, taking $ z=1 $ we have $ x^2+y^2+1=kxy $ so if $ k=1, xy=x^2+y^2+1\ge\ 2xy+1 $ which is not true so it must be $ k=3 $ or $ x^2+y^2+1=3xy $

I like andrejilievski's solution best. Sometimes when I am doing vieta jumping, I get x2 = something, and then I go blank like "so what". Also, we can manipulate to get sqrt {k^2b^2 - 4b^2 - 4} = (b^2+1-a^2)/a, and I get nothing out of it.

andrejilievski wrote: Using the lemma $ x^2+y^2+z^2=kxyz $ has solutions only for $ k=1 $ or $ k=3 $, taking $ z=1 $ we have $ x^2+y^2+1=kxy $ so if $ k=1, xy=x^2+y^2+1\ge\ 2xy+1 $ which is not true so it must be $ k=3 $ or $ x^2+y^2+1=3xy $ Quite a hefty lemma, using the theory of Markoff (and Hurwitz) equations ...

See also: http://www.artofproblemsolving.com/community/c6h1080455p4744999

Nice proofs all. They are all about the same to me. But what is "Vieta Jumping"? This is the first time I hear.

See Project Pen for a nice Vieta Jumping explanation. Could someone read my solution and see if it is correct? Consider the set $$S(k) = \left\{ (x,y)\in\mathbb{Z}^+\times\mathbb{Z}^+ : \dfrac{x^2+y^2+1}{xy}=k.\right\}.$$By AM-GM, we see that $k>2$. Now if $x=y$, then $2x^2+1=kx^2\implies \dfrac{1}{k-2} = x^2$, and hence $x=y=1, k=3$. Now let $(x,y)$ be the element of $S(k)$ with minimal $x+y$ and assume WLOG that $x>y$. Now consider the quadratic in $X$ $$X^2+y^2+1=kyX \iff X^2-kyX+y^2+1=0.$$One solution is obviously $X=x$; let the other one be $X = x_1$. Then by Vieta's formulas, $$ x_1 = ky-x = \dfrac{y^2+1}{x}.$$The first equality tells us that $x_1$ is integer, while the second one tells us that $x_1$ is positive; hence, $x_1\in\mathbb{Z}^+$ and $(x_1,y)\in S(k).$ However, note that $x^2-y^2=(x-y)(x+y)>1$, and thus $$x^2>y^2+1 \implies x>\dfrac{y^2+1}{x} = x_1,$$a contradiction to the minimality of $(x,y)$ in $S(k)$. Therefore, $x=y$, from which our conclusion follows.

hamup1, your solution is correct , but I don't know if you can take the minimal x+y, because don't you have to rigorously prove for all x,y? Please correct me if I am wrong, I am a novice at this Also found here, but they solve with infinite descent: https://en.wikipedia.org/wiki/Vieta_jumping#Constant_descent_Vieta_jumping

Peter wrote: Let $x$ and $y$ be positive integers such that $xy$ divides $x^{2}+y^{2}+1$. Show that \[\frac{x^{2}+y^{2}+1}{xy}=3.\] Use Vieta jumping

Apart from the solution we can show, that there exist infinitely many such $(x,y)$ where $x\le y$. Take $x_0=1\wedge y_0=2$. Then the equation holds. Define $(x_{n+1},y_{n+1})=(y_n,3y_n-x_n)$ for $n\ge 0$. Since $x_n\le y_n$ then by induction hypothesis $x_n,y_n\in Z_+\wedge x_n\le y_n\implies \left(x_{n+1}\ge x_n\wedge y_{n+1}\ge 2y_n>y_n\right)$ so we constructed infinitely many solutions.

mavropnevma wrote: andrejilievski wrote: Using the lemma $ x^2+y^2+z^2=kxyz $ has solutions only for $ k=1 $ or $ k=3 $, taking $ z=1 $ we have $ x^2+y^2+1=kxy $ so if $ k=1, xy=x^2+y^2+1\ge\ 2xy+1 $ which is not true so it must be $ k=3 $ or $ x^2+y^2+1=3xy $ Quite a hefty lemma, using the theory of Markoff (and Hurwitz) equations ... Rip