the problem is not too hard because the method of vieta-jumping is much more known today than 20 years before. first we take \[\frac{a^{2}+b^{2}}{ab+1}=k.\] so we can write instead \[a^{2}+b^{2}-kab-k=0.\] we assume $a>b$ ($a=b$ implies $\frac{2a^{2}}{a^{2}+1}\in\mathbb{Z}$ so $a=b=1$ and $k=1$.) we take the equation as a quadratic function in $a$. if we assume that there is a solution $a,b$ (otherwise the problem would be trivial), we have the solutions $a_{1}, a_{2}$ with $a_{1}=a$. wmls the equations $a_{1}+a_{2}=kb$ and $a_{1}a_{2}=b^{2}-k$ hold (vieta). we want to show $a_{2}\geq 0$. so we assume $a_{2}<0$ so $b^{2}-k<0$ so $k>b^{2}$. since $a_{1}>kb>b^{3}$ we get $ab+1>b^{4}+1$ which is a contradiction to \[ab+1 | b^{2}(a^{2}+b^{2})=b^{4}+1+(ab-1)(ab+1).\] so i can assume $a_{2}\geq 0$. because of $a_{1}a_{2}=b^{2}-k<b^{2}$ and $a_{1}>b$ $a_{2}<b$ holds. so we have created a new solution $a,b$ with $a<b$. We can re-do this until one of the numbers is $0$. Since the $k$ doesn't change we get $k=\frac{a^{2}}{1}=a^{2}$ which is a perfect square. Naphthalin

Peter wrote: Let $ a$ and $ b$ be positive integers such that $ ab + 1$ divides $ a^{2} + b^{2}$. Show that \[ \frac {a^{2} + b^{2}}{ab + 1} \] is the square of an integer. And futhermore Try *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 belongs to a finite set They are nice,do you think that P/s,I posted here but no reply http://www.mathlinks.ro/viewtopic.php?t=150373

Well let us assume that a^2 + b^2 / 1 + ab = q (where q is some non square integer) converting it into a linear form we get a^2 + b^2 -qab -q =0 Now let us consider the equation x^2 + y^2 - qxy - q =0.......(1) Let (m,n) be a soln. for which m^2 + n^2 is the least . (m>n without any loss of generality) Clearly m is the soln of the quadratic x^2 + n^2 -qxn - q =0.........(2) let us consider the other root of (2) say p. Now we have, p + m = qn pm = n^2 - q > n^2 Now we have m > n Thus clearly p <n Therefore p^2 + n^2 < 2n^2 < m^2 + n^2 Which is clearly a contradiction of our original assumption. Therefore the given expression with the given conditions must be a square. P.S. - The method is known as infinite descent in case. And sorry for the crude typing!

A K Bhattacharyya wrote: pm = n^2 - q > n^2 Shouldn't it be pm = n^2 - q < n^2 But that's ok with the part that follows... Rather it would be wrong otherwise. A K Bhattacharyya wrote: Thus clearly p <n Therefore p^2 + n^2 < 2n^2 < m^2 + n^2 Why $ p < n$ implies $ p^2 < n^2$ $ p$ could be negative, so you might have to prove it otherwise like Naphthalin did... EDIT: ... doesn't that show that there's actually no such solutions... square or not... ??

No it can't. We are dealing with both positive roots remember ? Yeah. it has to be < . Typing error there Sorry.

A K Bhattacharyya wrote: We are dealing with both positive roots remember ? We are dealing with positive $ m,n$, but $ p$ could be negative, which is the other root of that quadratic, you don't yet have both of it's roots positive, or am I missing something ??

My solution is the following. Let k = (a^2 + b^2) / (ab + 1) , and let k isn’t the square of an integer. The equality is equivalent to a^2 + b^2 - kab = k . The equality is symmetric by a and b, so we may accept, that a>=b. The equality is a quadratic equation by a . Let a and a1 are its roots. From Viet’s theorem a1 + a = kb , therefore a1 is integer. If a1=0 , then k = b^2 (contradiction). If a1 < 0 , then a > kb , and therefore a^2 + b^2 - kab = b^2 + a (a - kb) >= b^2 + a > b^2 + kb > k , which is impossible. Thus a1 > 0. From Viet’s theorem a1 = (b^2 - k) / a <= (a^2 - k) / a < a. If we take b1 = b, then (a1, b1) is a new solution of equation, for which a1 + b1 < a + b. If we do the same for (a1, b1), we will get a new solution (a1, b2) st a2 + b2 < a1 + b1. Continuing the operation, we will get (a, b), (a1, b1) , (a2, b2),.... infinitely many solutions, for which a + b > a1 + b1 > a2 + b2 > .... , which is impossible, because ai, bi (i = 1, 2, 3, ....) are positive integers. Therefore k is a perfect square (Q. E. D.). We can notice that a=2, b=8 (or a=b^3) numbers satisfy to equality.

a brilliant solution by someone else (not by me) without loss of generality we may assume that $ a < b$ let $ n = \frac {a^2 + b^2}{ab + 1}$ define the infinite integer sequence $ \cdots,x_{ - 1},x_0,x_1,x_2,\cdots$ as follows: \[ x_0 = a,\quad x_1 = b,\quad x_{i + 1} = nx_i - x_{i - 1}\mbox{ for }i = 1,2,\cdots,\mbox{n} \] it's easy to prove by mathematical induction that \[ x_i^2 + x_{i - 1}^2 = n(x_ix_{i - 1} + 1)\mbox{ for }k = \cdots, - 1,0,1,2,\cdots \] and by mathematical induction that \[ \cdots < x_{ - 1} < x_0 < x_1 < x_2 < \cdots,\mbox{(i.e. }x_i\mbox{ is a strictly increasing sequence)} \] the key to proving this problem is that one of the term $ x_i$ must be $ 0$ to prove this, we take $ x_m$ to be the smallest positive term in the sequence such that $ x_m > 0$, by defination that this term is smallest, we have $ x_{m - 1}\leq 0$ suppose that $ x_{m - 1} < 0$, then in the following equation $ (*)$, L.H.S. is greater than 0 while R.H.S. is less than or equal to 0, which is a contradiction \[ x_m^2 + x_{m - 1}^2 = n(x_mx_{m - 1} + 1)\qquad(*) \] therefore $ x_{m - 1} = 0$, substitute into equation $ (*)$ we have $ n = x_m^2 = \frac {a^2 + b^2}{ab + 1}$ Q.E.D.

i think this problem is overanalyzed: by definition, since ab+1 divides a^2+b^2, then a^2+b^2-abc-c=0, for c the number that results when ab+1 divides a^2+b^2. by considering the factorization of this into (a-x)(a-y), it is easily seen that xy= b^2-c x+y= b*c Now, since this deals with positive integers, b^2-c is less than or equal to b^2 b*c is less than or equal to b This gives the inequalities 0 less than or equal to c less than or equal to 1. since the quotient of the two is an integer by definition, that means c is either 0 or 1, both of which are squares. I dont see the need to prove it is a square when it suffices to show it only equals 0 or 1

braudiap, why is $b \cdot c \le b$?

Naphthalin wrote: the problem is not too hard because the method of vieta-jumping is much more known today than 20 years before. first we take \[\frac{a^{2}+b^{2}}{ab+1}=k.\] so we can write instead \[a^{2}+b^{2}-kab-k=0.\] we assume $a>b$ ($a=b$ implies $\frac{2a^{2}}{a^{2}+1}\in\mathbb{Z}$ so $a=b=1$ and $k=1$.) we take the equation as a quadratic function in $a$. if we assume that there is a solution $a,b$ (otherwise the problem would be trivial), we have the solutions $a_{1}, a_{2}$ with $a_{1}=a$. wmls the equations $a_{1}+a_{2}=kb$ and $a_{1}a_{2}=b^{2}-k$ hold (vieta). we want to show $a_{2}\geq 0$. so we assume $a_{2}<0$ so $b^{2}-k<0$ so $k>b^{2}$. since $a_{1}>kb>b^{3}$ we get $ab+1>b^{4}+1$ which is a contradiction to \[ab+1 | b^{2}(a^{2}+b^{2})=b^{4}+1+(ab-1)(ab+1).\] so i can assume $a_{2}\geq 0$. because of $a_{1}a_{2}=b^{2}-k<b^{2}$ and $a_{1}>b$ $a_{2}<b$ holds. so we have created a new solution $a,b$ with $a<b$. We can re-do this until one of the numbers is $0$. Since the $k$ doesn't change we get $k=\frac{a^{2}}{1}=a^{2}$ which is a perfect square. Naphthalin sorry what is the vieta jumping method

mozilla wrote:

wow. forget vieta jumping, this one is the best. do you know who gave this solution?

Naphthalin wrote: we want to show $a_{2}\geq 0$. so we assume $a_{2}<0$ so $b^{2}-k<0$ so $k>b^{2}$. since $a_{1}>kb>b^{3}$ we get $ab+1>b^{4}+1$ which is a contradiction to \[ab+1 | b^{2}(a^{2}+b^{2})=b^{4}+1+(ab-1)(ab+1).\]so i can assume $a_{2}\geq 0$. because of $a_{1}a_{2}=b^{2}-k<b^{2}$ and $a_{1}>b$ $a_{2}<b$ holds. so we have created a new solution $a,b$ with $a<b$. We can re-do this until one of the numbers is $0$. Since the $k$ doesn't change we get $k=\frac{a^{2}}{1}=a^{2}$ which is a perfect square. Naphthalin How do you get $a_{1}>kb>b^{3}$?

Can someone answer the above?

Peter wrote: Let $a$ and $b$ be positive integers such that $ab+1$ divides $a^{2}+b^{2}$. Show that \[\frac{a^{2}+b^{2}}{ab+1}\]is the square of an integer. Use Vieta jumping

This is also the legendary problem 6, from the 1988 IMO. https://artofproblemsolving.com/community/c6h57282_one_of_my_favourites_problems_yeah

its also solvable by finding quadratic equation roots.

Vieta jump on \[a^{2}+b^{2}-kab-k=0.\]

I tried to solve this problem by using elementary algebra. However, by assumption X = a^2 if b=0, the solve for x is a^2 = -2 https://documentcloud.adobe.com/link/track?uri=urn:aaid:scds:US:ed532489-a454-48eb-b48e-128f876d3a24 https://twitter.com/globeconomia/status/1453053910353072133?s=20

Peter wrote: Let $a$ and $b$ be positive integers such that $ab+1$ divides $a^{2}+b^{2}$. Show that \[\frac{a^{2}+b^{2}}{ab+1}\]is the square of an integer. Look here