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This question was originally asked in the 1988 IMO. It has been answered many times and in almost as many ways. I have not been able to find a complete and logical set of solutions this problem.

Two solution sets are known.

1) K=B^2,A=B^3.; & 2) K=M^2/4,B=M^3/8 .

I know how Set 1) was derived. How was Set 2) derived?

jllbones
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    @DonlansDonlans I think this is implicit in the fact that the OP says $$\frac{A^2+B^2}{AB+1}=K$$ and $K$ is integral. – Dave Aug 26 '19 at 23:38
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    It appears what you are saying is that you have seen many solutions and understand none. Meanwhile, I put a careful proof at http://math.stackexchange.com/questions/2188567/showing-that-m2-n21-is-a-square/2193043#2193043 The reference for this is Hurwitz (1907) http://zakuski.utsa.edu/~jagy/Hurwitz_A_1907.pdf – Will Jagy Aug 26 '19 at 23:52
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    LEMMA

    Given integers $$ M \geq m > 0, $$ along with positive integers $x,y$ with $$ x^2 - Mxy + y^2 = m. $$ Then $m$ is a square.

     – Will Jagy Aug 26 '19 at 23:54
  • The fraction to the left does not always yield an integer. What conditions on A, B are required? Example: A=4, B=5. Also for A=1, B=5 K is not an integer. Please add the constraints or describe the problem in full or at least mention a reference. – NoChance Aug 27 '19 at 00:08
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    Yes, the complete solution is one of the first results if you google the question. The Wikipedia page for Vieta Jumping has a decent proof that you can work through. There are tons of questions already on the site. A simple search will allow you to browse them – D.R. Aug 27 '19 at 00:15
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    http://math.stackexchange.com/questions/2344722/proof-check-on-qn-6-1988-imo and https://math.stackexchange.com/questions/645079/prove-quotient-is-a-perfect-square and https://math.stackexchange.com/questions/28438/alternative-proof-that-a2b2-ab1-is-a-square-when-its-an-integer and https://math.stackexchange.com/questions/2533928/does-this-proof-for-question-6-have-a-flaw and https://math.stackexchange.com/questions/2186630/simple-solution-to-question-6-from-the-1988-math-olympiad and https://math.stackexchange.com/questions/1841959/proving-that-a-particular-number-is-the-square-of-an-integer – Gerry Myerson Aug 27 '19 at 01:40
  • Also, https://math.stackexchange.com/questions/2078129/proof-question-of-integers and https://math.stackexchange.com/questions/1992951/vieta-jumping-related-to-imo-problem-6-1988-if-ab-1-divides-a2-b2-t and surely half-a-dozen others. Suggest you have a look at these, and any others linked to them, and see whether they answer your question. – Gerry Myerson Aug 27 '19 at 01:43
  • https://mathoverflow.net/questions/250172/when-is-fa-b-fraca2b21ab-a-perfect-square-rational-number/250300#250300 – individ Aug 27 '19 at 04:22

2 Answers2

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I've seen many solutions, but I think this solution was the most interesting.

https://en.m.wikipedia.org/wiki/Vieta_jumping

The Wikipedia answer is incomplete. See edited version of original question.

jllbones
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Capablanca
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I would like to give a simple solution based on basic algebraic principles of quadratic equations.

Let, $\frac{A^2 +B^2}{AB+1} = K$ and B = XA, where, X is a rational number (as both A and B are integers).

Then, $\frac{A^2 +B^2}{AB+1}$ = $\frac{A^2 + A^2X^2}{XA^2 +1}$ = $K$

This gives the following quadratic equation in X,

$A^2X^2$ -$A^2KX$ + $A^2 - K$ = $0$

As X is a rational number, sum ($A^4K^2$ - $4A^2(A^2-K)$ ) is either a perfect square or equal to zero.

(If the solution of a quadratic equation, $ax^2$ + $bx$ + $c$ is a rational number, then the term $b^2$ - $4ac$ must be either a perfect square or equal to zero)

Case 1: ($A^4K^2$ - $4A^2(A^2-K)$ ) is equal to zero

$\Rightarrow$ $A^4K^2$ + $4A^2K$ -$4A^4$ = $ 0$

$\Rightarrow$ $K$ = $ \frac{-2 \pm 2 \sqrt{2}}{A^2} $

This is not a desired solution, as K is an integer.

Case 2: ($A^4K^2$ - $4A^2(A^2-K)$ ) is a perfect square

$A^4K^2$ - $4A^2(A^2-K)$ is a quadratic equation in $A^2$

Therefore, $(K^2 - 4)A^2$ +$4KA^2$ is a perfect square,

$\Rightarrow$ Either $16K^2$ = $0$ or $4K$ = $0$(If the quadratic equation, $ax^2$ + $bx$ + $c$ is a perfect square, then one possibility is ($b^2$ - $4ac$) is equal to zero, the other possibility is both $b$ and $c$ are equal to zero)

$\Rightarrow$ $K$ = $0$ ; This is not a desired solution.

Reconsidering the equation $(K^2 - 4)A^2$ +$4KA^2$ and writing it in the following way,

$A^4K^2$ - $4A^2(A^2-K)$

Now, as we have exhausted the other alternatives, this equation is a perfect square only if the second term is equal to zero, or $4A^2(A^2-K)$ = $0$.

$\Rightarrow$ $A^2$ = $K$ $\Rightarrow$ $B$ = $A^3$ (This completes the proof of the preliminary question!)

Let M be an even integer such that M = 2A,

$\Rightarrow$ $K$ = $\frac{M^2}{4}$ and $B$ = $\frac{M^3}{8}$ (Hence both the solution sets as proposed by jllbones are the same!)

Vijay Honkalaskar
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