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IMO 1988 question No. 6

I have a confusion in the question.The question is as follows:

$a$ and $b$ are positive integers and $ab+1$ is a factor of $a^2+b^2$. Prove that $\displaystyle\frac{a^2+b^2}{ab+1}$ is a perfect square.

My confusion: According to me, $a$ and $b$ have only one value ie $a=b=1$. What are the other possible values of $a$ and $b$?

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Snehil Sinha
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    I think you should expand on the "According to me" part and show the community your justification for why $a=b=1$. – yurnero Jun 09 '16 at 19:00
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    For example, $a=2,b=8$ giving $17$ as a factor of $68$. Or $a=8,b=30$ giving $241$ as a factor of $964$. – almagest Jun 09 '16 at 19:03
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    https://en.wikipedia.org/wiki/Vieta_jumping#Standard_Vieta_jumping – Will Jagy Jun 09 '16 at 21:28
  • http://math.stackexchange.com/questions/483771/imo-1988-problem-6 – Behrouz Maleki Jun 20 '16 at 20:17
  • there is a counter-example, so the statement of problem 6 is wrong: proof: take a=2 and b=9, then aa+bb=4+81=85. a*b+1=18+1=19 which divides with result = 5, which is not a square. q.e.d. – Hermann Voellinger Aug 24 '19 at 14:32
  • $19\cdot5=95$.There is no error. A past contestant described the question to me in mid 90s, and I also solved it then (took me a couple of day though :-). Proofs can probably be found be searching. – Jyrki Lahtonen Aug 24 '19 at 14:35

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Let us assume $(a^2 + b^2)/(ab + 1) = k \in \mathbb{N}$. We then have $a^2 - kab + b^2 = k$. Let us assume that $k$ is not an integer square, which implies $k \ge 2$. Now, we observe the minimal pair $(a, b)$ such that $a^2 - kab + b^2 = k$ holds. We may assume, without loss of generality, that $a \ge b$. For $a = b$, we get $k = (2 - k)a^2 \le 0$, so we must have $a > b$.

Let us observe the quadratic equation $x^2 - kbx + b^2 - k = 0$, which has solutions $a$ and $a_1$. Since $a+ a_1 = kb$, it follows that $a_1 \in \mathbb{Z}$. Since $a > kb$ implies $k > a + b^2 > kb$ and $a = kb$ implies $k = b^2$, it follows that $a < kb$ and thus $b^2 > k$. Since $aa_1 = b^2 - k > 0$ and $a>0$, it follows that $a_1 \in \mathbb{N}$ and $a_1 = (b^2 - k)/a < (a^2 - 1)/a < a$. We have thus found an integer pair $(a_1, b)$ with $0 < a_1 < a$ satisfies the original equation. This is a contradiction of the initial assumption that $(a, b)$ is minimal. Hence, $k$ must be an integer square.

user345872
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  • I found no way to show that $ a > k b $ implies $ k > a + b ^ 2 $. I just observed that assuming $ a $ is a positive integer, $ a > k b $ implies $ a \ge k b + 1 $ which implies $ a ( a - k b ) \ge a $ and thus $ k \ge a + b ^ 2 $. Of course this doesn't change anything important about the whole proof but it's worth noticing. – Mohsen Shahriari Mar 14 '17 at 18:41