# 1988 IMO Problem 6 International Mathmatical Olympiad --- ## Problem Let $a, b \in {\displaystyle \mathbb {N}}$ s.t. $(ab + 1)|(a^{2} + b^{2})$. Show that $\frac {a^{2} + b^{2}}{ab + 1}$ is the square of an integer. ---- ## Solve Let $a', b'$ = $min(a+b)$ and $a > b$, $k = \frac {a^{2} + b^{2}}{ab + 1}$ is not a square of an integer. We have $a'^2 + b'^2 = k(a'b' + 1)$ ---- ## Solve $a'^2 + b'^2 = k(a'b' + 1)$ $\implies a'^2 - (kb')a' + (b'^2-k) = 0$ ---- ## Solve Since we have $a'^2 - (kb')a' + (b'^2-k) = 0$, Consider this quadratic equation: $x^2 - (kb')x + (b'^2-k) = 0$ Let the solution other than $a'$ be $x'$ ---- ## Solve By Vieta's Formula, another root $x' = kb' - a' = \frac{b'^2 - k}{a'}$ And we have $x' = \frac{b'^2 - k}{a'} ≤ \frac{b'^2}{a'} < a'$ $\implies x' + b' < a' + b'$ (Contradict!) ---- # Solve $k$ can't be a number that is not square of an integer. $\implies k$ is square of an integer. Q.E.D. --- # The End ---