IMO-Auswahlklausur

1. Lemma: If $k, l, a$ are positive integers with $a = (k^{2} + l^{2} + 1)/(k l)$, then $a = 3$ follows.

Proof of the lemma: For a given value of $a$ we consider a pair $(k, l)$ with $k^{2} + l^{2} + 1 = a k l$ and minimal sum $k + l$ among all such pairs. Without loss of generality let $k \leq l$.

First suppose $k < l$. By assumption, $k$ satisfies the quadratic equation
$$x^{2} + l^{2} + 1 - a x l = 0,$$
for whose 2nd solution $k'$ it follows by Vieta's theorem that: $k + k' = a l$ and $k k' = l^{2} + 1$. Thus $k' = a l - k$ is an integer, it is positive because $k' = (l^{2} + 1)/k$, and since $k \geq l + 1$ and $k k' = l^{2} + 1 < (l + 1)^{2}$ we have $k' \leq l$, so $(l, k')$ is a pair with smaller sum $l + k' < k + l$, contrary to the choice of $(k, l)$.

Hence $k = l$ holds, so $k^{2}(a - 2) = 1$, from which $k = 1$ and $a = 3$ follow.

Since $m \geq 1$ and $n \geq 1$, the expression $p^{m-1} q^{n-1} = (p^{2} + q^{2} + 1)/(p q) + 2$ is an integer, so by the lemma $p^{m-1} q^{n-1} = 5$ and thus $p^{m-1} = 1, q^{n-1} = 5$ or $p^{m-1} = 5, q^{n-1} = 1$.

In the first case $n = 2$ and $q = 5$; $p$ satisfies equation (1) for $l = 5, a = 3$ with solutions $2$ and $13$, from $p^{m-1} = 1$ it follows in each case that $m = 1$, and by substitution one confirms the solution quadruples $(1, 2, 2, 5)$ and $(1, 2, 13, 5)$. The second case yields correspondingly $(2, 1, 5, 2)$ and $(2, 1, 5, 13)$.

All summands of the original equation except for $q^{2} + 1$ are multiples of $p$, hence $p \mid q^{2} + 1$ holds. Thus the cases $q = 1, 2, \ldots, 6$ can be treated by direct checking, yielding the solution quadruples from Solution 1. For $p \geq q \geq 7$ (analogously: $p \geq p \geq 7$) there are no solutions:

Case 1: $m > 1$: Then $p^{m} q^{n} \geq p^{2} q > 6 p^{2} \geq (p + q)^{2} + 1$.

Case 2: $n > 2$: Then, because $p \leq q^{2} + 1$: $p^{m} q^{n} \geq p q^{3} \geq p^{2} q - p q > 5 p^{2} > (p + q)^{2} + 1$.

Case 3: $m = n = 1$: For this, $p^{2} + p q + q^{2} + 1 = 0$ would have to hold - impossible for $p > 0, q > 0$.

Case 4: $m = 1, n = 2$: The quadratic equation $p q^{2} = (p + q)^{2} + 1$ in $p$ has discriminant $D = q^{4} - 4 q^{3} - 4$, which for $q \geq 7$ lies between the consecutive square numbers $(q^{2} - 2q - 3)^{2}$ and $(q^{2} - 2q - 2)^{2}$, so $\sqrt{D}$, and hence also $p$, is not an integer.