IMO 1996 SL N2

Origin: RUS | Category: Number Theory

Problem

The positive integers $a$ and $b$ are such that the numbers

$15a + 16b$ and $16a - 15b$ are both squares of positive integers. What is

the least possible value that can be taken on by the smaller of these two

squares?

Solution

Let $15a + 16b = x^2$ and $16a - 15b = y^2$, where $x, y \in \mathbb{N}$. Then we obtain

$$x^4 + y^4 = (15a+16b)^2 + (16a-15b)^2 = (15^2 + 16^2)(a^2 + b^2) = 481(a^2 + b^2).$$

In particular, $481 = 13 \cdot 37 \mid x^4 + y^4$. We have the following lemma.

Lemma. Suppose that $p \mid x^4 + y^4$, where $x, y \in \mathbb{Z}$ and $p$ is an odd prime, with $p \not\equiv 1 \pmod{8}$. Then $p \mid x$ and $p \mid y$.

Proof.

Since $p \mid x^8 - y^8$ and by Fermat’s little theorem $p \mid x^{p-1} - y^{p-1}$, we deduce that $p \mid x^d - y^d$, where $d = \gcd(p-1, 8)$. But $d \neq 8$, so $d \mid 4$. Thus $p \mid x^4 - y^4$, which implies that $p \mid 2y^4$, i.e., $p \mid y$ and $p \mid x$.

In particular, we can conclude that $13 \mid x, y$ and $37 \mid x, y$. Hence $x$ and $y$ are divisible by $481$. Thus each of them is at least $481$.

On the other hand, $x = y = 481$ is possible. It is sufficient to take $a = 31 \cdot 481$ and $b = 481$.

Second solution.

Note that $15x^2 + 16y^2 = 481a^2$. It can be directly verified that the divisibility of $15x^2 + 16y^2$ by $13$ and by $37$ implies that both $x$ and $y$ are divisible by both primes. Thus $481 \mid x, y$.