20. N2 (RUS) $\mathrm{IMO}$ The positive integers $a$ and $b$ are such that the numbers $15 a+165$ and $16 a-155$ are both squares of positive integers. What is the least possible value that can be taken on by the smaller of these two squares?
20. Let $15 a+16 b=x^{2}$ and $16 a-15 b=y^{2}$, where $x, y \in \mathbb{N}$. Then we obtain $x^{4}+y^{4}=(15 a+16 b)^{2}+(16 a-15 b)^{2}=\left(15^{2}+16^{2}\right)\left(a^{2}+b^{2}\right)=481\left(a^{2}+b^{2}\right)$.
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, where $p \neq 1(\bmod 8)$. Then $p \mid x$ and $p \mid y$.
Proof. Since $p \mid x^{s}-y^{s}$ and by Fermat's theorem $p \mid x^{p-1}-y^{p-1}$, we deduce that $p \mid x^{d}-y^{d}$, where $d=(p-1,8)$. But $d \neq 8$, so $d \mid 4$. Thus $p \mid x^{4}-y^{4}$, which implies that $p \mid 2 y^{4}$, 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 - 481 and $b=481$.
Second solution. Note that $15 x^{2}+16 y^{2}=481 a^{2}$. It can be directly verified that the divisibility of $15 x^{2}+16 y^{2}$ by 13 and by 37 implies that both $x$ and $y$ are divisible by both primes. Thus $481 \mid x, y$.