Balti Tee 1995 · Ülesanne 3

Solution:

Since $a$ and $c$ are odd, $b$ must be even. We have $a^{2} = c^{2} - b^{2} = (c + b)(c - b)$. Let $d = \operatorname{gcd}(c + b, c - b)$. Then $d$ divides $(c + b) + (c - b) = 2c$ and $(c + b) - (c - b) = 2b$. Since $c + b$ and $c - b$ are odd, $d$ is odd, and hence $d$ divides both $b$ and $c$. But $b$ and $c$ are relatively prime, so $d = 1$, i.e., $c + b$ and $c - b$ are also relatively prime. Since $(c + b)(c - b) = a^{2}$ is a square, it follows that $c + b$ and $c - b$ are also squares. In particular, $b + c$ is a square as required.