67th Romanian Mathematical Olympiad

Let us suppose that $a^2=b^2+c^2$. The numbers $b$ and $c$ could not be both odd (the sum of two odd numbers is $4k+2$ which is not a square). Then at least one is an even number and thus, the product $bc$ is even.

On the other hand, the discriminants of the two equations are ${\Delta }_1=a^2-4d$ and ${\Delta }_2=a^2+4d$. If we define $d=\frac{bc}{2}$ we have $${\Delta }_1=a^2-4d=b^2+c^2-4\frac{bc}{2}=(b-c{)}^2,$$ and thus, the roots of the first equation are $x_{1,2}=\frac{a\pm (b-c)}{2}$. It is clear that $x_{1,2}$ are integers. (If both $b,c$ are even, then $a$ is also an even number, and if $b,c$ have different parities, then $a$ is odd, as $b-c$ it is.) Similarly we can show that the second equation has integer roots.

Conversely, we suppose that the equations have only integer roots. Then their discriminants are squares. Let ${\Delta }_1=u^2$ and ${\Delta }_2=v^2$. We have $$\begin{array}{rl}{a}^2-4d& =u^2,\\ a^2+4d& =v^2.\end{array}$$

It is clear that $u,v$ have the same parity. If we add the two equalities we get $$a^2=\frac{u^2+v^2}{2}={\left(\frac{u+v}{2}\right)}^2+{\left(\frac{u-v}{2}\right)}^2.$$ Define now $b=\frac{u+v}{2}$ and $c=\frac{u-v}{2}$. The numbers $b$ and $c$ are integers and $a^2=b^2+c^2$.