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
$$\begin{aligned} a^2 - 4d &= u^2, \\ a^2 + 4d &= v^2. \end{aligned}$$

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 $a^2 = b^2 + c^2$.