Some squares

The idea is that if $a \ne 0,$ then there is a corresponding $c$ exactly half the time (as expected, heuristically), and in that case there are two such $c,$ so the total number of such solutions is $70 \cdot 70 = 4900.$ Then it's easy to count the solutions when $a=0$ : there are $2 \cdot 70+1 = 141$ (the lone $1$ comes from $a=b=c=0$ ). So the total answer is $\fbox{5041},$ which happens to equal $71^2.$

The only thing we haven't justified is that first sentence, so let's write down a formal proof. First of all, $2$ is a square mod $71,$ so the equation is equivalent to $a^2+b^2 \equiv d^2$ mod $71.$ Now let $x=d-b, y = d+b.$ The condition that $a \ne 0$ corresponds to $x,y \ne 0.$ Also, there is a one-to-one correspondence between pairs $(x,y)$ and pairs $(b,d),$ so we can count $(x,y)$ instead.

Our equation now is $a^2 \equiv xy$ mod $71.$ This has two solutions if $xy$ is a square, and none if $xy$ is not. A fancier way to write that number of solutions is that it's $\left( \frac{xy}{71} \right) + 1,$ where $\left( \frac{\cdot}{71} \right)$ is the Legendre symbol . This equals $\left( \frac{x}{71} \right) \left( \frac{y}{71} \right) + 1,$ so the total number of solutions with $a \ne 0$ equals $\sum_{x,y \ne 0} \left( \left(\frac{x}{71}\right)\left(\frac{y}{71}\right) + 1\right) = 4900 + \sum_{x,y \ne 0} \left(\frac{x}{71}\right)\left(\frac{y}{71}\right) = 4900 + \sum_{x \ne 0} \left(\frac{x}{71}\right) \sum_{y \ne 0} \left(\frac{y}{71}\right).$ But both those sums on the right are equal to $0,$ because there are $35$ nonzero squares and $35$ nonzero non-squares mod $71.$

Maybe there is a cleaner way to do this? I'd be interested to see it.