Let $x=a+c, y=a-c, w=b+d$, and $z=b-d$. Then $$\left(w^{2}+x^{2}+y^{2}+z^{2}\right)^{2}=4\left(x^{2}-w^{2}\right)\left(y^{2}+z^{2}\right)$$ and since $\left|x^{2}+w^{2}\right| \geq\left|x^{2}-w^{2}\right|$ it follows that $w=0$ or $y=z=0$. Now $y=z=0$ implies $a=b=c=d=0$. Now $w=0$ gives $b=-d$. Then for equality to hold $x^{2}=y^{2}+z^{2}$. This is equivalent to $a c=b^{2}$, which includes the previous case. It suffices to count the number of triples $(a, b, c)$ that satisfy the equation. When $b=0$, either $a$ or $c$ is zero, which gives $11+11-1=21$ triples. When $b= \pm 1$, we have $|a|=|c|=1$ and $a, c$ have the same sign, for $2 \cdot 2=4$ triples. When $b= \pm 2$, we have $(a, c)=(1,4),(2,2),(4,1)$ or their negatives, for $2 \cdot 6=12$ triples. When $b= \pm 3, \pm 4, \pm 5$, we have $|a|=|b|=|c|$ and $a, c$ have the same sign, for $6 \cdot 2=12$ triples. So in total there are $21+4+12+12=49$ solutions.