No solution exists. Note that $$y^2\equiv\left\{ \begin{array}{cll} 0 \pmod{8},&\qquad if\ y\equiv 0 &\pmod{4}\\ 1 \pmod{8},&\qquad if\ y\equiv \pm 1&\pmod{4}\\ 4 \pmod{8},&\qquad if\ y\equiv 2&\pmod{4} \end{array} \right. $$

Then, for any integers $y$ and $z$, we have $$ y^2 + z^2\equiv\left\{ \begin{array}{cll} 0, 1, 4 \pmod{8},&\qquad if\ y\equiv 0 &\pmod{4}\\ 1, 2, 5  \pmod{8},&\qquad if\ y\equiv \pm 1&\pmod{4}\\ 0, 4, 5 \pmod{8},&\qquad if\ y\equiv 2&\pmod{4} \end{array}\right. $$

Now, given $x$, $x+1$, $x+2$, and $x+3$ form a complete residue system modulo $4$. Let's assume $$\left\{ \begin{array}{lclc} x   &\equiv 0 \pmod{4},&\implies x^2 + a^2    &\equiv 0, 1, 4\pmod{8}\\ x+1 &\equiv 1 \pmod{4},&\implies (x+1)^2 + b^2&\equiv 1, 2, 5\pmod{8}\\ x+2 &\equiv 2 \pmod{4},&\implies (x+2)^2 + c^2&\equiv 0, 4, 5\pmod{8}\\ x+4 &\equiv 3 \pmod{4},&\implies (x+3)^2 + d^2&\equiv 1, 2, 5\pmod{8} \end{array}\right. $$

However, the intersection of $\{0, 1, 4\}$, $\{1, 2, 5\}$, and $\{0, 4, 5\}$ is empty. This means no solution will satisfy all these relations. Hence, the given system is insolvable.