Looks like Pythagoras, not it is not.

For now let's let $a,b$ be positive or negative, and we can restrict to positive integers at the end.

In the ring of Gaussian integers , this factors as $(a+bi)(a-bi) = 11^2(1+i)(1-i)(12+7i)^2(12-7i)^2,$ where the right side is factored into primes. To factor $a+bi$ we need to choose one prime from each conjugate pair, so wlog $a+bi = u \cdot 11 (1+i) p_1p_2,$ where $u$ is a unit (equal to $\pm 1$ or $\pm i$ ), and $p_i$ are primes equal to $12 \pm 7i.$ So there are three cases.

Case 1: $a+bi = u(11+11i)(12+7i)(12+7i) = u(11+11i)(95+168i) = u(-803+2893i)$ leads to the representation $2 \cdot 2123^2 = (-803)^2 + 2893^2.$

Case 2: $a+bi = u(11+11i)(12+7i)(12-7i) = u(11+11i)(193)$ leads to the representation $2 \cdot 2123^2 = 2123^2 + 2123^2.$

Case 3: $a+bi = u(11+11i)(12-7i)(12-7i) = u(11+11i)(95-168i) = u(2893-803i)$ leads to the representation $2 \cdot 2123^2 = 2893^2 + (-803)^2.$

Note that there are twelve solutions to the equation in all, because each representation can be permuted and sign-adjusted by multiplying by the unit $u$ : multiplying by $-1$ switches the signs of $a$ and $b,$ multiplying by $i$ moves $a^2+b^2$ to $(-b)^2+a^2,$ and multiplying by $-i$ moves $a^2+b^2$ to $b^2+(-a)^2.$

Restricting our attention to positive $a,b,$ there are three total representations: $2 \cdot 2123^2 = 803^2 + 2893^2 = 2123^2+2123^2 = 2893^2+803^2.$ So the answer is $\fbox{2893}.$