Does the equation have $a^2+d^2+4=b^2+c^2$ where $d<c<b<a$ have any integer solutions? This isn't a homework problem, but I need to know for a separate problem I'm doing. Wolfram Alpha isn't very helpful.
$6^2+1^2+4=5^2+4^2$. Note the difference between the even squares $\bmod 8$.
Above equation shown below has parameterization:
For, $m=5$, we get:
Hence, the integer $4$ can be represented by sum difference of four squares.
Above equation shown below has solutions:
For $(k,w)=(4,-1)$ we get, $(a,b,c,d)=(16,13,10,3)$
For $(k,w)=(0,-1)$ we get, $(a,b,c,d)=(21,5,26,16)$
Where, $k$ is a parameter's