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:






Where, $w=[1/(k^2-4k-1)]$

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

  • $\begingroup$ Good to see a parametric family, but the OP wants to enforce the ordering $d<c<b<a$. Place suitable restrictions on $k$ maybe? How does this formula derive the solution $(6,5,4,1)$? $\endgroup$ – Oscar Lanzi Apr 1 at 9:56

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