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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.

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$6^2+1^2+4=5^2+4^2$. Note the difference between the even squares $\bmod 8$.

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Above equation shown below has parameterization:

$a^2+d^2+4=b^2+c^2$

$(2m)^2+(m-2)^2+(2)^2=(m+2)^2+(2m-2)^2$

For, $m=5$, we get:

$(10,3,2)^2=(7,8)^2$

Hence, the integer $4$ can be represented by sum difference of four squares.

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Above equation shown below has solutions:

$a^2+d^2+4=b^2+c^2$

$a=8w(k-2)$

$b=w(k^2-2k+5)$

$c=2w(k^2-6k+13)$

$d=w(k^2-10k+21)$

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

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  • $\begingroup$ 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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