Does the equation $a^2+d^2+4=b^2+c^2$ have any solutions?

Does the equation have $$a^2+d^2+4=b^2+c^2$$ where $$d 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:

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

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

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.

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

• 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 $(6,5,4,1)$? – Oscar Lanzi Apr 1 at 9:56