Existence of integer solutions to $a^2 + b^2 + c^2 = d^2$ for specified $d$ Previous questions contained cubes instead of squares, that was my mistake, sorry.
How to tell if $a^2 + b^2 + c^2 = d^2$ has any integer, non-zero solutions for specified positive integer $d$? Is there any criterium or algorithm for this?
EDIT: To clarify, $a, b, c$ are non-zero integers and $d$ is specified and it is a positive integer.
 A: 1) As regards the original post: existence of integer solutions to
$a^3 + b^3 + c^3 = d^3$ for a given $d$.
Note that the following identity holds:
$$(9t^4)^3+(1-9t^3)^3 +(3t-9t^4)^3=1.$$
Therefore, given a positive integer $d$ there is always an integer solution,
$$(9dt^4)^3+(d(1-9t^3))^3+(d(3t-9t^4))^3 = d^3.$$
For more details, take a look at this page (with references): https://ckrao.wordpress.com/2012/04/10/integers-equal-to-the-sum-of-three-cubes/
2) As regards the new post: existence of integer solutions to
 $a^2 + b^2 + c^2 = d^2$ for a given $d$.
Note that by a theorem of Legendre, if we allow $a,b,c$ to be zero, then
$$a^2 + b^2 + c^2=N$$
iff $N$ is NOT of the form $4^n(8m+7)$. 
For numbers that are the sum of three non zero squares see the OEIS sequence A000408. 
In the list we find the squares 9, 36,49, 81 (but not 4,25,64,100).
Since
$$t^2+(t+1)^2+(t(t+1))^2=(t^2+t+1)^2$$
it follows that if $d$ has a divisor of the form $t^2+t+1$ with $t\geq 1$ (like 3, 7, 13,  31, 43) then there is a solution.
According to this link,  there is a finite set $T$ such that any positive integer is a sum of three non-zero squares unless $n$ is of the form $4^n(8m+7)$ or of the form $4^am$ where $m\in T$.
It seems that
$$T=\{1,2,5,10,13,25,37,58,85,130\}$$
therefore $a^2 + b^2 + c^2=d^2$ has a solution iff $d\not= 2^n$ or $d\not= 2^n\cdot 5$.
