What are the integer solutions of the system $a^2+b^2=c^2$, $a^3+b^3+c^3=d^3$?

Question

How to solve these equations to find the integer numbers (a, b, c, and d)? $$a^2+b^2=c^2\tag{1}$$ $$a^3+b^3+c^3=d^3\tag{2}$$
I know one of solutions which is $a=3, b=4, c=5, d=6$, but I don't know if there are others solutions

Answer 1

$(3,4,5,6)$ is the only non-trivial solution involving strictly positive integers (by non-trivial I mean where all common factors are removed).

Proof:

$(a, b, c)$ is a Pythagorean triple (EDIT: and WLOG is a primitive triple since $k|a,b,c\implies k|d$), so it can be parametrised as $a = x^2 - y^2$, $b = 2xy$, $c = x^2 + y^2$ with $x > y \ge 1$ and $GCD(x, y) = 1$ (proof left as an exercise).

Then $d^3 = a^3 + b^3 + c^3 = \left(x^2-y^2\right)^3 + \left(2xy\right)^3 + \left(x^2+y^2\right)^3=2x^2\left(x^4+4xy^3+3y^4\right)$.

We first note that this is even, and so $d^3$ is even and hence divisible by 8. Then either $x$ is even (so that $2x^2$ provides the factor of 8) or $x^4+4xy^3+3y^4$ is divisible by 4. Note that these possibilities are not simultaneously possible, since that would imply that $y$ is even, contradicting the coprimality of $x$ and $y$.

If $x=2m$, then $d^3=8m^2\left(16m^4+8my^3+3y^4\right)$, which then implies that $m|16m^4+8my^3+3y^4$, and hence $m|y$ (i.e. $m$ divides evenly into $y$). If $m\neq 1$, then that would mean $x$ and $y$ share a common factor, which we've stated isn't the case, so $m=1$ which then gives $x=2$, and since $x>y\geq 1$ we must have $y=1$, which when you put it all into everything gives the $(3,4,5,6)$ solution.

On the other hand, if $x$ is odd (and greater than 1), then we must be able to pull a factor of $x$ out of $x^4+4xy^3+3y^4$ to make a perfect cube to equal $d^3$, which implies that $x|3y^4$. But since $x$ and $y$ are coprime, the only possibility is $x=3,y=1$. Putting these back into the parametrisation gives $a=8,b=6$ which is going to not give us a primitive/non-trivial solution.

We can relax our restriction down to $x\geq y\geq 1$, in which case $x=1,y=1$ is also a possibility, which gives $a=0,b=2$ which doesn't work for $a,b,c,d>0$ but does lead to the $\left(1,0,-1,0\right)$ solution with a bit of wrangling.

I suspect, but have no proof as yet, that these are the only two non-trivial solutions across all integers. The above proof may extrapolate neatly to negative values, or it may not.