Systems of Diophantine Equations

Question

Find all ordered 4-tuples of integers $(a,b,c,d)$ that satisfy:

$$a^n+b^n=c^n+d^n$$

for ALL positive integers $n$.

Trivial solutions are $(k,p,k,p)$ and $(k,p,p,k)$ for any integers $k$ and $p$. But does there exist any non-trivial solutions?

Answer 1

We assume $a+b=c+d$ and $a^2+b^2=c^2+d^2$. Let $r=(a+b)/2=(c+d)/2$, $s=(a-b)/2$, $t=(c-d)/2$; then $a=r+s$, $b=r-s$, $c=r+t$, $d=r-t$. Then $a^2+b^2=c^2+d^2$ becomes $2(r^2+s^2)=2(r^2+t^2)$, so $s=\pm t$, and $(a,b,c,d)$ is either $(a,b,a,b)$ or $(a,b,b,a)$.

Note that we didn't need to assume $a,b,c,d$ integers, or even real numbers; everything works fine even if they are allowed to be arbitrary complex numbers.

Answer 2

Hint:-

$a+b=c+d \implies (a-c)=(d-b)\tag{1}$

$\begin{align}a^2+b^2=c^2+d^2 & \implies \left(a^2-c^2\right) \left(d^2-b^2\right)\\&\implies (a+c)(a-c)=(d-b)(d+b)\\&\implies (a+c)=(b+d)\\&\implies(a-b)=(d-c)\tag{2}\end{align}$