GCD of $a^n + b^n$ and $c^n + d^n$ Prove or disprove that there does not exists any integers $a,b,c,d > 1$ such that $a,b,c,d$ are pairwise coprime, and  $a^n + b^n$ and $c^n + d^n$ are also coprime for all integer  $n > 1$.
I came up with this problem after asking about a related problem this afternoon. I was thinking something along the line of Euler's Theorem or F.L.T, though I am completely lost on how to even start tackling this problem.
 A: This is not the case if at least one of $a,b$ and at least one of $c,d$ is not a perfect square:
Assume that $b,d$ are not perfect squares. Note that for an integer $a\geq 1$ and a prime $p \not \mid a$: $$a^\frac{p-1}{2} \equiv \left(\frac{a}{p}\right) \bmod p.$$ Therefore it is enough to give a prime $p$ such that the system of equations $$ \begin{eqnarray}\left( \frac{a}{p}\right) = & 1&  =\left( \frac{c}{p} \right) \\ \left( \frac{b}{p}\right) = & -1&  =\left( \frac{d}{p} \right)\end{eqnarray}$$ holds.
Now use multiplicativity of the Legendre Symbol and quadratic reciprocity to transform this into a system of modular equations for $p$ as follows: For each prime $q \mid a$ we may require $\left(\frac{q}{p} \right) = 1$. We do the same for primes dividing $b,c,d$ except that we choose exactly one prime $q_1 \mid b$ such that $q_1$ has an odd power in $b$ (exists because $b$ is not a perfect square) and we do the same for some $q_2 \mid d$. For those we require $\left(\frac{q_1}{p} \right) = -1 = \left(\frac{q_2}{p} \right)$. 
Now if we haven't done it until now, we choose any value for $\left(\frac{2}{p}\right)$ and translate the equations of Legendre symbols into a system of modular equations in $p$ using quadratic reciprocity, which is possible because we have specified $\left(\frac{2}{p}\right)$. Now the Chinese Remainder Theorem gives an arithmetic progression of solutions to these equations (because $a,b,c,d$ are coprime). This progression contains a prime $p$ by Dirichlet. For this prime, $$p \mid \gcd(a^{\frac{p-1}{2}}+b^{\frac{p-1}{2}}, c^{\frac{p-1}{2}}+d^{\frac{p-1}{2}}).$$
