Assume the contrary. Let $S(n) = a^{n}+b^{n}-c^{n}$. Then the set $E= \{ p \text{ prime }; \exists n \in N, \ p \ | \ S(n) \}$ is finite - or empty (*), in which case the result is trivial by (**) -. Note also that $|S(n)| \rightarrow +\infty$ (**) since $c \notin \{a,b\}$. We also set the number: $K_{M}=M \prod_{p_{i} \in E} {(p_{i}-1)}$ Now there are a few cases: 1) $\forall i$, $p_{i}$ does not divide $abc$. Then by Fermat theorem, $\forall p_{i} \in E, \ S(K_{M}) = 1 \ mod \ p_{i}$ whence as $M \rightarrow +\infty$, $S(K_{m})$ will have a prime factor that does not belong to $E$ by (**). Absurd. 2) $\exists i$, $p_{i}$ divides c but does not divide a nor b. Then the same proof 1) still holds. 3) $\exists i$, $p_{i}$ divides a and b but not c. Then 1) still holds since $p_{i}$ will never divide $S(K_{m})$. 4) $p_{i}$ divides a or b (but not a and b) and p does not divide c. Let's suppose it divides a (without loss of generality). Then 1) does not apply anymore, and since $p_{i}$ divides some $S(n)$, there exists a smallest positive integer $T$ such that p divides $S(T)$. For such a T, we note $U$ the maximal integer such that $p^{U}$ divides $S(T)$. We also have in particular $p^{U}$ divides $b^{T}-c^{T}$. Then if $p$ also divides $S(X)$, i.e if $p_{i}$ also divides $b^{X}-c^{X}$, then $p$ also divides $b^{X-T}-c^{X-T}$, and by induction and minimality of T we must have $T$ divides $X$. But then $b^{zT}-c^{zT} = (b^{T} - c^{T})((b^{T})^{z-1} + \dots + (c^{T})^{z-1}$ and since we have $b_{T} = c_{T} \ mod \ p_{i}$ we also have the second factor $= z(c^{T})^{z-1} mod \ p_{i}$. Whence $n = p_{i}T$ is the smallest possible (candidate) positive integer such that $p_{i}^{U+1}$ divides $S(n)$. By induction, $n = p_{i}^r*T$ is the smallest positive integer such that $p_{i}^{U+r}$ divides $S(n)$. But now if you pick $m$ such that no $p_{i}$ divides $m$, then the product of powers of $p_{i}$ dividing $K_(m)$ will be bounded, and then you can conclude by taking $m$ large enough. 5) $\exists i$, $p_{i}$ divides a, b and c. Then we can refer to 1), 2), 3) or 4) by factoring out such $p_{i}$'s in $S_{n}$. -- Julien Santini