Not quite Fermat's Last Theorem

Question

Prove that the equation n^a + n^b = n^c, with a,b,c,n positive integers, has infinite solutions if n=2, and no solution if n≥3.

Answer 1

So this is fermats last theorem upside down? It occurs to me if we have two binary numbers we may add them to get another power of two,

   1000000
   1000000
+ --------
  10000000

but if we had two numbers in base 3, say

  1000000
  1000000
+ -------
  2000000

we would not have so much luck.

Answer 2

Wlog assume $a \le b$. Dividing by $n^a$ yields $1 + n^{b-a} = n^{c-a}$ => $a=b$ (else $n|1$) => $ n = 2, c = a+1$.

Answer 3

If $n=2$ we can take $a=k, b=k, c=k+1$ for any $k \in \mathbb{N}$.

Let $n \ge 3$. We can assume that $a, b, c \ge 0$ because if not we could multiply left and right side by $n^k$ to make them positive.

Now it's clear that $c \ge a$ and $c \ge b$. Then we have $n^a | n^c$, hence $n^a | n^a + n^b$ and $a \le b$. In the same way $b \le a$. So $a = b$. Hence $2n^a = n^c$ and $n=2$.

Answer 4

(assume b≥a) $n^b<n^a+n^b<n^{b+1}$.