Not quite Fermat's Last Theorem

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 $n\ge3$.

Solution 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.

Solution 2:

Wlog $\,a \le b$. Dividing by $n^a$ yields $\,1 + n^{b-a} = n^{c-a}$ $\Rightarrow$ $b=a\ $ (else $\,n\mid1)\,$ $\Rightarrow$ $\, n = 2,\, c = a\!+\!1$.

Solution 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 $n=2$.

Solution 4:

Assuming $b>a$:

$$n^b<n^a+n^b<n^{b+1}$$