Diophantine Equations (Problem $8$ )

We have that $n \in \mathbb{Z}$ and $n^n+n=(2n)^n$ .

Before we start proper, we consider the case when $n<0$ .

$-1 \leq n^n <0$ , from which $n-1 \leq n^n + n < n$ but $-1<(2n)^n<0$ , so no solution exists in this range.

We can rewrite the righthand side as $2^n n^n$ and do some simple factorising to get: $n(n^{n-1} - 1) = n(2^n n^{n-1})$ This is clearly true if $n=0$ , so consider the case when $n\neq 0$ .

We can divide through by $n$ to get $n(n^{n-1} - 1) = n(2^n n^{n-1})$ . If $n>1$ , then $n$ divides the lefthand side but not the right, which means no solutions exist for the range $n>1$ .

All that needs to be done is to check if $n=1$ satisfies the original equation. $1^1 + 1 = 2^1$ $2=2$

The only solutions are $n=0$ and $n=1$ . In other words, there are two solutions.

$0^0 + 0 = (0 + 0)^0 \rightarrow 1 + 0 = (0)^0 \rightarrow 1 + 0 = 1 \rightarrow 1 = 1$ or $0^0 + 0 = (0 + 0)^0 \rightarrow 0 + 0 = (0 + 0)^0 \rightarrow 0 + 0 = 0 \rightarrow 0 = 0$

$1^1 + 1 = (1 + 1)^1 \rightarrow 1 + 1 = (2)^1 \rightarrow 2 = 2^1 \rightarrow 2 = 2$

Therefore, the answer is $\fbox 2$