For any odd integer $n > 2$, show that there isn't any positive integer $x$, such that $x^n + (x+1)^n = (x+2)^n$.

Question

For any odd integer $n > 2$, show that there isn't any positive integer $x$, such that: $$x^n + (x+1)^n = (x+2)^n$$

Writing it using Newton's binom, we obtain: $$x^n = \sum_{i=1}^{n} \binom{n}{i} \cdot x^{n-i} \cdot (2^i - 1)$$

I don't know how to continue the problem. Can you help me, please? Thanks!

Answer 1

Claim 1: If $x$ is even then the left side is odd but right side is even so no even solution.

Claim 2: If $x$ is odd, then $x \equiv \pm 1 \pmod{4}$. In which case (with $n$ being odd) we have left side is either $1+2^n \pmod{4}$ or $-1 \pmod{4}$, whereas the right side is $-1 \pmod{4}$ or $1 \pmod{4}$ respectively. For $n>2$ the two sides are not equal. Hence no solution.

Answer 2

Hint: Consider the cases $x=3k$, $x=3k+1$, $x=3k-1$ separately. You should find the remainders on division by $3$ are incompatible in all three cases.

Answer 3

The reminders by $x$ is equal, then 　

$2^n=kx+1$

$LHD=x^n+(x+1)^n=x*(A(x)+n)+1$

$RHD=(x+2)^n=x(B(x)+2n)+kx+1=x(B(x)+2n+k)+1$

Since $n≠2n+k$, there doesn't such integers.