Problem 4

Determine all positive integers

n

for which

x^n+(2+x)^n+(2-x)^n=0

has an integer solution.

Step 5 of 6: Factor the sum of the two odd powers

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(1+y)^n+(1-y)^n=2T,\qquad T=\sum_{k=0}^{n-1}(-1)^k(1+y)^{n-1-k}(1-y)^k\text{ is odd}

Detailed analysis

For odd $n$ , use $a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+\cdots+b^{n-1})$ with $a=1+y$ and $b=1-y$ . Since $y$ is even, both $a$ and $b$ are odd; the second factor $T$ is a sum of $n$ odd terms, and $n$ is odd, so $T$ is odd. The equation becomes $y^n=-2T$ .