MathLabs

Problem 4

Determine all positive integers nn for which xn+(2+x)n+(2−x)n=0x^n+(2+x)^n+(2-x)^n=0 has an integer solution.
Step 3 of 6: Force x to be even for odd n>1
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n>1 odd⟹x is even,x=2yn>1\text{ odd}\Longrightarrow x\text{ is even},\qquad x=2y
Detailed analysis

For odd n>1n>1, if xx were odd then 2+x2+x and 2−x2-x would also be odd, so the sum of the three odd powers would be odd and could not equal zero. Hence xx is even; write x=2yx=2y. Dividing the equation by 2n2^n gives yn+(1+y)n+(1−y)n=0y^n+(1+y)^n+(1-y)^n=0.