Find all positive integers n for which xn + (x+2)n + (2-x)n = 0 has an integral solution.
Solution
Answer: n = 1.
There are obviously no solutions for even n, because then all terms are non-negative and at least one is positive. x = -4 is a solution for n = 1. So suppose n is odd n and > 3.
If x is positive, then xn + (x+2)n > (x+2)n > (x-2)n, so xn + (x+2)n + (2-x)n > 0. Hence any solution x must be negative. Put x = -y. Clearly x = -1 is not a solution for any n, so if x = -y is a solution then (x+2) = -(y-2) ≤ 0 we have (y+2)n = yn + (y-2)n. Now 4 = ( (y+2) - (y-2) ) divides (y+2)n - (y-2)n. Hence 2 divides y. Put y = 2z, then we have (z+1)n = zn + (z-1)n. Now 2 divides (z+1)n - (z-1)n so 2 divides z, so z+1 and z-1 are both odd. But an - bn = (a - b)(an-1n-2b + an-3b2 + ... + bn-1). If a and b are both odd, then each term in (an-1n-2b + an-3b2 + ... + bn-1) is odd and since n is odd there are an odd number of terms, so (an-1n-2b + an-3b2 + ... + bn-1) is odd. Hence, putting a=z+1, b=z-1, we see that (z+1)n - (z-1)n = 2(an-1n-2b + an-3b2 + ... + bn-1) is not divisible by 4. But it equals zn with z even. Hence n must be 1.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002