Use a polinomial identity: a^n+b^n=(a+b)(a^(n-1)-a^(n-2)·b+a^(n-3)·b^2-…+b^(n-1)) [n: any pos. odd int.]··· Ex.1 (if n is even, this identity doesn't hold)

Let A=2^n+5^n, B=3^n+5^n, so A+B=2^n+3^n+4^n+5^n ··· Ex.2 From Ex.1 we have:

A=2^n+5^n=(2+5)(2^(n-1)-2^(n-2)·5+2^(n-3)·5^2-…+5^(n-1))

B=3^n+4^n=(3+4)(3^(n-1)-3^(n-2)·4+3^(n-3)·4^2-…+4^(n-1))

Let P1=(2^(n-1)-2^(n-2)·5+2^(n-3)·5^2-…+5^(n-1)), and P2=(3^(n-1)-3^(n-2)·4+3^(n-3)·4^2-…+4^(n-1))

So that, A=(2+5)(P1)=7(P1), B=(3+4)(P2)=7(P2), so A+B=7(P1+P2) ⇒ (A+B)/7=(P1+P2) ··· Ex.3

Since both polinomials, P1 and P2, consist of 7 integer-terms each, P1 and P2 are integers. So that,(P1+P2) is also an integer. From Ex.2, (A+B) is integer as well, so that (A+B) is divisible by 7.

From Ex.2 and Ex.3, (2^n+3^n+4^n+5^n)/7=(P1+P2), (P1+P2): integer

Therefore, 2^n+3^n+4^n+5^n is divisible by 7 for any possitive odd integer n. QED.