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P(n) = n^5 - n

n(n-1)(n^3+n+1)

when n = 5

5 * 4* 131 = 620

620 is a factor of 5. therefore true for n=5

assume true n=k

P(k) = k^5 - k

when n = k+1

P(k+1) = (k+1)(k+1-1)((k+1)^3 + k+2)

= (k+1)(k)(k^3 + 3k^2 + 3k + 1 + k + 2)

= (k+1)(k)(k^3 + 3k^2 + 4k + 3)

= (k^2+k)(k^3 + 3k^2 + 4k + 3)

= k^5 + 4k^4 + 7k^3 + 7k^2 + 3k

what shall I do from there?

thanks xxx