30in5 - N.docx

Factor it .

n5−n=n(n+1)(n−1)(n2+1)n5−n=n(n+1)(n−1)(n2+1) 1.) Three consecutive integers will be divisible by 2 and 3 . 2.)If n2−4n2−4 is divisible by 5 then n2+1n2+1is also divisible by 5. 3.)Put n2−4n2−4 in place of n2+1n2+1 4.) We get

n5−n=(n+1)(n)(n−1)(n−2)(n+2)n5−n=(n+1)(n)(n−1)(n−2)(n+2) Which is a product of 5 consecutive integers . And Hence it is divisible by 5 as well .

Here's another approach, just to feed the imagination for similar problems. For the factors 22 and 33 use @Jared's factorisation. For the factor 55 note that the squares modulo 55 are 0,±10,±1 and use n(n2−1)(n2+1)n(n2−1)(n2+1) (or little Fermat, because 55 is prime). The point was to draw attention to the squares, since these are sometimes useful and/or significant.