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Old 2007-07-08, 19:14   #5
fivemack
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Feb 2006
Cambridge, England

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Proofs by working to a single modulus M are, surely, precisely proofs by induction; just with M distinct base cases and the induction rule being n -> n+M.

But in fact it seems to work for me by straight n->n+1 induction:

3(x+1)^5 = 3x^5 + 15x^4+30x^3+30x^2+15x+3
5(x+1)^3 = 5x^3 + 15x^2 + 15x + 5
7(x+1) = 7x + 7

so the sum is 3x^5 + 5x^3 + 7x (ex hypothesi divisible by 15)
+ lots of things which are in form a multiple of 15
+ 3 + 5 + 7 which equals 15
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