Trying to find a simpler way to calculate this function

by SeventhSigma
Tags: function, simpler
SeventhSigma is offline
Nov18-12, 12:36 PM
P: 250
let P(n) = n^4 + an^3 + bn^2 + cn

M(a,b,c) returns largest m that divides P(n) for all n

then let function S(N) return the sum of all M(a,b,c) for 1 <= a,b,c <= N

I am trying to understand a simpler way to calculate S(N) so I don't have to actually process every single combination of a,b, and c but I am having trouble finding patterns to take advantage of on a broad scale.

So far I know from trying all sorts of values that M(a,b,c) tends to return values of form 2^i * 3^j where i,j>=0.
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mfb is offline
Nov18-12, 02:01 PM
P: 10,840
If m divides P(n) for all n, it also divides all differences: m divides P(2)-P(1) = 15+7a+3b, for example.
Using more values for n, you can eliminate more variables. This could help to reduce testing.
coolul007 is offline
Nov21-12, 07:32 AM
coolul007's Avatar
P: 234
P(n) can be looked at as a base 'n' number. This would make a,b,c digits and a max value for that part.

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