Question

Can you proof the following?

For each integer `n in N`
`5n^3+7n^5-=0` (mod 12)

Answer 1

See below.

Explanation:

Given

`5n^3+7n^5 = n(5 n^2+7 n^4)`

we have that a typical polynomial such that

`p(n,a) equiv 0 mod 12` is given by

`p(n,a) = 7(n+a)(n+a+1)^2(n+a+2)`

then making

`5n^2+7n^4 equiv p(n,a) mod 12`

we have

`{(7a^4+28a^3+35a^2+14a equiv 0 mod 12),(28a^3+84a^2+70a+14 equiv 0 mod 12),(42a^2+84a+30 equiv 0 mod 12),( 28a+28 equiv 0 mod 12):}`

so choosing `a = -1` we have

`{(0 equiv 0 mod 12),(0 equiv 0 mod 12),(12 equiv 0 mod 12),(0 equiv 0 mod 12):}`

or

`5n^2+7n^4 equiv p(n,-1) equiv 0 mod 12`

where

`p(n,-1) = 7n^4-7n^2`

then as `7n^4-7n^2 equiv 0 mod 12 rArr 5n^2+7n^4 equiv 0 mod 12 rArr 5n^3+7n^5 equiv 0 mod 12`

Answer 2

See explanation...

Explanation:

For any integer `n`, note that exactly one of `n`, `n-1`, `n+1` is divisible by `3`. So `n(n-1)(n+1)` is always divisible by `3`.

If `n` is even then `n^2` is divisible by `4`

If `n` is odd then `(n-1)(n+1)` is divisible by `4`

Hence for any integer `n` we have `n^2(n-1)(n+1)` is divisible by `4`.

Hence `n^2(n-1)(n+1)` is divisible by `3*4 = 12`

Then modulo `12` we have `5 -= -7`

So:

`0 -= 7n^3(n-1)(n+1) = 7n^3(n^2-1) = 7n^5-7n^3 -= 5n^3+7n^5`