# how to prove that for an arbitary even n the number N=n³+20n is divisible by 48?

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### 1 Answer

Show that for n even, `N=n^3+20n` is divisible by 48, or `48|(n^3+20n)`

To show `48|(n^3+20n)` it suffices to show that `16|(n^3+20n)` and `3|(n^3+20n)` (see link).

(1) Since n is even, let n=2k for some `kinNN` (k a natural number).

(2)We need to show that `3|(n^3+20n)` or `3|((2k)^3+20(2k))=8k^3+40k=8k(k^2+5)`

k is in one of the following forms: 3p-1,3p,3p+1 for some `p inNN` :

(a) If k=3p then `3|8(3p)((3p)^2+5)` (`a|ab` ) so `3|(n^3+20n)`

(b) If k=3p-1 then `8k(k^2+5)=8(3p-1)((3p-1)^2+5)`

`=8(3p-1)(9p^2-6p+6)=8(3p-1)(3)(3p^2-2p+2)` so `3|(n^3+20n)`

(c) If k=3p+1 then `8k(k^2+5)=8(3p+1)((3p+1)^2+5)`

`=8(3p+1)(9p^2+6p+6)=8(3p+1)(3)(3p^2+2p+2)` so `3|(n^3+20n)`

Thus `3|(n^3+20n)` for all even `n in NN`

(3) We need to show that `16|(n^3+20n)=8k(k^2+5)` :

k is of the form 2p or 2p+1:

(a) If k=2p then `8k(k^2+5)=16p(4p^2+5)` and `16|16p(p^2+5)` so `16|(n^3+20n)`

(b) If k=2p+1 then `8k(k^2+5)=8(2p+1)((2p+1)^2+5)`

`=8(2p+1)(4p^2+4p+6)=16(2p+1)(2p^2+2p+3)` so `16|(n^3+20n)`

Therefore `16|(n^3+20n)` for all even `n in NN`

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**Since `3|(n^3+20n)` and `16|(n^3+20n)` for all even `n in NN` , we have `48|(n^3+20n)` for all even `n in NN` **

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**Sources:**