# Show that 'n5' - 'n' is divisible by 30 for all integers 'n' 'n5' = 'n' to the power of 5

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### 1 Answer

Show `n^5-n` is divisible by 30 `(30|n^5-n)` for all `ninZZ` (N an integer)

`n^5-n=n(n^4-1)`

`=n(n^2-1)(n^2+1)`

`=n(n+1)(n-1)(n^2+1)`

(1) It is clear that every number of the form `n(n+1)(n-1)(n^2+1)` is even since either `n,n+1` are even. Thus the number is divisible by 2.

(2) Also, a number of the form `n(n+1)(n-1)(n^2+1)` is divisible by 3 since one of `n-1,n,n+1` is a multiple of 3.

(3) If the number is also a multiple of 5 then we are done.

Now every number is either a multiple of 5, 1 more than a multiple of 5, 2more, 3 more, or 4 more than a multiple of 5. Consider the cases:

- If n is a multiple of 5 then `n(n-1)(n+1)(n^2+1)` is also.

- If n is 1 more than a multiple of 5 then n-1 is a multiple of 5 and the product is also.

-If n is 2 more than a multiple of 5 then `n^2+1` is a multiple of 5.

-If n is 3 more than a multiple of 5 then `n^2+1` is a multiple of 5.

- If n is 4 more than a multiple of 5, then n+1 is a multiple of 5.

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**Thus every number of the form `n(n-1)(n+1)(n^2+1)` is a multiple of 2,3,and 5 and thus a multiple of 30 as required.**

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** Show a number that is 2 more than a multiple of 5 is a multiple of five when you square it and add 1:

`(5k+2)^2+1=25k^2+20k+4+1=25k^2+20k+5=5(5k^2+4k+1)`

** Show a number that is 3 more than a multiple of 5 is a multiple of 5 when you square it and add 1:

`(5k+3)^2+1=25k^2+30k+9+1=25k^2+30k+10=5(5k^2+6k+2)`