# Show that `n^5 - n` is divisible by 30 for all integers n.

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### 3 Answers

Note that a number is divisible by 30 if it is divisible by the prime factors of 30, which are 2, 3 and 5.

To show, let's factor out n. Then, evaluate each expression.

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

If n is any even integers,

>> `n^4` results to an even number.

>> `(n^4 - 1` ) results to an odd number.

>> And if we multiply,

`n(n^4-1)` =(even #) * (odd #) = even number.

>>Since the product is an even number, it is divisible by 2. Hence, it is also divisible by 30.

If n is any odd integers,

>> `n^4` will be an odd number.

>> `(n^4 - 1)` will be an even number.

>> And,

`n (n^4-1) ` = (odd#)*(even #) = even number

>> Thus, the product is divisible by 30.

If n is any integers that ends with 5,

>> `n^4` results to an odd number and it ends with 5.

>>`(n^4-1)` results to an even number,

>> So,

` n(n^4-1)` = even number that end with zero.

>> The product is divisible by 5. Hence, it is divisible by 30.

**Answer: For all integers n, n^5 - n is divisible by 30.**

Any number that is divisible by both 5 and 6 is divisible by 30.

To show that `n^5-n` is divisible by 30, we show that `5|n^5-n` ( is a factor of `n^5-n` , or 5 divides `n^5-n` ) and `6|n^5-n` :

`n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n+1)(n-1)(n^2+1)`

(a) Either n or n+1 is even, so 2 is a factor.

One of n-1,n,n+1 is a multiple of 3 so 3 is a factor.

Therefore, `6|n^5-n`

(b) If any of n-1,n,n+1 is a multiple of 5 we are done. So assume that none of n-1,n,n+1 is a multiple of 5. Then either n-1 is 1 more than a multiple of 5 (n-1=5k+1) or n-1 is 2 more than a multiple of 5 (n-1=5k+2)

**Ex if n=7 then none of 6,7,8 is a multiple of 5, and 6 is 5+1; if n=8 then none of 7,8,9 is a multiple of 5 and 7=5+2. **

(b1) Let n-1=5k+1. Consider `n^2+1` :`n-1=5k+1==>n=5k+2==>n^2+1=(5k+2)^2+1`

But `(5k+2)^2+1=25k^2+20k+4+1=25k^2+20k+5=5(5k^2+4k+1)` so `n^2+1` is a multiple of 5.

(b2) Suppose n-1=5k+2. Consider `n^2+1` :

`n-1=5k+2==>n=5k+3==>n^2+1=(5k+3)^2+1`

But `(5k+3)^2+1=25k^2+30k+9+1=5(5k^2+6k+2)` so `n^2+1` is a multiple of 5.

So one of n-1,n,n+1,`n^2+1` must be a multiple of 5 which implies that `5|n^5-n`

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**Since `5|n^5-n` and `6|n^5-n` we have `30|n^5-n` QED.**

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It has to be shown that `n^5 - n` is divisible by 30 for all integers.

`n^5 - n`

= `n(n^4 - 1)`

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

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

Of the three consecutive integers n - 1, n and n + 1, one of them is a multiple of 2 and another is a multiple of 3. This gives `(n - 1)*n*(n+1)` is a multiple of 6.

To show that `n^5 - n` is a multiple of 5 use mathematical induction.

If n = 2, `n^5 - n = 30` which is divisible by 5

Assume `n^5 - n` is divisible for n, or `n^5 - n = 5*k` , then it has to be shown that `(n +1)^5 - (n + 1)` is also divisible by 5.

`(n + 1)^5 - (n + 1)`

=> `n^5 + 5n^4 + 10n^3 + 10*n^2 + 5*n + 1 - n - 1`

=> `n^5 + 5n^4 + 10n^3 + 10*n^2 + 5*n - n`

=> `5*k + 5n^4 + 10n^3 + 10*n^2 + 5*n`

All the terms in the polynomial arrived have 5 as a factor and are therefore divisible by 5.

By mathematical induction this proves that `n^5 - n` is divisible by 5.

**As `n^5 - n` is divisible by 6 and also divisible by 5, it is divisible by 30 for all integers n.**