Gear Up For Divisibility Test

First note that by Fermat's Little Theorem $n^{5} \equiv n \pmod{5} \Longrightarrow n^{5} - n \equiv 0 \pmod{5}$ for any positive integer $n$ , i.e., $5$ divides $n^{5} - n$ .

Next, note that $n^{5} - n = n(n^{4} - 1) = n(n^{2} - 1)(n^{2} + 1) = n(n - 1)(n + 1)(n^{2} + 1)$ . Now as $n , n - 1, n + 1$ are 3 consecutive integers, at least one of these numbers must be divisible by $2$ , and precisely one must be divisible by $3$ , so $2 \times 3 = 6$ divides $n^{5} - n$ .

So finally as both $5$ and $6$ divides $n^{5} - n$ we can conclude that $5 \times 6 = 30$ divides $n^{5} - n$ , i.e., the statement in question is $\boxed{\text{True}}$ .

==============================================================================================================

Edit: In response to Justin Johnson's query in the comments below, note that we can also prove this by induction. The statement is true for $n = 1$ . Next. assuming it is true for some $n \gt 1$ , we can write

$(n + 1)^{5} - (n + 1) = n^{5} + 5n^{4} + 10n^{3} + 10n^{2} + 5n + 1 - n - 1 = (n^{5} - n) + 5n(n^{3} + 2n^{2} + 2n + 1) = (n^{5} - n) + 5n(n + 1)(n^{2} + n + 1)$ .

Now $n^{5} - n$ is divisible by $30$ by the induction hypothesis. Next, we note that $n(n + 1)$ must be even, i.e., divisible by $2$ . Also, one of $n, n + 1$ or $n^{2} + n + 1$ must be divisible by $3$ : if $n \equiv 0 \pmod{3}$ we are done; if $n \equiv 1 \pmod {3}$ then $n^{2} + n + 1$ is divisible by $3$ ; and finally if $n \equiv 2 \pmod{3}$ then $n + 1$ is divisible by $3$ . Thus $5n(n + 1)(n^{2} + n + 1)$ is divisible by $5 \times 2 \times 3 = 30$ .

Thus $30 | (n + 1)^{5} - (n + 1)$ given that $30 | n^{5} - n$ , completing the proof by induction.

The answer is true. Set $\begin{aligned} N &= n^5-n \\ &= n(n^4-1) \\ &= n(n^2-1)(n^2+1) \\ &= n(n-1)(n+1)(n^2+1) \end{aligned}$

In order to be divisible by 30, we must have $N\equiv 0\bmod 2$ , $N\equiv 0\bmod 3$ , and $N\equiv 0\bmod 5$ .

• Either $n\equiv 0\bmod 2$ or $n+1\equiv 0\bmod 2$ , so $N\equiv 0\bmod 2$ .
• Similarly, $n-1\equiv 0\bmod 3$ , $n\equiv 0\bmod 3$ or $n+1\equiv 0\bmod 3$ , so $N\equiv 0\bmod 3$ .
• Finally, we must show $N\equiv 0\bmod 5$ , which is slightly more complicated. If $n\equiv 0\bmod 5$ we're done. If $n\equiv 1\bmod 5$ then $n-1\equiv 0\bmod 5$ and we're done. Similarly, if $n\equiv 4\bmod 5$ then $n+1\equiv 0\bmod 5$ and again we're done. So assume $n\equiv 2\bmod 5$ or $n\equiv 3\bmod 5$ . In either case, $n^2\equiv 4\bmod 5$ and $n^2+1 \equiv 0\bmod 5$ . Thus, in all cases, $N\equiv 0\bmod 5$ , completing the proof.

The first step would be to factorize $n^5 - n$ :

$N = n^5 - n$

$= n(n^4 - 1)$

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

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

At this point it’s clear that $N$ has three consecutive factors, $n-1, n$ , and $n+1$ . This guarantees at least one even number and one multiple of three, so $N$ is definitely divisible by $6$ .

The next question is whether $N$ is also divisible by $5$ . Because a multiple of $5$ isn’t guaranteed in a set of three consecutive numbers, the $n^2 + 1$ factor has to be able to account for that $5$ .

While $n^2 + 1$ isn’t always divisible by $5$ , remember that we only need it to be so when the three consecutive numbers aren’t ; if $n-1, n,$ or $n+1$ can be divided by $5$ , factors, $n^2 + 1$ doesn’t matter.

If $n$ is in the form $5m-1, 5m$ , or $5m+1$ then $n + 1$ , $n$ , or $n-1$ would be divisible by $5$ respectively. Thus, we only need to check if $n^2 + 1$ is divisible by $5$ when $n =5m+2$ and $n =5m+3$ .

Plugging both into $n^2 + 1$ gets $25m^2+10m+5$ and $25m^2+15m+10$ respectively, both of which are multiples of $5$ for any integer $m$ .

Thus, $N$ will always be divisible by $2$ , $3$ , and $5$ , making it a multiple of $30$ .

n^5-n factorises as n(n^2+1)(n+1)(n-1)

As n-1, n and n+1 are consecutive numbers, so at least one of these must be divisible by 2, and at least one by 3. Therefore the factoral above is divisible by 6. If any of the factors is divisible by 5, then the expression is divisible by 30.

Remember that any numbers ending in 0 or 5 are divisible by 5.

Consider the final digit of n:

If n ends in 0, then n divides by 5. If n ends in 1, then n-1 ends in 0 and is divisible by 5. If n ends in 2, then n^2 ends in 4, n^2 + 1 ends in 5 and is divisible by 5. If n ends in 3, then n^2 ends in 9, n^2 + 1 ends in 0 and is divisible by 5. If n ends in 4, then n+1 ends in 5 and is divisible by 5. If n ends in 5, then n divides by 5. If n ends in 6, then n-1 ends in 5 and divides by 5. If n ends in 7, then n^2 ends in 9, n^2 + 1 ends in 0 and is divisible by 5. If n ends in 8, then n^2 ends in 4, n^2 + 1 ends in 5 and is divisible by 5. If n ends in 9, then n+1 ends in 0 and divides by 5.

Therefore for all positive integer values of n, n(n^2+1)(n+1)(n-1) divides by 2,3 and 5, hence n^5 - n divides by 30.

This is a simple solution which doesn't use Fermat's little theorem

First we write n^5 - n as (n - 1)n (n+1)(n^2 +1)

Since it contains (n-1)n (n+1) which is the product of 3 consecutive Numbers, it is divisible by both 2 and 3 and thus divisible by 6

Now we check the number's divisiblity by 5

We know that any n can be written as
5k + r where 0 <= r <5

Now for each n = 5k , 5k +1 , 5k + 2, 5k +3, 5k + 4 we can easily check that (n-1)n (n+1)(n^2+1) will always yield an multiple of 5

So n^5 - n is divisible 30

$n^5-n=(n-1)n(n+1)(n^2+1)$ Now that the expression is factored, it's easy to see that, as there are three consecutive terms, this number is divisible by 2 and 3, and therefore, by 6. As $30=5\cdot 6$ we only need to verify if its divisible by 5. If the last digit of $n$ is not 3 nither 7, ist obvious that the expression will be divisible by 5 ('cause one of the consecutive terms will have 5 or 0 as last digit). So, if the last digit of $n$ is 3 or 7, the last digit of $n^2+1$ is 0, thus its squares have 9 as last digit. So the expression is definitively divisible by 30.

Mathologer has helpfully provided a video on Fermat's Little Theorem.

https://www.youtube.com/watch?v=_9fbBSxhkuA

Since 30 = 2 3 5 and these are all pairwise coprime, it suffices that show that n^5 - n is divisible by 2, 3 and 5. Note that n^5 - n = n(n-1)(n+1)(n^2 +1).

If n is even, then n^5 - n is divisible by 2. If n is odd, then n+1 is even and so n^5 - n is divisible by 2.

In any set of 3 consecutive integers, exactly one of them is divisible by 3. Then, since n-1, n, n+1 are consecutive, one of which is divisible by 3. Thus, (n-1)n(n+1) is divisble by 3. Thus, n^5 - n is divisible by 3.

In any set of 5 consecutive integers, exactly one is divisible by 5. Consider n-2, n-1, n, n+1, n+2. Exactly one of them is divisible by 5. If n-1, n, or n+1 are divisible by 5, then clearly n^5 - n is divisible by 5. If n +/- 2 is divisible by 5, then n +/- 2 = 5k for some integer k n = 5k +/- 2 n^2 = 25k^2 +/- 10k + 4 n^2 + 1 = 25k^2 +/- 10k + 5 n^2 + 1 = 5(5k^2 +/- 2k + 1) Thus, n^2 + 1 is divisble by 5. Thus, n^5 - n is divisible by 5.

Therefore, n^5 - n is divisible by 30.

Proof by mathematical induction: let P(k) = $k^5$ - k

P(1) is true. Let P(k) be true.

30 | $k^5$ - k.

P(k+1) = $(k+1)^5$ - (k+1).

= $k^5$ + 5 $k^4$ +10 $k^3$ + 10 $k^2$ + 5k + 1 - k - 1

= ( $k^5$ - k) + 5 $k^4$ +10 $k^3$ + 10 $k^2$ + 5k

= ( $k^5$ - k) + 5k( $k^3$ + 2 $k^2$ + 2k + 1)

5 | P(k+1). Since P(k) is true, ignore ( $k^5$ - k) below.

k( $k^3$ + 2 $k^2$ + 2k + 1)

= k(k+1)( $k^2$ + k + 1)

2 | k(k+1);

3 | k(k+1)( $k^2$ + k + 1) since ( $k^2$ + k + 1) = k(k+1) +1, sum of 3 consecutive integers is divisible by 3;

Therefore, 30 | P(k+1).

P(k) implies (or =>) P(k+1).

• True.

Note that $n^5-n = n(n^4-1)$ and $30 = 2 \times 3 \times 5$ . Using Fermat's Little Theorem:

If $2 \nmid n$ , then $n^1 \equiv 1 \pmod{2} \Longrightarrow n^4 \equiv 1 \pmod{2} \Longrightarrow n^4-1 \equiv 0 \pmod{2}$ .

If $3 \nmid n$ , then $n^2 \equiv 1 \pmod{3} \Longrightarrow n^4 \equiv 1 \pmod{3} \Longrightarrow n^4-1 \equiv 0 \pmod{3}$ .

If $5 \nmid n$ , then $n^4 \equiv 1 \pmod{5} \Longrightarrow n^4-1 \equiv 0 \pmod{5}$ .

2^5 - 2 = 30. Not a coincidence

Interesting result. Here’s a simple proof. First, note that if d is the last digit of n, then d is also the last digit of n^5. Therefore, n^5 - n ends in 0, divisible by 10. Then note that for m = 0, 1, and 2, if n is congruent to m mod 3, n^5 - n is congruent to 0 mod 3. Thus n^5 - n is divisible by 3. Done.

If you do trial and error then it will always work

It was literally stated in another question I found earlier