Find all natural numbers $n$ such that $n^{17}-n$ is divisible by 10

Question

I've encountered this math problem and I don't know how to solve it. What math region,field(however you call it. English is not my first language) is used here ?

Find all natural numbers for which ($n^{17} - n $ ) is divisible by 10.

Any tips or advice you can give me and if you could share some websites where I can learn more about this.

See this. Observe that $n^4-1$ is a factor of $n^{12}-1$ as well as $n^{16}-1$. The answers by robjohn and Bill Dubuque in particular study generalizations of this. — Jyrki Lahtonen, Jul 15 '16 at 10:40
Actually, $ n^{17}-n $ is divisible by $2\cdot 3\cdot 5\cdot 17$ for all $n$. See this question. — lhf, Jul 15 '16 at 11:08
See also http://math.stackexchange.com/questions/1849603/the-greatest-common-divisor-of-all-numbers-of-the-form-na-nb. — Batominovski, Jul 15 '16 at 11:39

smcc · Accepted Answer · 2016-07-15T11:22:19.780

4

You can write $$n^{17}-n=(n-1)n(n+1)(n^2+1)(n^4+1)(n^8+1).$$ To satisfy the condition it is enough that one of these factors be divisible by 10. For example, any $n$ that is a multiple of $10$ satisfies the condition.

In fact any natural number larger than $1$ satisfies the condition. The first three factors are consecutive numbers. If none of those numbers is a multiple of $5$ then (as long as $n>1$) it must be that $n=5k+2$ or $n=5k+3$ for some integer $k\geq 0$. In the first case $$n^2+1=(5k+2)^2+1=5 (5k^2+4k+1),$$ while in the second case $$n^2+1=(5k+3)^2+1=5 (5k^2+6k+2).$$

edited Jul 15 '16 at 11:22

answered Jul 15 '16 at 10:33

smcc

5,694

Could it not also happen that one of those factors is divisible by two and another by five? – Jyrki Lahtonen Jul 15 '16 at 10:36
Yes, I meant that it is enough that one of the factors is divisible by 10 to satisfy the condition. I will edit. – smcc Jul 15 '16 at 10:37
Though there is always a 2 – Shailesh Jul 15 '16 at 10:37
Actually there is always a five, too :-) – Jyrki Lahtonen Jul 15 '16 at 10:41

Roby5 · Answer 2 · 2016-07-15T10:41:25.850

2

Hint:

$1)$ $n^{17}-n=n(n^4-1)(n^4+1)(n^8+1)$

$2)$ $\phi(10)=4$

edited Jul 15 '16 at 10:41

answered Jul 15 '16 at 10:34

Roby5

4,287

How did you rearrange $n^{17} - n$ into the above equations ? – Gigaxel Jul 15 '16 at 10:35
2

The hint may be too subtle for someone asking such an elementary question. Perhaps Euler Totient function and/or Euler's theorem is way beyond reach. – DonAntonio Jul 15 '16 at 10:37
1

@Gigaxel $n^{17}-n=n(n^16-1)$. Then, $n^{16}-1=(n^8-1)(n^8+1)$. Do you notice the pattern? – Roby5 Jul 15 '16 at 10:38
I think this is fine. I predict/hope that the question will be closed as a duplicate. Similar theme occurs so often. – Jyrki Lahtonen Jul 15 '16 at 10:41
Why was this downvoted? – Roby5 Jul 15 '16 at 11:27
@Roby5 Oh yes now I get it. Makes perfect sense. Thank you :D – Gigaxel Jul 15 '16 at 12:59

score 2 · Answer 3 · answered Jul 15 '16 at 10:37

A systematic approach: just check the cases of $2$ and of $5$ respectively.
It is clear that $n^{17}-n$ is even. Also we see $$n^{17}-n=(n^4)^{4}\times n-n\equiv 0\pmod5$$ since if $n$ is not divisible by $5,$ then $n^4-1$ is divisible by $5.$

Hope this helps.

score 2 · Answer 4 · answered Jul 15 '16 at 10:37

2

You can check all the remnants by $mod \ 10$ simply taking 17th power (there are 10 of them - $1,2, \ldots , 9$) or you can check whether the polynomial mentioned above takes its value divided by 10.

answered Jul 15 '16 at 10:37

Easter

21

score 0 · Answer 5 · answered Jul 15 '16 at 10:41

I don't have time to develop this, but it seems that all positive numbers satisfy your problem.

In Mathematica:

Reduce[Mod[n^17 - n, 10] == 0, n, Integers]

We have output:

C[1] \[Element] Integers && 
(n == 10 C[1] || n == 1 + 10 C[1] || n == 2 + 10 C[1] ||
n == 3 + 10 C[1] || n == 4 + 10 C[1] || n == 5 + 10 C[1] || 
n == 6 + 10 C[1] || n == 7 + 10 C[1] || n == 8 + 10 C[1] || 
n == 9 + 10 C[1])

Find all natural numbers $n$ such that $n^{17}-n$ is divisible by 10

5 Answers5