# Homework Help: Prove that n(n^4 - 1) = 10Q

Tags:
1. Dec 5, 2016

• Thread moved from the technical forums, so no Homework Help Template is shown.
For some values Q and n being integers, prove that n(n^4 - 1) = 10Q.

So I've tried this with induction, but it gets pretty messy pretty quickly. So I can see that the LHS will be even no matter what, but I'm not sure where to go beyond this.

2. Dec 5, 2016

### Buffu

Is induction compulsory for this question ?

You essential need to prove that $n(n^4 - 1)$ is divisible by 10. Can you show the induction step ?

3. Dec 5, 2016

### Ray Vickson

Your problem statements does not make much sense as written. Do you mean the following?

"For all integers n >= 1, show that the integer n(n^4-1) is divisible by 10 (that is, has zero remainder when divided by 10".

4. Dec 5, 2016

### Staff: Mentor

I would consider a few cases instead of induction.

5. Dec 5, 2016

### PeroK

Factorising $n^4-1$ might not be a bad idea.

6. Dec 5, 2016

How do I factorize it?

7. Dec 5, 2016

Induction isn't compulsory, but it's for all n belonging to the natural numbers.

8. Dec 5, 2016

### Buffu

$$x^n - y^n = (x-y)\left(\sum^{n-1}_{k =0} x^{n-1-k}y^k\right)$$

Last edited: Dec 5, 2016
9. Dec 5, 2016

### Staff: Mentor

Also known as "find (x-y) as factor and proceed from there". And it is not the last possible factorization step.

10. Dec 5, 2016

### PeroK

If you let $x = n^2$ then $n^4-1=x^2-1$ and hopefully you know how to factorise that.

11. Dec 5, 2016

### Buffu

$n^4-1=(n^2-1)(n^2 + 1)$

12. Dec 5, 2016

### Staff: Mentor

No. $n^4 - 1 = (n^2 - 1)(n^2 + 1)$

13. Dec 5, 2016

### Staff: Mentor

PeroK's version was a correct substitution. You did the factorization step but with the wrong variable.

14. Dec 5, 2016

### Buffu

Did not see the substitution.
Can you please delete my last post.

15. Dec 5, 2016

Hmmm, so now I have $n(n+1)(n-1)(n^2+1))$ So it's pretty clear that anything here will be divisible by 3... Also I can see that whether the parity of n is either even or odd, the entire term will come out to even. So the term must be divisible by 6, but I'm not seeing how to get that number up to 10...

Thanks to everyone so far, didn't even see the conjugate there.

16. Dec 5, 2016

### PeroK

Why not just crunch through that modulo 10?

17. Dec 5, 2016

I'm not sure what that means.

18. Dec 5, 2016

### PeroK

Every number is equal to 0-9 modulo 10. Just check each of these possibilities. For example $7^2+1=50$ so $n^2+1$ is a multiple of 10 for every number that equals 7 modulo 10.

19. Dec 5, 2016

OK, it took me a minute to figure out the modular arithmetic, but I do understand what you're saying.

So pretty much by stating that all numbers are 0-9 mod 10, then checking each case, we can effectively run through all the natural numbers.
For real, that helps a lot. Thanks.

20. Dec 5, 2016

### Staff: Mentor

You already know that the expression is even, checking 0 to 4 to prove divisibility by 5 is sufficient.

21. Dec 5, 2016

### Ray Vickson

Induction works just fine. If $f(n) = n(n^4-1)$, then it is not hard to find that $f(n+1) - f(n) = 5 M + 10N$, where the integers $M$ and $N$ are sums of some powers of $n$.

Last edited by a moderator: Dec 6, 2016
22. Dec 5, 2016

### PeroK

If you go back to the original expression:
$p(n) = n(n^4-1)$

As this is even, it is enough to show that it divisible by 5. But, also, $p(-n) = -p(n)$, so it is enough to check 0, 1, 2.

23. Dec 5, 2016

### haruspex

Or get there in one step using https://en.m.wikipedia.org/wiki/Fermat's_little_theorem