What is the proof for n(n^4 - 1) = 10Q for some values Q and n being integers?

[James Brady]

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.

 

[Buffu]

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.

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 ?

 

[Ray Vickson]

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.

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".

 

[mfb]

I would consider a few cases instead of induction.

 

[PeroK]

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

 

[James Brady]

How do I factorize it?

 

[James Brady]

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

 

[Buffu]

How do I factorize it?

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

 

[mfb]

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

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

 

[PeroK]

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

 

[Buffu]

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

 

[Mark44]

typo : $n^4-1=(x^2-1)(x^2 + 1)$

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

 

[mfb]

typo : $n^4-1=(x^2-1)(x^2 + 1)$

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

 

[Buffu]

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

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

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

 

[James Brady]

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.

 

[PeroK]

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.

Why not just crunch through that modulo 10?

 

[James Brady]

I'm not sure what that means.

 

[PeroK]

I'm not sure what that means.

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.

 

[James Brady]

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.

 

[mfb]

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

 

[Ray Vickson]

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.

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 $n$.

 

[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.

 

[haruspex]

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.

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