How to prove if $m,n\in \mathbb{Z}$,then $30\mid mn(m^4 -n^4)$

Question

I first thought I'd just have to do cases, i.e. if $m,n$ are even, $m=2l, n=2k$, where $k, l\in \Bbb Z$. But even in this case, alone, I wind up with $4kl(16l - 16k) = 64k(l^2) - 64l(k^2)\dots$ and that doesn't really get me anywhere.

Induction won't work, because I have no base case.

I feel as though this requires the use of mods, as being divisible by $30$ is tantamount to being equivalent to $0 \pmod {30}$, but where would I go from here?

Answer 1

Alright, let's see how to do this.

First, it clearly suffices to prove that the term is always divisible by 2, 3 and 5 to prove that it's divisible by $30=2 \cdot 3 \cdot 5$.

So why is it divisible by 2?

Assume that it's not i.e. it's odd. Then clearly all factors must be odd. In particular, $m$ and $n$ must be odd. But then clearly $m^4-n^4$ is even. So we derived a contradiction i.e. the term is indeed even.

Let's go on to a more difficult part: Why is it divisible by 3?

Assume that it's not. Then none of the factors can be divisible by 3.

In particular, $m$ and $n$ must not be divisible by 3. Also $m^4-n^4=(m^2+n^2)(m+n)(m-n)$ (it's really useful to know this factorization!) must not be divisible by 3.

But if $m,n$ are both not divisible by 3 then they either have the same remainder $\mod 3$ in which case $m-n$ is divisible by 3 or one of them is $1 \mod 3$ and the other $2 \mod 3$ in which case $m+n$ is divisible by 3. You see, in either case we derived a contradiction. Hence the term is indeed divisible by 3.

Now, we are left to prove that it's divisible by 5. This is surely the hardest part of the problem but maybe you can now think of a way to prove this on your own?

Answer 2

HINT:

For prime $p,$ either $p|a$ or $(p,a)=1$

For the second case, $a^{p-1}\equiv1\pmod p$

If $p\nmid mn, m^{p-1}\equiv n^{p-1}\equiv1\pmod p\implies p\mid mn(m^{p-1}-n^{p-1})$

Answer 3

Induction can work. First, note that you only need to prove for $m,n\geq 0$, since:

$$30\mid mn(m^4-n^4)\iff 30\mid |m|\cdot|n|\left(|m|^4-|n|^4\right)$$

Now, it is true for $(0,n)$.

If true for $(m,n)$ then $$\begin{align} (m+1)n((m+1)^4-n^4) &= (m+1)^5n - n^5(m+1)\\ &=(m^5n-n^5m) + (5m^4+10m^3+10m^2+5m+1)n-n^5\\ &=(m^5n-n^5m) + 5mn(m^3+2m^2+2m+1)-(n^5-n) \end{align}$$

So you need to prove that $30\mid n^5-n$ and $6\mid m(m^3+2m^2+2m+1)=m(m+1)(m^2+m+1)$.

These you can prove as lemmas by induction on the one variable.

Not a pretty answer, but you can do induction on this sort of question.