Prove: $33|mn(m^{10}-n^{10})\forall m,n\in\mathbb Z$

Question

Prove: $$33|mn(m^{10}-n^{10})\forall m,n\in\mathbb Z$$

Apart from the trivial cases: $$3|m\;\lor\;3|n\;\;\&\;\;11|m\;\lor\;11|n$$ If $3\nmid m\;\land 3\nmid n$

Let $x\in\mathbb Z$ s.t.

$$x\equiv 1\pmod{3}\implies x^{10}\equiv 1\pmod{3}$$ $$x\equiv 2\equiv-1\pmod{3}\implies x^{10}\equiv 1\pmod{3}$$ $$\implies m^{10}\equiv n^{10}\pmod{3}\;\forall m,n\in\mathbb Z\implies m^{10}-n^{10}\equiv 0\pmod{3}$$ $$\implies 3|mn(m^{10}-n^{10})$$

$11\nmid m\;\land\;11\nmid n$

Let $x,y\in\mathbb Z$ s.t.

For each of the cases: $$x\equiv y\pmod{11},\;y\in\{1,\ldots, 10\}$$ by the Fermat's little theorem: $$x^{10}\equiv y^{10}\equiv 1\pmod{11}$$ $$\implies m^{10}\equiv n^{10}\pmod{11}\implies m^{10}-n^{10}\equiv 0\pmod{11}$$ Finally, $$\implies 33|mn(m^{10}-n^{10}),\;\forall\;m,n\in\mathbb Z $$ $$Q.E.D.$$

Is this correct?

Answer 1

Even easier

I will show how $10$ has been chosen

$$mn(m^{10}-n^{10})=n\cdot m(m^{10}-1)-m\cdot n(n^{10}-1)$$

Now using Fermat's little theorem,

$$3|(a^3-a), 11|(a^{11}-a)$$ for any integer $a$

Then $[3,11]$ will divide $[a(a^2-1),a(a^{10}-1)]=a(a^{[2,10]}-1)=?$

Answer 2

While the proof looks correct to me, note that using lots of symbols often doesn't make for a very readable proof. Instead, try to write in full sentences.

Another thing is that many mathematical results already have names. By referring to these names instead of spelling out the full result, you can shorten your proof, while keeping it clear.

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For instance, you could start off with: By the Chinese Remainder Theorem, it is enough to show that $3,11\mid mn(m^{10}-n^{10})$ for all integers $m,n$.

Next, note that your're basically proving the same thing twice. The argument is: I have some prime $p$ (either $3$ or $11$ in your case) and either $p\mid nm$ or $m^{10}\equiv n^{10}\equiv 1\pmod p$. In both cases, it follows that $p\mid mn(n^{10}-m^{10})$.

I'd turn this into a simple lemma.

Lemma: Let $p$ be a prime, $k$ a positive integer with $p-1\mid k$. Then $p\mid mn(m^k-n^k)$ for all $m,n\in\mathbb{Z}$.

Proof: Let $m,n\in\mathbb{Z}$ be given. If $p\mid mn$, we are done. Otherwise, $m$ and $n$ can be regarded as elements of $(\mathbb{Z}/p\mathbb{Z})^\times$, which is a cyclic group of order $p-1$. This means that $m^{p-1}\equiv n^{p-1}\equiv 1\pmod p$. Write $k=(p-1)l$, then $$m^{k}-n^{k}\equiv (m^{p-1})^l-(n^{p-1})^l\equiv 1^l-1^l\equiv 0\pmod p,$$ so $p\mid m^{k}-n^{k}$. In both cases, $p\mid mn(m^k-n^k)$. Q.E.D.

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Now, if you ever get ten exercises like these on a test, you can write down the lemma once and solve all the exercises very quickly by simply referring to the lemma and Chinese Remainder Theorem.

In general, it can be very helpful to look at an exercise like this and the proof you've given for it and ask yourself how far you can take it. What if we take some other number instead of $33$? What if this new number is divisible by the square of some prime? Or any prime power? How will you have to change the lemma? Can you even change the lemma or are there cases in which an analogous statement doesnt hold?