Proof that $a^5 b - b^5 a$ is divisible by $30$ for any integers $a$ and $b$

Question

I am trying to prove that $a^5\times b - b^5\times a$ is divisible by $3$. The actual task is to prove divisibility by $30$ but I have managed to prove that the expression is divisible by $5$ and $2$. However, I am having difficulties proving it's divisible by $3$. I don't want a direct answer but rather some kind of a hint.

Answer 1

$a^5b-ab^5=ab(a^2+b^2)(a+b)(a-b)$. Now $a=3k$, $3k+1$ or $3k+2$ for some integer $k$ and $b=3l$, $3l+1$ or $3l+2$ for some integer $l$. Simply consider these cases.

Answer 2

Hint: for $p\in\{2,3,5\}$, if $a\not\equiv0\pmod{p}$ and $b\not\equiv0\pmod{p}$, show that $a^4\equiv b^4\pmod{p}$.

Answer 3

If I were to write the expression as $$ab(a^4-b^4)$$ could you complete the factorisation and use that to help you with the different possibilities for $a$ and $b$ modulo $3$?

Answer 4

We can write $\displaystyle F=a^5b-ab^5=b(a^5-a)-a(b^5-b)$

Using Fermat's little theorem, $\displaystyle c^5-c\equiv0\pmod5$ for all integer $c$

$\displaystyle\implies F\equiv0\pmod5$ as $a,b$ are integers

Again from this, $\displaystyle a^3-a=(a-1)a(a+1)$ is divisible by $3!=6$

$\displaystyle\implies a^3\equiv a\pmod6\implies a^5b=a^3\cdot a^2b\equiv a\cdot a^2b\pmod6\equiv a^3b\equiv ab$

Similarly, $b^5a\equiv ba\pmod6$

$\displaystyle\implies a^5b-b^5a\equiv?\pmod6$

Now as $F$ is divisible by $5,6$ so it will be divisible by lcm$(5,6)$

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Generalization:

Consider $\displaystyle a^nb-b^na=ab\left[\underbrace{a^{n-1}-1}-(\underbrace{b^{n-1}-1)}\right]$

It will be divisible by prime $p$ if $p$ divides at least one of $a,b$

Else $(a,p)=(b,p)=(ab,p)=1$

Now using Fermat's little theorem, each of $a^{n-1}-1,b^{n-1}-1$ will be divisible by $p$ if $\phi(p)=p-1$ divides $n-1$

So if $p-1$ divides $n-1,p$ will divide $\displaystyle a^nb-b^na$ for all integer $a,b,n$

If $p_i$s are distinct primes, $\prod_{1\le i\le r} p_i$ will divide $\displaystyle a^nb-b^na$ if $n-1$ is divisible by $p_i-1$ for each $1\le i\le r$

i.e., if $n-1$ is divisible by lcm$(p_i-1)$ for $1\le i\le r$

Here, we can safely set $p_1=2,p_2=3,p_3=5$ hence, $n-1$ needs to be divisible by lcm$(1,2,4)=4$

Answer 5

Hint $\ $ It is the special case $\ p_i = 2,3,5,\,\ n = 4\ $ of

Theorem $\ \ \ $ If $\ p_i\ $ are primes, $ $ and $\ p_i\!-\!1\mid n\,$ for all $\,i\,$ then $\ \prod p_i \mid ab(a^n-b^n)\, =\, :c $

Proof $\ $ Let $\,p = p_{\large i}.\,$ $\,\color{#c00}{n = (p\!-\!1)k}\,$ by $\,p\!-\!1\mid n.\,$ If $\,p\mid ab\,$ then $\,p\mid ab\mid c.\,$ Else, by little Fermat ${\rm mod}\,\ p\!:\,\ a\not\equiv0\,\Rightarrow\,a^{\large\color{#c00}n} = (a^{\large\color{#c00}{ p-1}})^{\color{#c00}{\large k}}\equiv 1^{\large k}\equiv 1.\,$ Similarly $\,b^n\equiv 1\,$ so $\,a^n-b^n\equiv0\,\Rightarrow\,c\equiv 0.\ $ Therefore $\,p_i \mid c\,$ in either case, hence $\,{\rm lcm}\,\{\,p_i\} = \prod p_i$ also divides $c.\ \ $ QED

Answer 6

Hint (as you asked): consider the equation over $\mathbb{Z}/3\mathbb{Z}$ (it is the same as taking modulo 3). We then have 9 possibilities. You can discard the case $a=\overline{0}$ or $b=\overline{0}$ (trivial). You, then, have this possibilities

$$a=\overline{1}, b=\overline{1}$$ $$a=\overline{1}, b=\overline{2}$$ $$a=\overline{2},b=\overline{1}$$ $$a=\overline{2}, b=\overline{2}$$

If you find $0$ (or a multiple of 3) for all possibilities, you will have you answer. You could discard the case $a=b$ as trivial as well. You'd have just 2 cases: $a=1$, $b=2$ and $a=2$, $b=1$.