<s>[INST] Proof that $a^5 b - b^5 a$ is divisible by $30$ for any integers $a$ and $b$ 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.
 [/INST] 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$(5,6)$

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$(1,2,4)=4$
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