No triple of primes Prove that $p^5+q^5+r^5$ is not divisible by $10pqr$ where $p,q,r$ are prime numbers.
I proved using orders that if $p>q>r$ then $r=2$ and $q|p+2$. 
I proved (it is easy) using orders that there are no primes $p,q,r$ for which $6pqr|p^3+q^3+r^3$ . 
And I tried to solve this one, but it couldn't solve it. 

We use Fermat little theorem:
$P^5+q^5+r^5 ≡ (p+q+r) \ mod (5)$
⇒ $(p^5-p)+(q^5-q)+(r^5-r)≡ 0 \ mod(5)$
It can be shown that $30 \big | n^5-n$; $n∈ \mathbb N$: If n is prime, then:
$n^5-n≡ 0 \ mod(5)$
$n^5-5=n(n^2-1)(n^2+1)$
$n^2-1≡ 0 \ mod(3)$
Also $n(n+1)$ is even, that is $n^5-n$ is divisible by $30$. (Notice that this fact is true for all natural numbers).
Now we can write:
$(p^5-p)+(q^5-q)+(r^5-r)=30 k$
⇒ $p^5+q^5+r^5≡ (p+q+r) \ mod (30)= 30 m +(p+q+r)$; $m∈ \mathbb N$.
For $pqr$  we do not see a common factor like $pqr$ on RHS of following identity(the fourth term does not have factor $pqr$):
$p^5+q^5+r^5=\frac{(p+q+r)^5}{6}-\frac{5(p+q+r)^3(p^2+q^2+r^2)}{6}+\frac{5(p+q+r)^2(p^3+q^3+r^3)}{6}+\frac{5(p^2+q^2+r^2)(p^3+q^3+r^3)}{6}$
Hence  $10 pqr $ does not divide $p^5+q^5+r^5$.
Moreover you may make your question stronger:
Prove that $p^5+q^5+r^5$ is not divisible by $30pqr$.
