1. product of primes II

Let $\displaystyle q_i, p$ be odd primes, $\displaystyle k \ge 2$ integer

If $\displaystyle \prod_{i=1}^{k}q_i = p^5+2$
then

$\displaystyle \prod_{i=1}^{k}(q_i-1) \not= p^5-p^4+2$

2. $\displaystyle \prod_{i=1}^{k}q_i = q_1 q_2 \cdot\cdot\cdot q_k = p^5+2$

$\displaystyle \prod_{i=1}^{k}(q_i-1) = (q_1-1) (q_2-1) \cdot\cdot\cdot (q_k-1) = p^5 - p^4 +2$

$\displaystyle \phi(q_1 q_2 \cdot\cdot\cdot q_k) = (q_1 - 1) (q_2 - 1) \cdot \cdot \cdot (q_k - 1)$ (I'll leave this proof to someone else)

So, let's consider the following, seeking contradiction

$\displaystyle \phi(p^5+2) = p^5 - p^4 + 2$ .

Quite honestly, I'm not sure if this is how you attack this. Fermat's little Thm. applied to $\displaystyle P^5$ sheds light on the fact that
$\displaystyle p^5 \equiv p (\mod 5) \implies p^5 - p \equiv 0 (mod 5) \implies (p-1)(p)(p+1)(p^2+1) \equiv 0 (mod 5)$
.

Here we observe that this $\displaystyle p^5 - p = (p-1)(p)(p+1)(p^2+1)$ bit is divisible by 2 (2 consecutive numbers multiplied together), divisible by 3 (one of three consecutive terms certainly is divisible by 3), divisible by 5 (by virtue of the 0(mod 5)), and divisible by 6 since that's 2*3.

Just offering my \$0.02. I'd be interested in a elegant solution to this.

-Andy