product of primes II

**wauwau** · December 27th 2009, 04:36 AM

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

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

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

**abender** · January 3rd 2010, 01:37 PM

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

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

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

So, let's consider the following, seeking contradiction

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

Quite honestly, I'm not sure if this is how you attack this. Fermat's little Thm. applied to $P^5$ sheds light on the fact that

$p^5 \equiv p (\mod 5) \implies p^5 - p \equiv 0 (mod 5) \implies (p-1)(p)(p+1)(p^2+1) \equiv 0 (mod 5)$

.

Here we observe that this $p^5 - p = (p-1)(p)(p+1)(p^2+1)$ 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

Thread: product of primes II

LinkBack

Thread Tools

Display

product of primes II

Similar Math Help Forum Discussions

Product of perfect square and primes

Prove that (product of primes between n and 2n) > 2^n

product of primes

Product of Primes

Product of primes using induction

Search Tags