# Advise on determining prime numbers such that the product of their geometric series is equivalent to another prime number raised to a power.

I am trying to find some set of prime numbers $p$ such that the below identity holds true. I was hoping to seek advice as to finding some stronger or more useful relations that may help find such a set $p$.

Let there exist some prime number $P$ raised to some positive integer $N$. Let there further exist some prime numbers $p_0,..., p_i \in p$ such that $p_a \ne p_b$ for any $a \ne b$, and positive integers $n_0,..., n_i \in n$ such that $n_a \leq n_b$ for any $a \gt b$, and such that they fulfill the below, equivalent identities. $$gcd(\prod_{j=0}^{i} \sum_{k=0}^{n_j} p_j^k,P^N)=P^N$$ $$\prod_{j=0}^{i} gcd(\sum_{k=0}^{n_i}p_j^k,P^N)=P^K,K \ge N$$

It may be useful to note the connection between this identity and multiperfect numbers, wherein this identity must hold for each factor $P^N$ in the prime factorization of some multiperfect number $x$. This becomes evident when considering the formula for the sum of the proper divisors of $x$ , $\sigma(x)=-x+\prod_{j=0}^{i} \sum_{k=0}^{n_j} p_j^k$, and recalling that $\frac{\sigma(x)}{x}$ must be an integer.

Any insight into the problem is appreciated.

Using the formula for summation of finite geometric sequences, we have: $$\prod\limits_{j=0}^i\sum\limits_{k=0}^{n_j}p_j^k = \prod\limits_{j=0}^i\frac{p_j^{n_j+1}-1}{p_j-1}$$

So you are asking for a set of distinct primes $p_0, \ldots, p_i$ and positive integers $n_0 \geq \ldots \geq n_i$ such that this quantity is divisible by $P^N$.

Here is one possible way to construct such a set (which works for primes $P \gt 2$).

First let $i = N-1$.

Now choose the $p_j$ to be distinct primes not equal to $P$ such that $p_j \not\equiv 1 \pmod P$.

This is possible because there are infinitely many primes in any of the congruence classes modulo $P$ with representatives that are coprime to $P$ (see Dirichlet's theorem on arithmetic progressions).

For each $j$, let $n_j = P-2$.

Then by Fermat's little theorem we have $p_j^{n_j+1} - 1 = p_j^{P-1} - 1 \equiv 0 \pmod P$, which means $p_j^{n_j+1}-1$ is divisible by $P$ for all $j$, and moreover $p_j - 1$ is not divisible by $P$ by the choice of $p_j$.

Therefore we obtain a product of $N$ terms, each of which is divisible by $P$, which means the product itself is divisible by $P^N$.

• Another possibility would be to let $i = 0$, $p_0 \neq P$ prime such that $p_0 \not\equiv 1 \pmod P$ and $n_0 = \varphi(P^N) - 1 = P^{N-1}(P-1) - 1$. Then you can use Euler's theorem. – Tob Ernack Mar 5 '18 at 21:15