# If $\frac{1}{8} \left(5^m-2\cdot 3^m+1\right)$ is prime, then $m=2p$ where $p$ is prime?

The following statements are easy to prove with elementary arguments:

• $X_m=\frac{1}{8} \left(5^m-2\cdot3^m+1\right)$ is an integer for all integers $m\ge 0$
• ($m \equiv 0 \mod 4$ or $m \equiv 1 \mod 4$ )$\Longrightarrow$ ($X_m \equiv 0 \mod 2$)
• ($m \equiv 3 \mod 4$ )$\Longrightarrow$($X_m \equiv 0 \mod 3$)
• ($m \equiv 6 \mod 12$ )$\Longrightarrow$($X_m \equiv 0 \mod 7$)
• ($m \equiv 10 \mod 20$ )$\Longrightarrow$($X_m \equiv 0 \mod 11$)
• ($m \equiv 6 \mod 20$ )$\Longrightarrow$($X_m \equiv 0 \mod 11$)
• ($X_m \equiv0 \mod 5$)$\Longrightarrow$($m\equiv0 \mod 2$ )
• $X_m$ is prime $\Longrightarrow$ $m=2k$ where $k$ is odd

Moreover, it seems that $X_{158}$ is prime.

$\frac{1}{8} \left(5^m-2\cdot 3^m+1\right)$ is prime $\Longrightarrow$ $m=2p$ where $p$ is prime

Is this true?

This would be analogous to $2^n-1$ is prime $\Longrightarrow$ $n$ is prime.

But here, I have doubts because sometimes when $7$ is the smallest odd prime factor of $\frac{m}{2}$, my numerical computations show me that the prime divisors of $\frac{1}{8} \left(5^m-2 \cdot 3^m+1\right)$ can be quite large.

EDIT 1. Actually $7$ is the first prime which is not a Sophie Germain prime. If $m=2(2k+1)$ and $2k+1$ has a prime divisor $p$ such that $2p+1$ is also prime then it is easy to see that $2p+1$ divides $X_m$, so that $X_m$ cannot be prime. The above conjecture then becomes :

$\frac{1}{8} \left(5^m-2\cdot 3^m+1\right)$ is prime $\Longrightarrow$ $m=2p$ where $p$ is prime but not a Sophie Germain prime

EDIT 2. The same can be shown for the Stirling Number of 2nd kind ${m+1\brace 3}=\frac{1}{2} \left(3^m-2\cdot2^m+1\right)$ and the corresponding conjecture would be

${m+1\brace 3}$ is prime $\Longrightarrow$ $m=2p$ where $p$ is prime but not a Sophie Germain prime

But It seems that no $m$ is known yet for which ${m+1\brace 3}$ is prime.

EDIT 3. These numbers have a general form $$q_{m,k}=\frac{\underset{j=0}{\overset{k}{\Sigma }}(-1)^{k-j} (2 j+1)^m \left( \begin{array}{c} k \\ j \\ \end{array} \right) }{2^k k!}$$ From https://oeis.org/A028491 there are at least 18 known values of $m$ for which $q_{m,1}=\frac{3^m -1}{2}$ is prime.

So far we have 2 prime values for $X_m=q_{m,2}$

$q_{159,3}$ is prime and it remains to be checked whether other prime $q_{m,3}$ can be found for reasonnably small $m$ $$q_{m,3}=\frac{1}{48} \left(7^m-3.5^m+3.3^m-1\right)$$

In the same line $${m\brace k}=\frac{\underset{j=0}{\overset{k}{\Sigma }}(-1)^{k-j} j^m \left( \begin{array}{c} k \\ j \\ \end{array} \right) }{k!}$$ For $k=2$, 44 of them are known to be prime (Mersenne primes)

For $k=3$, apparently none is known to be prime

For $k=4$, only 4 of them are known to be prime (https://oeis.org/A100958)

For $k\gt 4$, apparently none is known to be prime either.

These numbers grow with $m$ like $(2k+1)^m$ or $k^m$ : this is probably why they cannot be prime as often for large $k$ as for small $k$.

But it seems that being prime is more likely for $q_{m,k}$ than for ${m\brace 2k+1}$, I wonder why...

• $m=158$ and $m=6662$ are the only solutions for $m\le10^4$ so that $X_m$ is prime, and they are both of the form $2p$. – Lucian Nov 26 '15 at 3:02
• It would be nice if this were not just a coincidence. Is there a known factorisation for $x^m-2y^m+1$? like there is the cyclotomic factorisation for $x^m-1$? – René Gy Nov 26 '15 at 7:13
• No further primes upto $m=100,000$. Note, that for the two examples, $m-1$ is prime. – Peter Nov 30 '15 at 8:56
• @Peter; thanks for the numerical search. – René Gy Nov 30 '15 at 19:54
• @Peter; would you do the same search for $$q_{m,3}=\frac{1}{48} \left(7^m-3.5^m+3.3^m-1\right)$$ It is prime for $m=3* 53$. It would be nice to find a second $m$ for which $q_{m,3}$ is prime, and see whether is also three times a prime... – René Gy Nov 30 '15 at 19:58