In this post we denote the Euler's totient function that counts the number of positive integers $1\leq k\leq n$ such that $\gcd(k,n)=1$ as $\varphi(n)$, and the sum of divisors function $\sum_{1\leq d\mid n}d$ as $\sigma(n)$. As reference I add the Wikipedia Mersenne prime) that refers the well-known definition that a prime $p$ is a Mersenne exponent if $2^p-1$ is prime. This prime constellation corresponds to the entry A000043 from the On-Line Encyclopedia of Integer Sequences.

From previous definition and the calculations of previous arithmetic function $\varphi(n)$ and $\sigma(n)$ it is easy to check the proof of the following claim.

Claim. If $x=p$ is (a prime) such that $y=2^p-1$ is (also) prime, then $(x,y)$ solves the equation $$\sigma(x^{\sigma(y)-1})=\frac{1}{\varphi(x)}(x^{y+1}-1).\tag{1}$$

We propose the following conjecture inspired in previous claim, from the substitution $$y=2\cdot 2^{\varphi(x)}-1=2^{1+\varphi(x)}-1.\tag{2}$$

Conjecture. Let $x\geq 1$ be an integer that satisfies $$\sigma(x^{\sigma(2^{1+\varphi(x)}-1)-1})=\frac{1}{\varphi(x)}(x^{2^{1+\varphi(x)}}-1),\tag{3}$$

then $x$ is a Mersenne exponent.

Question. What work can be done with the purpose to prove or refute previous Conjecture? Many thanks.

Computational evidence. You can check in the web Sage Cell Server this line written in Pari/GP

for(x=1,16,if(sigma(x^(sigma(2*2^(eulerphi(x))-1 )-1))==(x^(2*2^(eulerphi(x))-1 +1)-1)/eulerphi(x) ,print(x)))

just copy and paste it to evaluate in the web choosing as Language the option GP.

  • $\begingroup$ One can to show that $x=\prod_{1\leq k\leq n}p_k$ has no repeated prime factors (using the same reasoning that the answer of the recent post MSE 3578715), thus $n=\omega(x)$ is the number of distinct prime factors $p_k$ dividing $x$. And (also from the reference) if there aren't typos $$\prod_{k=1}^{\omega(x)}(p_k^{\sigma(2\cdot 2^{\varphi(x)}-1)}-1)=-1+\prod_{k=1}^{\omega(x)}p_k^{\,2\cdot 2^{\varphi(x)}}.$$ I would like to know if it is possible to prove the Conjecture or to find a counterexample. $\endgroup$
    – user759001
    Mar 19, 2020 at 19:24
  • $\begingroup$ Of course if you want to denote $x=M$ or $x=E$ feel free to do it in your answer. On the other hand is very welcome if some person can to test our conjecture for a segment of integers greater than my Computational evidence. Many thanks. $\endgroup$
    – user759001
    Mar 19, 2020 at 19:35

1 Answer 1


The conjecture is true.

Proof :

$x=1$ does not satisfy $(3)$, and $x=2$ satisfies $(3)$. In the following, $x\ge 3$.

As you already noticed, $x$ has to be a square-free integer.

Then, letting $x=\displaystyle\prod_{k=1}^{n}p_k$ with $n=\omega(x)$, we get


Suppose here that $2^{1+\varphi(x)}-1$ is a composite number.

Here, using the following facts :

  • If $N$ is a composite number, then $\sigma(N)\ge 1+\sqrt N+N$.

  • If $N\ge 3$, then $\varphi(N)\ge 2$.

  • If $m\ge 2$ and $y\gt 0$, then $m^{2+y}-1\ge m^{1+y}$.

we get

$$\begin{align}-1&=\prod_{k=1}^{n}\bigg({p_k}^{\sigma(2^{1+\varphi(x)}-1)}-1\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}} \\\\&\ge \prod_{k=1}^{n}\bigg({p_k}^{1+\sqrt{2^{1+\varphi(x)}-1}+2^{1+\varphi(x)}-1}-1\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}} \\\\&=\prod_{k=1}^{n}\bigg({p_k}^{\sqrt{2^{1+\varphi(x)}-1}+2^{1+\varphi(x)}}-1\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}} \\\\&\ge\prod_{k=1}^{n}\bigg({p_k}^{\sqrt{2^{1+2}-1}+2^{1+\varphi(x)}}-1\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}} \\\\&\ge\prod_{k=1}^{n}\bigg({p_k}^{2+2^{1+\varphi(x)}}-1\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}} \\\\&\ge\prod_{k=1}^{n}\bigg({p_k}^{1+2^{1+\varphi(x)}}\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}}\end{align}$$ from which we have $$-1\ge \prod_{k=1}^{n}\bigg({p_k}^{1+2^{1+\varphi(x)}}\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}}$$ which is impossible since the RHS is positive.

So, we see that $2^{1+\varphi(x)}-1$ is a prime number.

Then, we get $$\prod_{k=1}^{n}\bigg({p_k}^{2^{1+\varphi(x)}}-1\bigg)=-1+\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}}$$

Suppose here that $n\ge 2$. Then, we get $$\begin{align}1&=\prod_{k=1}^{n}\bigg({p_k}^{2^{1+\varphi(x)}}\bigg)-\prod_{k=1}^{n}\bigg({p_k}^{2^{1+\varphi(x)}}-1\bigg) \\\\&\ge {p_n}^{2^{1+\varphi(x)}}\prod_{k=1}^{n-1}\bigg({p_k}^{2^{1+\varphi(x)}}\bigg)-\bigg({p_n}^{2^{1+\varphi(x)}}-1\bigg)\prod_{k=1}^{n-1}\bigg({p_k}^{2^{1+\varphi(x)}}\bigg) \\\\&=\prod_{k=1}^{n-1}\bigg({p_k}^{2^{1+\varphi(x)}}\bigg)\end{align}$$ from which we have $$1\ge \prod_{k=1}^{n-1}\bigg({p_k}^{2^{1+\varphi(x)}}\bigg)$$ which is impossible since the RHS is larger than $1$.

So, we have to have $\omega(x)=n=1$, and $x$ has to be a prime number.

Therefore, $x$ has to be a Mersenne exponent. $\quad\blacksquare$

  • 1
    $\begingroup$ Many thanks I am going to read it again, and study i in next days, but I would like to say that this seems to me as a little miracle. I would to emphasize that this is now your proposition as a characterization of Mersenne exponents, many thanks again for such nice proof and sharing your calculations. $\endgroup$
    – user759001
    May 17, 2020 at 7:36

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