# Can we prove that $t$ must be a Mersenne prime?

This question

Equations involving arithmetic functions, totatives and even perfect numbers

can be answered positive if the following can be proven.

Let $t\ge 1$ be an integer and denote $$s:=\sigma\left(\frac{t(t+1)}{2}\right)$$ $$p:=\varphi\left(\frac{t(t+1)}{2}\right)$$

$\sigma(n)$ is the divisor-sum-function and $\varphi(n)$ the totient-function.

Claim :

• If $s=4p+t+1$ holds, then $t$ is a Mersenne-prime.

• If $t^2-1=4p$ holds, then $t$ is a Mersenne-prime.

The claim is true for $t\le 10^6$

• I think I can prove that $t$ is Mersenne prime if both $s=4p+t+1$ and $t^2-1=4p$ hold. Judging by the other question you quoted, there is an "and" in between. Is this what you are after? – rtybase Apr 21 '18 at 11:20
• This is a useful partial result , but I think the formulation clearly indicates that I search a proof of each implication. Perhaps, $t^2-1=4p\implies t$ is Mersenne-prime is easier to show. – Peter Apr 21 '18 at 14:00
• Probably. Technically, propositions 1 to 3 use only $t^2-1=4p$ ... work in progress ... the only scaring factor is Wikipedia mentioning "It is unknown whether there is any odd perfect number ..." – rtybase Apr 21 '18 at 14:53

If both $$t^2-1=4p \tag{1}$$ $$s=4p+t+1 \tag{2}$$

hold. We can deduce:

Proposition 1. $t$-odd.

Easy to see from $(1) \Rightarrow t^2=4p+1$.

Proposition 2. If $t=2k+1$ then $$k(k+1)=\varphi(2k+1)\varphi(k+1)=p$$

Given $\gcd(t,t+1)=1$ and $\varphi(n)$ is multiplicative, which means $$p=\varphi\left(\frac{t(t+1)}{2}\right)=\varphi\left(t\right)\varphi\left(\frac{t+1}{2}\right)$$ then $(1)$ becomes $$t^2=4\varphi\left(t\right)\varphi\left(\frac{t+1}{2}\right)+1 \tag{3}$$ Substituting $$k(k+1)=\varphi(2k+1)\varphi(k+1)$$

Proposition 3. $t$-is square free

If $q$-prime has the property that $q^2 \mid t \Rightarrow q \mid \varphi(t)$. But then, from $(3) \Rightarrow q \mid 1$ - contradiction.

Proposition 4. $\frac{t(t+1)}{2}$ is a perfect number.

I.e. $t+1=s-4p$ and $$(t-1)(t+1)=4p \iff (t-1)(s-4p)=4p \iff t(s-4p)-s+4p=4p \iff \\ t=\frac{s}{s-4p}$$ But $t$-is odd (propositions 1 and 2) $$2k+1=\frac{s}{s-4k(k+1)} \iff 2k=\frac{4k(k+1)}{s-4k(k+1)} \iff \\ s-4k(k+1)=2(k+1) \iff s=4k(k+1)+2(k+1)=2(k+1)(2k+1)$$ or $$s=\sigma\left(\frac{t(t+1)}{2}\right)=2\cdot\frac{t(t+1)}{2}$$

Proposition 5. $t$ is a Mersenne prime.

From Proposition 1 and 3 $t$ is odd and square free, i.e. its prime factorisation is $t=q_1\cdot q_2\cdot ...\cdot q_r, q_i>2, q_i\ne q_j, i\ne j$. $\sigma(n)$ is multiplicative and $\gcd(t,t+1)=1$ thus $$\sigma\left(\frac{t(t+1)}{2}\right)=\sigma(t)\sigma\left(\frac{t+1}{2}\right)= \sigma\left(\frac{t+1}{2}\right)\prod\limits_{i=1}^r(q_i+1)=2^r\cdot Q$$

• if $r\geq 2$, from $(2)$, $4 \mid (t+1)$ and (from proposition 4) $\frac{t(t+1)}{2}$ is even perfect number, thus $t$ is Mersenne prime.
• if $r=1$, then $t$-is prime and $\varphi(t)=t-1=2k$ and from proposition 2 $$k(k+1)=\varphi(2k+1)\varphi(k+1)=2k\varphi(k+1) \Rightarrow 2\mid k+1=\frac{t+1}{2}$$ and (from proposition 4) $\frac{t(t+1)}{2}$ is even perfect number, thus $t$ is Mersenne prime.

Done. It might be more difficult (if possible at all) to deduct this from individual $(1)$ or $(2)$.

Remark: This book, page 72, has a short proof of

Theorem 1.51. An even positive integer $n$ is perfect if and only if $n = 2^{k−1}M_k$ for some positive integer $k$ for which $M_k$ (Mersenne number) is a prime.