4
$\begingroup$

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 $t\le 10^6$

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – rtybase Apr 21 '18 at 11:20
  • 1
    $\begingroup$ 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. $\endgroup$ – Peter Apr 21 '18 at 14:00
  • $\begingroup$ 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 ..." $\endgroup$ – rtybase Apr 21 '18 at 14:53
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.