On characterizations for near-square primes and Fermat primes in terms of equations involving arithmetic functions

Question

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 $\sigma(n)$.

We propose the first conjecture inspired in the form of near-square primes, that are the primes of the form $n^2+1$ corresponding to the sequence A002496 from the OEIS, and the second conjecture from the form of the primes $2^{2^n}+1$, that are the known as Fermat primes A019434 from the OEIS. As general reference I add the Wikipedia articles related to Landau's problems and for Fermat number.

Conjecture 1. Let $x> 1$ be an integer for which there exists a positive integer $y\geq 1$ such that the identity $$y^{x-1-\sigma(x)}=\frac{1}{\varphi(x)}\tag{1}$$ holds, then $x$ is a near-square prime.

As remark notice that we exclude the (input) case $x=1$ that corresponds to $x=0^2+1$, that is the specialization of $n^2+1$ for $n=0$, an integer that isn't a prime number.

Conjecture 2. Let $x\geq 1$ be an integer for which there exists a positive integer $y\geq 1$ such that the identity $$-(\sigma(x)-x+1)^y=\log_2\left(\frac{1}{\varphi(x)}\right)\tag{2}$$ holds, then $x$ is a Fermat prime greater than $3$.

As remark the equation $(2)$ excludes (as output) the first Fermat prime that corresponds to the specialization of $2^{2^n}+1$ for $n=0$.

Question. What work can be done with the purpose to prove or refute previous conjectures? Can you find counterexamples? Many thanks.

Computational experiments. You can check in the web Sage Cell Server this (or similar ones) line written in Pari/GP

for(x=2, 10000, for(y=1, 100, if(y^(x-1-sigma(x))==1/eulerphi(x),print(x," ",y))))

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

And for the second of our conjectures we've the following or similar

for(x=1, 10000, for(y=1, 100, if((sigma(x)-x+1)^y==log(eulerphi(x))/log(2),print(x," ",y))))

Answer 1

The two conjectures are true.

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Conjecture 1 is true.

Proof :

$(1)$ is equivalent to $$y^{\sigma(x)-x+1}=\varphi(x)$$

Suppose that $x$ is a composite number. Then, there exists a divisor $d$ of $x$ such that $\sqrt x\le d\lt x$, so we get $\sigma(x)\ge 1+\sqrt x+x$. So, we have $$\varphi(x)=y^{\sigma(x)-x+1}\ge 2^{\sqrt x+2}$$ Here, let us prove that $2^{\sqrt x+2}\gt x$ for $x\gt 1$.

Let $f(x)=2^{\sqrt x+2}-x$. Then, we have $f'(x)=\frac{g(x)}{\sqrt x}$ where $g(x)=2^{\sqrt x+1}\ln 2-\sqrt x$. We have $g'(x)=\frac{h(x)}{2\sqrt x}$ where $h(x)=2^{\sqrt x+1}(\ln 2)^2-1$. Since $h(x)$ is increasing with $h(1)=\ln(4e)\ln(\frac 4e)\gt 0$, we get $h(x)\gt 0$ from which $g'(x)\gt 0$ follows with $g(1)=\ln\frac{16}{e}\gt 0$. Since $g(x)\gt 0$, we see that $f'(x)\gt 0$ with $f(1)=7\gt 0$ from which $f(x)\gt 0$ follows.$\quad\square$

So, we get $$\varphi(x)=y^{\sigma(x)-x+1}\ge 2^{\sqrt x+2}\gt x\gt \varphi(x)$$ which is impossible.

So, $x$ has to be a prime number, and we get $y^{2}=x-1$.

It follows that $x$ has to be a near-square prime.$\quad\blacksquare$

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Conjecture 2 is true.

Proof :

$(2)$ is equivalent to $$(\sigma(x)-x+1)^y=\log_2\varphi(x)$$ The LHS is a positive integer, so there has to be a positive integer $k$ such that $\varphi(x)=2^k$. So, we see that $x$ has to be either of the form $x=2^m$ or of the form $$x=2^m\prod_{i=1}^{n}(2^{a_i}+1)$$ where $2^{a_1}+1,2^{a_2}+1,\cdots, 2^{a_n}+1$ are distinct primes.

In the former, we get $$(2^{m})^y=m-1$$ which is impossible since the LHS is larger than the RHS.

In the latter, suppose that $m\ge 1$. Then, we have $$\begin{align}m-1+a_1+a_2+\cdots +a_n&=\bigg(1+(2^{m+1}-1)\prod_{i=1}^{n}(2^{a_i}+2)-2^m\prod_{i=1}^{n}(2^{a_i}+1)\bigg)^y \\\\&\gt (2^{m+1}-1)\prod_{i=1}^{n}(2^{a_i}+2)-2^m\prod_{i=1}^{n}(2^{a_i}+1) \\\\&\gt (2^{m+1}-1)\prod_{i=1}^{n}(2^{a_i}+1)-2^m\prod_{i=1}^{n}(2^{a_i}+1) \\\\&= (2^m-1)\prod_{i=1}^{n}(2^{a_i}+1)\end{align}$$ from which we have $$m-1+a_1+a_2+\cdots +a_n\gt (2^m-1)\prod_{i=1}^{n}(2^{a_i}+1)$$ which is impossible. So, we have to have $m=0$.

Suppose that $n\ge 2$. Then, we have $$\begin{align}a_1+a_2+\cdots+a_n&=\bigg(1+\prod_{i=1}^{n}(2^{a_i}+2)-\prod_{i=1}^{n}(2^{a_i}+1)\bigg)^y \\\\&\gt \prod_{i=1}^{n}(2^{a_i}+1+1)-\prod_{i=1}^{n}(2^{a_i}+1) \\\\&\gt \sum_{i=1}^{n}(2^{a_i}+1)\end{align}$$ from which we have $$a_1+a_2+\cdots+a_n\gt \sum_{i=1}^{n}(2^{a_i}+1)$$ which is impossible. So, we have to have $n=1$.

So, we have to have $x=2^{a_1}+1$ and $2^y=a_1$.

It follows that $x$ has to be a Fermat prime greater than $3$.$\quad\blacksquare$