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While I was trying to create a problem using the sum of divisors function $$\sigma(n)=\sum_{d\mid n}d$$ and the Euler's totient function $\varphi(n)$ from this pattern $$\sigma(\text{a positive integer})-\sigma(\text{a different integer})\cdot\varphi(\text{a different integer})=\text{something},$$ I wondered next question.

Question. Are there infinitely many different primes solving the RHS as $$\sigma(n^2)-\varphi(n)\sigma(n)=\text{a prime number}\tag{1}$$ when $n$ runs over the positive integers $\geq 1$?

Example 1. For $n=4$ one has that $$\sigma(16)-\varphi(4)\cdot\sigma(4)=31-2\cdot7=17=\text{ a prime number solving the RHS of our equation}.$$

Example 2. See with Wolfram Alpha online calculator codes for different upper limits

table IsPrime(sigma(n^2)-sigma(n)*phi(n)), for n=1 to 5000

table sigma(n^2)-sigma(n)*phi(n), for n=1 to 1000

Thus our sequence of prime numbers solving the RHS for some integer $n$ of our condition $(1)$ starts (as I said we ommit repetead primes) as $$5, 7, 11, 13, 17, 19\ldots$$


I don't know if this question was in the literature I tried find strings like as sigma(n^2)-sigma(n) or sigma(n^2)-sigma(n) in The On-Line Encyclopedia of Integer Sequences. Of course, if you can identify our sequence and if previous Question was in the literature refers it, and I try find the information.

Then, from this information what are you saying? Many thanks.

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  • $\begingroup$ One can to study for the cases (maybe this can be rule out easily) $\gcd(p,n)=1$, where $p$ is $\text{ the prime number}$ solving $(1)$. Multiplying $(1)$ by this prime, one has $p\sigma(n^2)-p\varphi(n)\sigma(n)=(\sigma(n^2)-\varphi(n)\sigma(n))^2$ (or if you prefer multiply by a prime power to improve my approach). Then the strategy is rewrite in LHS terms as (the first term of LHS can be written as) $(p+1-1)\sigma(n^2)-\sigma(pn^2)$, and try to apply knowledges about the size of those arithmetic functions (for example one has for $n>1$ that $\sigma(n^2)\geq 1+(\sigma(n)-1)+n^2$). $\endgroup$ – user243301 Oct 10 '17 at 10:12
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Edit: Mistake spotted, the answer was changed.

Let us look at the case $n=p^k$. We have $$\sigma(p^{2k})-\phi(p^k)\sigma(p^k)=1+p+p^2+..+p^{2k}-(p^{k-1})(p-1)(1+p+...+p^{k})\\ =1+p+p^2+..+p^{2k}-(p^{k-1})(p^{k+1}-1)\\ =1+p+p^2+...+p^{k-1}+p^{k+1}+p^{2k-1}=\frac{p^{2k}-1}{p-1}-p^k$$

In general it is hard to decide if there are infinitely many primes of a certain form. If $k=1$, your question becomes equivalent to the twin prime conjecture.

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  • $\begingroup$ Many thanks for your reasoning. $\endgroup$ – user243301 Sep 25 '17 at 14:33

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