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This question was previously posted to MSE here.

I noticed something with the totient function $ \phi(n) $ and sum of divisors function $ \sigma(n) $ when $n > 1$.

It seems than :

$ \sigma(4n^2-1) \equiv 0 \pmod{\phi(2n^2)}$ only if $ 2n - 1 $ is a prime number.

for example :

$ \sigma(4 \cdot 6^2-1) \equiv 0 \pmod{\phi(2 \cdot 6^2)} $ and $ 2 \cdot 6 - 1 = 11$ and $11$ is a prime number.

$ \sigma(4 \cdot 7^2-1) \equiv 0 \pmod{\phi(2 \cdot 7^2)} $ and $ 2 \cdot 7 - 1 = 13$ and $13$ is a prime number.

I found this sequences of primes : $``3,5,11,13,19,29,31,53,67,83,103,113,131,139,193,233,251,271,313,383,389 ..."$

I've checked until $n = 1000000$ and another MSE user checked to $n = 10^9$ and didn't find any counterexamples.

This sequence is not on OEIS.

I would like to know why some primes are here and some primes are not here. This is only an coincidence or not ? And if not, is there a way to prove it ?

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    $\begingroup$ I do not see any easy way to prove that this can only occur when $2n-1$ is prime, but I can give a partial explanation of what you are seeing. If $2n-1$ is prime, then $\sigma(2n-1)=2n$, and since $\sigma$ is multiplicative, $\sigma(4n^2-1)=\sigma(2n+1)\sigma(2n-1)=2n\sigma(2n+1)$. Notice that $\phi(2n^2)$ is always divisible by n, and if $n$ is even, then $\phi(2n^2)$ is divisible by 2n. So having $2n-1$ prime is an easy way to get a high chance of $\phi(2n^2)| \sigma(4n^2 -1)$. $\endgroup$
    – JoshuaZ
    Commented Dec 29, 2023 at 1:34
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    $\begingroup$ $n=2625=3\cdot5^3\cdot7$, with $2n-1=29\cdot181$ is very close to a counterexample, where $\phi(2n^2) = 2^43^25^57^1 \nmid \sigma(4n^2-1)=2^53^45^37^1{13}^1$, which I believe shows many approaches couldn't work. $\endgroup$
    – Daniel Weber
    Commented Dec 29, 2023 at 5:27
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    $\begingroup$ I encourage you to contribute it to the OEIS! $\endgroup$
    – Robert Israel
    Commented Dec 29, 2023 at 6:46
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    $\begingroup$ Another close counterexample is $n=13376=2^6\cdot11\cdot19$ with $2n-1=3\cdot37\cdot241$, where $\phi(2n^2)=2^{14}3^2 5^1 11^1 19^1 \nmid \sigma(4n^2-1) = 2^{14}3^3 11^2 19^1$. It seems quite hard to find any simple strengthening, which makes me feel like it's just a coincidence and it's very unlikely. $\endgroup$
    – Daniel Weber
    Commented Dec 29, 2023 at 9:35
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    $\begingroup$ Do you prefer $\phi$ to $\varphi$? (To me, $\phi$ looks too much like $\varnothing$ or $\emptyset.$) $\qquad$ $\endgroup$
    – Michael Hardy
    Commented Dec 29, 2023 at 18:48

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