I noticed something interesting 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 I 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 ?
This question is certainly difficult to solve.
Let $x=2n$. As @Lucid has stated in the comments, it is easy to show that $x\mid\phi\left(\frac{x^2}{2}\right)$, so by transitivity $\phi\left(\frac{x^2}{2}\right) \mid \sigma(x-1)\sigma(x+1)$. As $\sigma(x-1)=x$ if $x-1$ is prime, to validate the OP conjecture it would suffice to show that $x \nmid \sigma(x-1)\sigma(x+1)$ unless $\sigma(x-1)=x$. But this task seems quite difficult.