If $b^2 \equiv 1 \pmod 3$, is it possible to have $\sigma(b^2) \equiv b^2 \pmod 3$?

Question

The title says it all.

Let $\sigma(N)$ denote the sum of the divisors of the positive integer $N$.

To paraphrase my question:

If $3 \mid \left(b^2 - 1\right)$, is it possible to have $3 \mid \left(\sigma(b^2) - b^2\right)$?

As a follow-up question:

If $X = \nu_{3}(b^2 - 1)$ and $Y = \nu_{3}(\sigma(b^2) - b^2)$, which of the following is true? $$X \leq Y$$ or $$Y \leq X$$

Update: It turns out that, yes it is possible, and indeed many integers satisfy the conditions. I am hoping somebody has some bright ideas on how to approach the second part of the problem.

Answer 1

Not a complete answer. Let's take primes $b>3$, due to Fermat's little theorem, $b^2 \equiv 1 \pmod{3}$. Then $\sigma(b^2)= 1 + b + b^2$ and $\sigma(b^2) - b^2 = b+1$. All the primes $b>3$ are either of $b=6q+1$ or $b=6q+5$ form. Obviously for those of form $b=6q+5$ we have $\sigma(b^2) - b^2 = b+1=6q+6$ which is divisible by 3. This is not true for the primes of $b=6q+1$ form. E.g. $b=7$ or $b=13$.

Answer 2

Since $3 \mid (b^2 - 1)$ is equivalent to $3 \nmid b$ (as noted in the comments), then it appears (from a few quick computations for $b < 100$) that, for $3 \nmid b$:

(a) If $3 \mid \left(\sigma(b^2) - b^2\right)$, then we have $$\nu_{3}(b^2 - 1) = X = Y = \nu_{3}(\sigma(b^2) - b^2).$$ (b) If $3 \nmid \left(\sigma(b^2) - b^2\right)$, then we have $$\nu_{3}(b^2 - 1) = X > Y = \nu_{3}(\sigma(b^2) - b^2).$$

So, it appears plausible that, in general, we have $$\nu_{3}(b^2 - 1) = X \geq Y = \nu_{3}(\sigma(b^2) - b^2),$$ but I have no proof.

Added June 2 2016 The difficulty in pursuing a proof for $Y \leq X$ lies in the obstruction of trying to rule out $0 < X < Y$.