If $\sigma(N)$ is odd, $N = 2y^2$, and $y$ is not a power of two, does it follow that $\gcd(2,y) = 1$?

Question

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

It is well-known that $$\sigma(N) \equiv 1 \pmod 2 \iff \left\{\{N = x^2\} \lor \{N = 2y^2\}\right\}.$$

Here is my question:

If $\sigma(N)$ is odd, $N = 2y^2$, and $y$ is not a power of two, does it follow that $\gcd(2,y) = 1$?

I think the answer is NO, but I cannot think of a quick counterexample right away.

Answer 1

Actually, you have most of the work already there - we just need to approach the problem from the other direction.

Letting $y$ be even (and positive), we have $\gcd(2,y)=2$ and additionally $N=2y^2$ has the property that $\sigma(N) \equiv 1 \pmod 2$. The requirement that $y$ not be a power of $2$ is a bit of a red-herring.

For some examples, we can take $y=6$ and $y=10$, which correspond to $N=72$ and $N=200$ respectively.