Possible divisors of $s(2s+1)$

Question

I write $\psi(s) = s(2s+1)$ and let $d$ be the divisor function. If $s$ is prime then 4 divides $d(\psi(s))$. For example if $s=37$ then $d(\psi(s)) = d(2775) = 12$ and $4|12$. Is this trivial? I am not sure how to attack and prove this. Here is my approach:

Now I am thinking $d$ is multiplicative so we need to determine $d(2s+1)$.

case 1: If $2s+1$ is prime then $d(\psi(s))=d(s(2s+1))=d(s)*d(2s+1) = 2*2 =4$ and $4|4$.

case 2: $2s+1$ is not prime. This is where I am stuck. Now, $d(2s+1)$ is either even or odd. If it is even then $d(\psi(s))=d(s(2s+1))=d(s)*d(2s+1) = 2*2k =4k$ and $4|4k$ for some positive integer $k$. But if it's odd I am not sure of the next move. If it is odd and squarefree then $2s+1$ has an even number of divisors?

Is this the right approach or is there a better way to go about this? Also is the statement even true, is there a quick counter example? I want to go a step further and claim that $d(\psi(s))$ is equal to $4, 8, 12, 16, 20$ or $24$

Answer 1

Note that $s$ and $2s+1$ are relatively prime. Thus $d(s(2s+1))=d(s)d(2s+1)$.

Now suppose that $s$ is an odd prime. Then $d(s)$ is even.

We need to show that $d(2s+1)$ is even. Recall that $d(m)$ is odd if and only if $m$ is a perfect square.

If $s$ is odd, then it has shape $2k+1$, so $2s+1$ has shape $4k+3$. But no number of the shape $4k+3$ is a perfect square. Thus $2s+1$ cannot be a perfect square, and we are finished.

Answer 2

$d(n)$ is odd if and only if $n$ is a perfect square. (Because if $a|n$ then $(n/a)|n$ so the divisors form distinct pairs ($a$, $n/a$). These pairs are distinct so $d(n)$ is odd if and only if there is a factor such that $n/a = a$ i.e. $n = a^2$.)

If $2s + 1$ is a perfect square then $2s = a^2 - 1 = (a -1)(a+1)$ so as $s$ and $2$ are prime either:

$a-1 = 1; 2s = a+1 \implies a = 2; s = 1 1/2$ which is impossible.

$a-1 = 2; s = a+1 \implies a = 3; s = 4$ which is impossible as $s$ is prime.

$a-1 = s; 2 = a+1 \implies a = 1; s = 0$ which is impossible.

or $a-1 = 2s; 1 = a + 1 \implies a = 1; s = - 1/2$ which is impossible.

So $2s+1$ is not a perfect square and $d(2s+1)$ is even and $d(s(2s+1)) = 2d(2s+1)$ is divisible by 4.

As for the second $797161$ is prime. $2*797161 + 1 = 3^{13}$ So $d(797161*(2*797161 + 1)) = 2d(3^{13}) = 2*14 = 28$. It was the third counter example I tried.