# Is the asymptotic density of positive integers $n$ satisfying $\gcd(n, \sigma(n^2))=\gcd(n^2, \sigma(n^2))$ equal to zero?

(This post is an offshoot of this MSE question.)

Let $$\sigma(x)$$ denote the sum of divisors of $$x$$. (https://oeis.org/A000203)

QUESTION

Is the asymptotic density of positive integers $$n$$ satisfying $$\gcd(n, \sigma(n^2))=\gcd(n^2, \sigma(n^2))$$ equal to zero?

I tried searching for examples and counterexamples to the equation $$\gcd(n, \sigma(n^2))=\gcd(n^2, \sigma(n^2))$$ via Sage Cell Server, it gave me this output for the following Pari-GP script:

for(x=1, 100, if(gcd(x,sigma(x^2))==gcd(x^2,sigma(x^2)),print(x)))


All positive integers from $$1$$ to $$100$$ (except for the integer $$99$$) satisfy $$\gcd(n, \sigma(n^2))=\gcd(n^2, \sigma(n^2))$$.

Generalizing the first (counter)example of $$99$$ is trivial.

If $${3^2}\cdot{11} \parallel n$$, then $$11 \parallel \gcd(n,\sigma(n^2))$$ and $$11^2 \parallel \gcd(n^2,\sigma(n^2))$$. So the asymptotic density in question is less than $$1-\frac{2}{3^3}\cdot\frac{10}{11^2} = \frac{3247}{3267} \approx 0.993878.$$

Also, if $$3 \parallel n$$, then with probability $$1$$ there exist two distinct primes $$y$$ and $$z$$ congruent to $$1$$ modulo $$3$$ such that $$y \parallel n$$ and $$z \parallel n$$. In this case, we get $$3 \parallel \gcd(n,\sigma(n^2))$$ and $$3^2 \parallel \gcd(n^2,\sigma(n^2))$$. So the asymptotic density in question is less than $$1-\frac{2}{3^2} = \frac{7}{9} \approx 0.\overline{777}.$$

The real open problem is whether the asymptotic density is $$0$$.

I think the density does go to zero, but quite slowly. If $$p \equiv 1 \bmod 6$$ is prime then there are two solutions $$0 of $$x^2+x+1=0 \bmod p$$
If $$p\parallel n$$ then, with probability $$1,$$ there are two distinct primes $$x$$ and $$y,$$ each congruent to $$r \bmod p,$$ with $$x \parallel n$$ and $$y \parallel n.$$ ( Either or both could be congruent to $$s$$ as well.)
Then $$p \parallel \gcd(n,\sigma(n^2))$$ while $$p^2 \parallel \gcd(n^2,\sigma(n^2)).$$ So the asymptotic density for this not to happen is $$1-\frac{p-1}{p^2}<1-\frac{1}{p+2}$$
If we can argue that the chance that none of these events happen is asymptotically $$\prod(1-\frac{p-1}{p^2})$$ over the primes congruent to $$1 \bmod 6,$$ then that asymptotic density is $$0.$$
• Thank you for your answer, @AaronMeyerowitz. Assuming $$1 - \frac{p-1}{p^2} < 1 - \frac{1}{p-1}$$ is correct, then we obtain $$\frac{1}{p-1} < \frac{p-1}{p^2}$$ which implies that $$p^2 < (p-1)^2 = p^2 - 2p + 1$$ resulting in the contradiction $$p < \frac{1}{2}.$$ Hence, I am led to conclude that there must be a typo in your upper bound for $$1 - \frac{p-1}{p^2}.$$ – Arnie Bebita-Dris Aug 19 at 1:41
• OK, I fixed it. Though I hadn't used it. The idea was, roughly $1-\frac{1}{p}$ – Aaron Meyerowitz Aug 19 at 2:03