For what odd (positive) integers $x$ is this number-theoretic equation true?
$$\gcd(x^2, \sigma(x^2)) = 2x^2 - \sigma(x^2)$$
Here, $\gcd(a, b)$ is the greatest common divisor of $a$ and $b$, and $\sigma$ is the classical sum-of-divisors function.
My attempt:
For $x = q^r$ where $q$ is (an odd) prime and $r \geq 1$, I have
$$1 = \gcd(q^{2r}, \sigma(q^{2r})) = 2{q^{2r}} - \sigma(q^{2r}) = \frac{2(q - 1){q^{2r}} - (q^{2r + 1} - 1)}{q - 1} = \frac{q^{2r + 1} - 2{q^{2r}} + 1}{q - 1}$$
WolframAlpha says that
$$\frac{q^{2r + 1} - 2{q^{2r}} + 1}{q - 1}$$
has a local minimum at $5$ (which occurs for $q = 3$ and $r = 1$). I'm not sure if this means that
$$\frac{q^{2r + 1} - 2{q^{2r}} + 1}{q - 1} \geq 5.$$
If so, then this is a contradiction. Therefore, we conclude that if the number-theoretic condition
$$\gcd(x^2, \sigma(x^2)) = 2x^2 - \sigma(x^2)$$
is true, then $x$ must be composite (i.e., $\omega(x) \geq 2$, where $\omega(x)$ is the number of distinct prime factors of $x$).
My question is as follows: Is it possible to prove that the condition
$$\gcd(x^2, \sigma(x^2)) = 2x^2 - \sigma(x^2)$$
never holds, for all positive integers $x$?
A computer search reveals that the condition $\gcd(x^2, \sigma(x^2)) = 2x^2 - \sigma(x^2)$ holds for $x=3003$: both sides equal $819$. (It almost holds for $x=13167$ as well: the two sides are negatives of each other.) This is the only solution up to $10^7$.
In general, $\gcd(a,b)=2a-b$ holds exactly when $a=kg$ and $b=(2k-1)g$ for some integers $k$ and $g$ (to see this, let $g=\gcd(a,b)$ and divide through by $g$). So you're looking for odd solutions $x$ to $$ \frac{\sigma(x^2)}{x^2} = 2 - \frac1k. $$ Equivalently, you want $$\frac{x^2}{2x^2-\sigma(x^2)}$$ to be a positive integer, that is, you want $2x^2-\sigma(x^2)$ to divide $x^2$.