Sum of divisors of perfect square

Question

Let $c,d$ be natural numbers of same parity (both odd or both even) and $\sigma$ be sum of divisors function. Is it known whether or under what conditions $\sigma (c^{2})$=$\sigma (d^{2})$ ? I am guessing that it can happen but unsure if certain things must hold

Answer 1

As pointed out in the comments, starting from the simple example $$\sigma (4^2)=\sigma (5^2)$$ we can generate infinitely many by multiplying by a factor prime to $10$. Thus, $$\sigma(12^2)=(1+2+2^2+2^3+2^4)\times (1+3+3^2)=31\times 13=403$$ $$\sigma(15^2)=(1+3+9)\times (1+5+25)=13\times 31=403$$

and so on.

Answer 2

It looks to me like $\sigma(627^2)=\sigma(749^2)$.

Feel free to check my work. (Or Python's work!)

And here's an even-even pair: $\sigma(740^2)=\sigma(878^2)$.