Exercise and proof given in a Number Theory textbook 
Prove that $(\forall m\in\mathbb N)(\exists$ $x,y \in \mathbb N)$, s.t. $x-y \geq m$ and $\sigma(x^2)=\sigma(y^2)$
$\sigma(x):=\displaystyle\sum_{i=1}^kd_i,\;d_i\mid x,\;\forall i\in\{1,\ldots,k\},\;k\le x$

Proof (in text book):
Let $n \in \mathbb{N}$ with $n > m$ and $(n,10)=1$ . For $x=5n$, $y=4n$, we have $x-y=n>m$ and $\sigma(x^2)=\sigma(y^2)=31 \sigma(n^2)$
I don't think this is true. For example, let $n=17$, $m=12$

General framework. It's interesting to study natural numbers $\ s<t\ $ such that $\ \sigma(s)=\sigma(t).\ $ In particular, it's a tough challenge to find natural numbers $\ a<b\ $ such that $\ \sigma(a^2)=\sigma(b^2).\ $ Indeed, this last equation is the difficult part of the given exercise.
Once you have such $\ a<b\ $ then routine considerations solve the problem. Namely, you can always find a prime number $\ p\ $ such that it divides neither $\ a\ $ nor $\ b. $ Every prime $\ p>b\ $ would do. Then we would have:
$$ \sigma(a^2\cdot p^2)\ =\ \sigma(b^2\cdot p^2) $$
In particular, $\ b^2\cdot p^2 - a^2\cdot p^2\ $ can be arbitrarily large
-- just take appropriately huge prime $p$.
Thus, the main challenge is finding solutions for
$\ \sigma(a^2)=\sigma(b^2)\ $ (where $a\ne b$). The textbook mentioned
in OP's question provided
$$ \sigma(4^2)\ =\ \sigma(5^2)\ =\ 31 $$
When you play with the perfect and multi-perfect numbers (I call them
baroque numbers) then you run into more examples.
