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I have to prove that there exist infinite number of pairs of distinct integers $(m,n)$ such that $\sigma(m^2)=\sigma(n^2)$ ($\sigma$ is sum of divisors). I tried solving it by using the fact that $\sigma$ is multiplicative function. Then if I can find a pair $(x^2,y^2)$ such that $\sigma(x^2)=\sigma(y^2)$ , then $\sigma(a^2x^2)=\sigma(a^2y^2) \forall (a^2,x^2)=(a^2,y^2)=1$ . But I can't find such pair.

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2 Answers 2

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First we show there is one pair. Let $m=5$ and $n=4$. Then $\sigma(m^2)=31$ and $\sigma(n^2)=31$.

Now let $p$ be a prime other than $2$ or $5$. Then $\sigma(25p^2)=\sigma(25)\sigma(p^2)=31\sigma(p^2)$ and $\sigma(16p^2)=\sigma(16)\sigma(p^2)=31\sigma(p^2)$.

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I'm probably wrong, but I'm gonna give it a go.
Suppose there exist only $n$ finitely many pairs of distinct integers $(m, n)$ such that $\sigma(m^2) = \sigma(n^2)$.
One such pair is $(5, 6)$ because $1 \times 5 = 5$ and $1 \times 2 \times 3 = 6$, and $5 = 2 + 3$.
Out of all the pairs, select the pair that contains the largest integer, call it $(a, b)$, where without loss of generality we can assume that $a$ is the largest integer. Note that $a \ge 6$.
Observe also that $a^2$ and $b^2$ have exactly the same prime divisors as $a$ and $b$ respectively, so it is also true that $\sigma((a^2)^2) = \sigma((b^2)^2)$.
Therefore the pair $(a^2, b^2)$ is valid.

But $a^2 > a$, which means $a$ is not the largest integer in our collection of $n$ pairs, giving a contradiction.
Therefore there are infinitely many pairs of distinct integers $(m, n)$ such that $\sigma(m^2) = \sigma(n^2)$.

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  • $\begingroup$ $\sigma(5)=6\neq\sigma(6)=12=1+2+3+6$ $\endgroup$
    – zed111
    Mar 15, 2015 at 6:48
  • $\begingroup$ Hmm, I was assuming you were referring to prime divisors. $\endgroup$
    – Vizuna
    Mar 15, 2015 at 6:49

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