# On odd perfect numbers and a GCD - Part III

Let $$m = q^k n^2$$ be an odd perfect number with special prime $$q$$ satisfying $$q \equiv k \equiv 1 \pmod 4$$ and $$\gcd(q,n)=1$$.

It is known that $$\gcd(\sigma(q^k),\sigma(n^2)) = \frac{(\gcd(n,\sigma(n^2)))^2}{\gcd(n^2,\sigma(n^2))}$$ and therefore that $$\gcd(\sigma(q^k),\sigma(n^2)) = \gcd(n^2,\sigma(n^2)) = \frac{\sigma(n^2)}{q^k} \geq 3$$ if and only if $$\gcd(n,\sigma(n^2))=\gcd(n^2,\sigma(n^2))$$.

Lastly, it is known that $$\gcd(\sigma(q^k),\sigma(n^2))=1$$ implies $$k=1$$.

Here is my:

QUESTION: Under what conditions is it true that $$\gcd(n,\sigma(n^2))=\gcd(n^2,\sigma(n^2))?$$

I know that the GCD equation is true, for example, when $$\sigma(n^2) = q^k n$$ (and therefore, $$\sigma(q^k) = 2n$$). Are there other conditions under which the GCD equation is true?

• Care to explain the downvote? Commented Jul 27, 2022 at 3:37

Let $$p^s Q^2$$ be an odd perfect number with special prime $$p$$ satisfying $$p \equiv s \equiv 1 \pmod 4$$ and $$\gcd(p,Q)=1$$.

I did some more digging on when the equations

$$\gcd(Q^2, \sigma(Q^2)) = \gcd(\sigma(Q^2), \sigma(p^s))$$ $$\gcd(Q, \sigma(Q^2)) = \gcd(Q^2, \sigma(Q^2))$$ $$\gcd(\sigma(Q^2), \sigma(p^s)) = \gcd(Q, \sigma(Q^2))$$

simultaneously hold. Note that we have the identity

$$\gcd(\sigma(Q^2), \sigma(p^s)) \gcd(Q^2, \sigma(Q^2)) = \left(\gcd(Q, \sigma(Q^2))\right)^2.$$

Hence, when exactly one of the three equations above holds, then the other two equations follow.

In particular, note that $$\gcd(Q^2, \sigma(Q^2)) = \gcd(\sigma(Q^2), \sigma(p^s))$$

is equivalent to $$\frac{Q^2}{\sigma(p^s)/2} = \frac{\left(\gcd(\sigma(p^s)/2, Q)\right)^2}{\sigma(p^s)/2}$$

which, in turn, is equivalent to

$$Q = \gcd(\sigma(p^s)/2, Q).$$

This last GCD equation holds if and only if $$Q \mid \sigma(p^s)/2$$.

Furthermore, in particular, note that $$\gcd(Q, \sigma(Q^2)) = \gcd(Q^2, \sigma(Q^2))$$

is equivalent to

$$\left(\frac{Q}{\sigma(p^s)/2)}\right)\cdot\gcd(\sigma(p^s)/2, Q) = \frac{Q^2}{\sigma(p^s)/2}$$

which, in turn, is equivalent to

$$\gcd(\sigma(p^s)/2, Q) = Q.$$

This last GCD equation holds if and only if $$Q \mid \sigma(p^s)/2$$.

Lastly, in particular, note that $$\gcd(\sigma(Q^2), \sigma(p^s)) = \gcd(Q, \sigma(Q^2))$$

is equivalent to $$\frac{\left(\gcd(\sigma(p^s)/2, Q)\right)^2}{\sigma(p^s)/2} = \left(\frac{Q}{\sigma(p^s)/2}\right)\cdot\gcd(\sigma(p^s)/2, Q)$$

which, in turn, is equivalent to

$$\gcd(\sigma(p^s)/2, Q) = Q.$$

This last GCD equation holds if and only if $$Q \mid \sigma(p^s)/2$$.

Thus, if we set $$G = \gcd(\sigma(Q^2), \sigma(p^s))$$ $$H = \gcd(Q^2, \sigma(Q^2))$$ $$I = \gcd(Q, \sigma(Q^2))$$

then we get the biconditional

$$G = H = I \iff Q \mid \sigma(p^s)/2.$$

Of course, as a sanity check, when $$\sigma(p^s) = 2Q$$, then we obtain the conjunction

$$Q \mid \sigma(p^s)/2$$

and

$$\sigma(p^s)/2 \mid Q,$$

which by Conjunction Elimination yields

$$Q \mid \sigma(p^s)/2$$

and hence, that

$$G = H = I.$$

• A suggestion: the huge parentheses might be easier to read with \big rather than \Bigg; for example, $\dfrac{\gcd\bigl(n\sigma(n^2), i(q)\bigr)}{i(q)}$ \dfrac{\gcd\bigl(n\sigma(n^2), i(q)\bigr)}{i(q)} rather than $\dfrac{\gcd\Biggl(n\sigma(n^2), i(q)\Biggr)}{i(q)}$ \dfrac{\gcd\Biggl(n\sigma(n^2), i(q)\Biggr)}{i(q)}. Commented Jul 27, 2022 at 0:35
• Thank you for your suggestion, @LSpice! Incorporating these changes in the post now. Commented Jul 27, 2022 at 0:48