the product of an odd perfect number and some even perfect number is perfect

Question

If $a$ were an odd perfect number ,does there exist an even perfect number $b$ such that $ab$ is a perfect number?

Answer 1

It is known that the EVEN perfect numbers all have the form

$$2^{n-1}(2^n-1)$$, where $2^n-1$ is a Mersenne-prime.

So, the answer to your question is no.

Answer 2

No.

Suppose $a,b$ are perfect, and $ b$ is even. If $a$ and $b$ are coprime, then $$\sigma(ab)=\sigma (a)\sigma(b)=(2a)(2b)\ne 2ab$$ So $ab$ is not perfect. Hence, we can assume that $ a $ and $ b$ have a common factor. Since $b$ must be of the form $$2^{n-1}(2^n-1)$$ where $2^n-1$ is prime, and since $a$ is odd, the common factor must be $2^n-1$. So $$ab=2^{n-1}(2^n-1)^2c$$ where $c$ is odd. This is clearly not of the correct form for an even perfect number.

Answer 3

All even perfect numbers are of the form $b=2^{k-1}\left(2^k-1\right)$ (with $2^k-1$ prime, but not needed here.)

Then in $ab=2^{k-1}\left(\left(2^k-1\right)a\right)$ we have $\left(\left(2^k-1\right)a\right)$ must be of the form $2^k-1$, impossible, since $a>1$.

Answer 4

Hint: In general, the product of two perfect numbers (one odd, the other even) is not perfect, since every nontrivial multiple of a perfect number is abundant.

Can you take it from here?