the product of an odd perfect number and some even perfect number is perfect If $a$ were an odd perfect number ,does there exist an even perfect number $b$ such that $ab$ is a perfect number?
 A: 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.
A: 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 $a>1$.
A: 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?
A: 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.
