Perfect numbers (prime number theory)

Question

Calculate x and y, so that $2 ^ x. 3 ^ y$ be a perfect number.

I tried to produce using this formula of the sum of the divisors of a number: Let $a ^ x, b ^ y,$ and $c ^ z$ prime factors of a number: $\frac{ a^{x+1} -1}{ a-1}. \frac{b^{y-1} -1}{b-1}. \frac{c^{c +1} -1}{c-1}$

It replaces $ 2 ^ x = a$ and $ 3^y = b $, but it can't produce much else: $ a.b $ = $ \frac {(2a-1) (3b-1)} {2} $

Answer 1

The sum of all divisors of $2^x 3^y$ is $(2^{x+1}-1)(3^{y+1}-1)/2$. If $2^x 3^y$ is perfect, it means that:

$$(2^{x+1}-1)(3^{y+1}-1)/2=2\cdot 2^x3^y=2^{x+1}3^y$$

or

$$(2^{x+1}-1)(3^{y+1}-1)=2^{x+2}3^y$$

Now, this means that both $2^{x+1}-1$ and $3^{y+1}-1$ are products of only twos and threes; however, $2^{x+1}-1$ is not divisible by $2$ and $3^{y+1}-1$ is not divisible by $3$, so it follows that all the twos in the right-hand side ($2^{x+2}3^y$) "go" into $3^{y+1}-1$ and all the threes "go" into $2^{x+1}-1$, in other words:

$$\begin{array}{c}2^{x+1}-1=3^y\\3^{y+1}-1=2^{x+2}\end{array}$$

Now if you set $a=2^x, b=3^y$, this leads to the system of equations:

$$\begin{array}{c}2a-1=b\\3b-1=4a\end{array}$$

which has a unique solution $a=2, b=3$, so the only solution is $2^x 3^y=ab=6$.

Answer 2

An even number is perfect if and only if it is of the form $2^{n-1}(2^n-1)$ for some $n$ with $2^n-1$ being prime. And $2^n-1=3^m$ is prime only for $n=2$ and $m=1$. So we obtain $2^1\cdot 3^1=6$.