<s>[INST] Perfect numbers (prime number theory) 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} $
 [/INST] 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 $2^x 3^y=ab=6$.
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