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Prove that the only perfect number of the form $2^x 3^y$ is $6$

My proof

A number $n$ is perfect if and only if:

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

Since $3 \nmid 3^{y+1}-1$ and $2\nmid 2^{x+1}-1$ we'll have that $2^{x+2} \mid 3^{y+1}-1$ and $3^y\mid 2^{x+1}-1$. This means that:

$$\frac{3^{y+1}-1}{2^{x+2}},\frac{2^{x+1}-1}{3^y} \in \Bbb{Z}$$

If $2$ numbers are integer, also their product is integer:

$$\frac{3^{y+1}2^{x+1}-3^{y+1}-2^{x+1}+1}{2^{x+2}3^y} \in \Bbb{Z}$$ $$\frac 32 - \frac{3^{y+1}+2^{x+1}-1}{2^{x+2}3^y} \in \Bbb{Z}$$ $$\frac 12 \left[3-\frac{3^{y+1}+2^{x+1}-1}{2^{x+1}3^y}\right]\in \Bbb{Z}$$

$$\left[3-\frac{3^{y+1}+2^{x+1}-1}{2^{x+1}3^y}\right]\in \Bbb{Z} \ \ (\equiv 0 \pmod{2}) $$

$$z=\frac{3^{y+1}+2^{x+1}-1}{2^{x+1}3^y} \ \ \in \Bbb{Z} \ \ (\equiv 1 \pmod{2})$$

Since the denominator grows more quickly (if $y\neq 0$) this reduces the exercise to a finite number of computations. We can also exclude the cases in which $x=0 \vee y=0$ because:

$$\sigma(2^x)=2^{x+1}-1<2^{x+1}$$ $$\sigma(3^y)=\frac{3^{y+1}-1}{2}<2\times 3^{y}$$

So our computations are reduced to:

$$(x,y)=(1,1) \Rightarrow z=1$$

With the successive couples the denominator becomes bigger than the denominator so it's useless to check. This completes the proof.

Is this correct?

Thank you for your time

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    $\begingroup$ The Euclid-Euler theorem ought to help tremendously. $\endgroup$ – Arthur May 15 at 15:12
  • $\begingroup$ @Arthur I think that the proof has to be done withouth that superpowerful theorem $\endgroup$ – Eureka May 15 at 15:14
  • $\begingroup$ It is not too difficult to show that the EVEN perfect numbers must be of the form $$2^{n-1}\cdot (2^n-1)$$ where $2^n-1$ is a Mersenne prime. If this is done, the only remaining cases are the powers of $3$ $\endgroup$ – Peter May 15 at 15:18
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    $\begingroup$ @Eureka It's not superpowerful. See the proof. It's rather elementary. $\endgroup$ – Arthur May 15 at 15:18
  • $\begingroup$ I agree Arthur which does not rule out a short and easy proof with another approach. $\endgroup$ – Peter May 15 at 15:20
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If $(2^{x+1}-1)(3^{y+1}-1)=2^{x+2}3^y$, you have equality, not just divisibility.

$3^{y+1}-1=2^{x+2}$ (because those are the only even numbers on both sides of the equation), so $2^{x+1}-1=3^y$.

Following your proof, that means:

$$\frac{(3^{y+1}-1)(2^{x+1}-1)}{2^{x+2}3^y}=\frac12 \left[3-\frac{3^{y+1}+2^{x+1}-1}{2^{x+1}3^y}\right]=1.$$

Thus,

$$\frac{3^{y+1}+2^{x+1}-1}{2^{x+1}3^y}=1 \Rightarrow 3^{y+1}+2^{x+1}=2^{x+1}3^y+1 \Rightarrow 3^{y+1}=2^{x+1}(3^y-1)+1.$$

I think it's relatively easy to see from this final equation that $x=1$ is the only possible solution, which forces $y=1$. I had a harder time seeing why your fraction $z$ couldn't have other solutions.

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Here is a relatively shorter solution to this problem:

Since $N = 2^{p-1}(2^p - 1)$ is an even perfect number whenever $2^p - 1$ is a Mersenne prime, it follows that $\omega(N)=2$ where $N$ is even perfect and $\omega(x)$ is the number of distinct prime factors of $x$.

In our case, $N = 2^x 3^y$. Since $\omega(N)=2$, then $x \geq 1$ and $y \geq 1$.

Suppose that $x > 1$. Then $6 \mid N$ and $6 < N$, which implies that $2 = I(6) < I(N)$, where $I(N) = \sigma(N)/N$ is the abundancy index of $N$ and $\sigma(N)$ is the sum of divisors of $N$. This implies that $N$ is abundant, contradicting the hypothesis that $N$ is perfect.

Similarly, if $y > 1$, then $6 \mid N$ and $6 < N$, which implies that $2 = I(6) < I(N)$. This implies that $N$ is abundant, contradicting the hypothesis that $N$ is perfect.

So it cannot happen that $x > 1$ or $y > 1$. This implies that $x = 1$ and $y = 1$, that is, $N = 6$ is the only perfect number of the form $2^x 3^y$.

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  • $\begingroup$ This argument works since every (nontrivial) multiple of a perfect number is abundant. $\endgroup$ – Jose Arnaldo Bebita-Dris Jun 12 at 8:43

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