# Perfect numbers of the form $2^x 3^y$ , is my proof correct?

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?

• The Euclid-Euler theorem ought to help tremendously. – Arthur May 15 at 15:12
• @Arthur I think that the proof has to be done withouth that superpowerful theorem – Eureka May 15 at 15:14
• 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 $3$ – Peter May 15 at 15:18
• @Eureka It's not superpowerful. See the proof. It's rather elementary. – Arthur May 15 at 15:18
• I agree Arthur which does not rule out a short and easy proof with another approach. – Peter May 15 at 15:20

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

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

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.

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 $$2^x 3^y$$.

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