<s>[INST] Show that no number of the form $2^a3^b$ is $3$-perfect. Show that no number of the form $2^a3^b$ is $3$-perfect.
A number is $3$-perfect if $\sigma(n)=3n$. I assumed that there exists a number $n=2^a3^b$ such that $\sigma(n)=\sigma(2^a3^b)=\sigma(2^a)\sigma(3^b)=3\cdot2^a3^b=3n$. Then I had $(1+2+\cdots +2^a)(1+3+\cdots+3^b)=2^a3^{b+1}$ and so $(2^{a+1}-1)({1\over 2}(3^{b+1}-1))=2^a3^{b+1}$. and through some algebra on the LHS I convinced myself that these two things are not equal to each other, contradiction.
I'm confident there is a better and more elementary way to do this problem. Any hints or solutions are greatly appreciated.
 [/INST] Multiplying both sides of $(2^{a+1}-1)(\frac{3^{b+1}-1}{2})=2^a3^{b+1}$ by $2$, we get $(2^{a+1} - 1)(3^{b+1} - 1) = 2^{a+1} 3^{b+1}$.
But this is impossible, because $2^{a+1}-1 < 2^{a+1}$ and $3^{b+1}-1 < 3^{b+1}$.
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