# Perfect numbers and perfect powers

This was asked earlier at MSE.

The observation that 28 = 27 + 1 shows that it is possible to have consecutive perfect numbers and perfect powers. However, this must be extremely rare. Is it unique?

Questions: (1) Is there another example known of such a consecutive pair?

(2) Is there a rigorous or heuristic proof showing that there are only finitely many such pairs?

Thanks

• What do you mean by a perfect power? Any $n^k$ where $k > 1$? May 2, 2018 at 17:56
• For (1), the list of known perfect numbers is quite short. Have you tried taking each of them, subtracting 1, and checking if the result is a perfect power? If not, how many of them have you tried, so that people will know which ones remain to be checked. May 2, 2018 at 18:42
• There is 5,6 and 7. This question might have a reasonable combinatorial proof, in that even perfect numbers are limited in their expression and odd perfect numbers may have too many factors to be close to a power. Start an answer with a partial proof, and see where others take it. Gerhard "Faint Heart Ne'er Won Argument" Paseman, 2018.05.02. May 2, 2018 at 18:49

Using Joe Silvermann’s notations, and under the assumption that any perfect number is even, we are trying to find the integer solutions of $$2^{p-1}(2^p-1)=x^m+1,$$ where $x \geq 2, m \geq 2, p\geq 2$ and $2^p-1$ (hence $p$) is prime. We know that $m$ is odd, from Joe Silvermann’s answer as well.

We infer that $x$ is odd, and therefore $$1 < \frac{x^m+1}{x+1}=x^{m-1}-x^{m-2}...-x+1$$ is odd. Since $2^p-1$ is prime, then $x+1$ must be equal to $2^{p-1}$, thus $$x^3+1 \leq x^m+1 =2^{p-1}(2^p-1) \leq 4*(2^{p-1})^2=4(x+1)^2.$$

This fails when $x > 5$, and since $x+1$ is a power of $2$, $x=3$, hence $3^m+1 \leq 4(3+1)^2=64$.

This gives only 27 and 28 as the answers, when you consider only even perfect numbers.

If you accept the conjecture that there are no odd perfect numbers, then the finiteness follows quite easily from the $ABC$ conjecture. Thus an example has the form $$2^{p-1}(2^p-1) = x^m + 1.$$ First note that we can't have $m=2$, since otherwise $x^2\equiv-1\pmod{4}$. So we may assume that $m\ge3$.

The $ABC$ conjecture says that $$\max\bigl\{ x^m, 2^{p-1}(2^p-1) \bigr\} \le C_\epsilon\bigl(2x(2^p-1)\bigr)^{1+\epsilon}.$$ Replacing $2^p-1$ with $2^p$ (I'll let you make that rigorous) yields (with a new constant) $$\max\{x^m,4^p\} \le C_\epsilon (x2^p)^{1+\epsilon}.$$ I am going to take $\epsilon=1/6$, so $$\max\{x^m,4^p\} \le C_\epsilon (x2^p)^{7/6}.$$ From this we find that $$(x^m)^5\cdot(4^p)^7 \le (C(x2^p)^{7/6})^{12} = C x^{14}\cdot 2^{14p}.$$ Hence $$x^{5m-14} \le C,$$ which shows that $x$ is bounded (note $m\ge3$), and then $p$ is also bounded.

Actually a stronger result (for cube powers) is proved here:

Luis H. Gallardo, "On a remark of Makowski about perfect numbers"
Elem. Math. 65 (2010) 121 – 126

The only even perfect number that is also a sum of two cubes is 28.

The paper is easy (and free) to read. I believe you will enjoy it!