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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

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  • $\begingroup$ What do you mean by a perfect power? Any $n^k$ where $k > 1$? $\endgroup$ – Todd Trimble May 2 '18 at 17:56
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    $\begingroup$ 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. $\endgroup$ – Joe Silverman May 2 '18 at 18:42
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    $\begingroup$ 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. $\endgroup$ – Gerhard Paseman May 2 '18 at 18:49
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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.

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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.

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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!

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