I was looking at perfect numbers and came across something that might serve a little interesting.

Denote by $P_n$ the $n^\text{th}$ perfect number, then there appears to always exist $k\in\mathbb{W}$ and $l\in\mathbb{Z}$ such that


for which $\mathbb{W}:= \mathbb{N}\cup\big\{0\big\}=\big\{0,1,2,3,\ldots\bigr\}$; id est, the whole numbers.

Here is what I have discovered thus far, given that $(6,28,496)=(P_1,P_2,P_3)$ respectively:

$$\begin{align}6^2+5^2+2^6&=5^2+10^2 \\ 28^2+5^2+2^0&=27^2+9^2 \tag*{$\begin{pmatrix}(k,l)\,\,\,=&(6,10)&\text{resp.}\\ (k,l)\,\,\,=&(0,9)&\text{resp.} \\ (k,l)\,\,\,=&(3,32)&\text{resp.}\end{pmatrix}$}\\ 496^2+5^2+2^3&=495^2+32^2.\end{align}$$ This is as far as I have tested. I know that for every perfect number $P_n$, there exists a prime number $p$ such that $P_n=2^{p-1}(2^p-1)$ thus far (or at least, for every even perfect number), but this does not help me prove the conjecture. It is also conjectured that every perfect number is the sum of three cubes, but supposing this is valid does not help me prove this conjecture either.

And, writing the equation like the following does not help as well. $$2^k+(P_n+l)(P_n-l)=(P_n+4)(P_n+6).$$

Can it be verified whether or not my conjecture holds water?

Thank you in advance.

  • 2
    $\begingroup$ It is unknown whether each perfect number is of the form $2^{p-1}(2^p-1)$. $\endgroup$ – Hagen von Eitzen May 26 '18 at 14:08
  • $\begingroup$ @HagenvonEitzen well, you are right actually. If it was known, then we wouldn't be looking for odd perfect numbers. I will add that in. Thanks for the note :) $\endgroup$ – Mr Pie May 26 '18 at 14:10
  • $\begingroup$ Wait, $P_n-1$ is not a typo for $P_{n-1}$? $\endgroup$ – Hagen von Eitzen May 26 '18 at 14:10
  • $\begingroup$ @HagenvonEitzen no it is not a typo. Is that a common typo going around? $\endgroup$ – Mr Pie May 26 '18 at 14:10

You were unlcucky to stop testing where you did.

Note that $$ P^2+5^2+2^k=(P-1)^2 +l^2$$ is equivalent to $$\tag1P^2-(P-1)^2=2P-1 =l^2-5^2-2^k $$ Now let $P$ be an even perfect number, $P=2^{p-1}(2^p-1)$. After handling small cases manually, we may assume $p>5$.

If $k=0$, equation $(1)$ becomes $$ 2^{p}(2^p-1)=(l+5)(l-5)$$ where $l$ must be odd and exactly one of the factors on the right is a multiple of $4$. So $l\pm5 = 2^{p-1}u$ and $l\mp5=2v$ with $uv=2^p-1$. As the difference of the factors is $\pm10$, we conclude $2^{p-2}u-v=\pm5$. If $u=1$, this leads to $$\pm5=2^{p-2}-(2^p-1)=1-3\cdot 2^{p-2}<-5\qquad\text{contradiction}.$$ And $u\ge 3$ leads to $$\pm5 \ge 3\cdot2^{p-2}-\frac 13(2^p-1)>(3-\tfrac43)2^{p-2}\ge \frac 53\cdot 32\qquad\text{contradiction}.$$ We conclude that $k\ge 1$. Then $(1)$ tells us that $l$ must be even. Then $2^k=l^2-24-2^p(2^p-1)$ is a multiple of $4$, hence $k\ge 2$. Then $$\tag22^{k-2}=(l/2)^2-6-2^{p-2}(2^p-1).$$ If $l/2$ is odd, the right hand side of $(2)$ is $\equiv 3\pmod 8$, whereas the left side is $\equiv 1$, $2$, or $0\pmod 4$. Hence $l/2$ is even, so that the right hand side of $(2)$ is $\equiv 2\pmod 4$. We conclude $k=3$ and $l=4m$ for some $m$. So now $(2)$ becomes $$ 2^{p-4}(2^p-1)+2=m^2.$$ For $p>5$, the left is still even, but not a multiple of $4$, hence cannot be a perfect square.

  • $\begingroup$ Ah... I did not subtract both sides by $(P_n-1)^2$.... oh and by the way, $P$ is not odd if $P=2^{p-1}(2^p-1)$. The contradictions, though, are very nice, and your explanation was thorough. Of course if $m^2$ is even, then it must be a multiple of $2^2=4$... but it is not! Well done!! Unfortunately, I have reached my daily voting limit and have to wait $7$ hours before I can upvote again. Nonetheless, congratulations! $$\color{green}{\checkmark}$$ $\endgroup$ – Mr Pie May 26 '18 at 16:02
  • 1
    $\begingroup$ @user477343 Ooops, odd perfect number of course. I doubt I will ever write any thing non-trivial beginning with "Let $P$ be an even perfect number ... " $\endgroup$ – Hagen von Eitzen May 26 '18 at 16:04
  • $\begingroup$ hahahah, no problem. It was a great answer! :) $\endgroup$ – Mr Pie May 26 '18 at 16:06

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