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I believe that this first claim will be well known in the literature about amicable numbers because it is easy to deduce

Claim 1. The primes $p,q$ and $r$ in Thābit ibn Qurra theorem satisfy that have the form, respectively, $p$, $2p+1$ and $2p^2+4p+1$.

There are simple consequences of this Claim 1 (for example one can write each of one of the amicable numbers (see the form in previous Wikipedia article, same section) $N:=2^npq$ and $M:=2^n r$ in a amicable pair in terms only of $p$ and $2^n$, or state a closed-form also in terms of $p$ and $n$ for the difference $M-N$) and I tried combine previous claim with Wilson-Lagrange theorem to show the following

Claim 2. Let $p$ the prime in Thābit ibn Qurra theorem, then Wilson-Lagrange theorem implies that there exist postive integers $c,c'$ and $c''$ satisfying

$$c'(2p+1)-1=2p(2p-1)(2p-2)\cdots(p+1)p(cp-1)$$ $$\quad\quad\quad c''(2p^2+4p+1)-1=(2p^2+4p)(2p^2+4p-1)\cdots(2p+1)(c'(2p+1)-1)$$

Question. I would like to know if my claims were rights, especially (only is required a proof of) Claim 2. As remark, to deduce Claim 2, I've combined with Claim 1 and three applications of Wilson-Lagrange theorem. Then, can you provide me your calculations to do a comparison with mine? Many thanks.

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    If in comments you want criticize my approach (the form of the primes in the algorithm+Wilson-Lagrange theorem), improve it or state new claims as a companion of your answer feel free to do it. – 2017-02-22

1 Answers 1

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The claims are right and easy to check ...

Claim 1 $$p = 3\cdot2^{n − 1} − 1\\ q = 3\cdot2^n − 1=2\cdot(3\cdot2^{n-1}-1)+1=2\cdot p+1\\ r = 9\cdot2^{2n − 1} − 1=2\cdot9\cdot2^{2n − 2} − 1=2\cdot (3\cdot2^{n-1})^2-1=2\cdot (p+1)^2-1$$

Claim 2 from Wilson's theorem $$(p-1)!\equiv -1 \pmod{p} \Leftrightarrow \color{#C00}{(p-1)!=c\cdot p-1} \\ (2p)!\equiv -1 \pmod{2p+1} \Leftrightarrow \color{#C0C}{(2p)!=c'\cdot(2p+1)-1}\\ (2p^2+4p)!\equiv -1 \pmod{2p^2+4p+1}$$ or $$(2p)(2p-1)...(p+1)p\color{#C00}{(p-1)!}\equiv -1 \pmod{2p+1}\\ (2p^2+4p)(2p^2+4p-1)...(2p+2)(2p+1)\color{#C0C}{(2p)!}\equiv -1 \pmod{2p^2+4p+1}$$ or $$(2p)...p\color{#C00}{(c\cdot p-1)}\equiv -1 \pmod{2p+1}\\ (2p^2+4p)...(2p+1)\color{#C0C}{(c'\cdot(2p+1)-1)}\equiv -1 \pmod{2p^2+4p+1} $$

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    Many thanks then perfect. Your calculations seems much more clean and understandable than mine. Many thanks also for your effort to do your proof nice. – 2017-02-23