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In Ramanujan's Notebooks, Vol IV, p.20, there is the rather curious relation for primes of form $4n-1$,

$$\sqrt{2\,\Big(1-\frac{1}{3^2}\Big) \Big(1-\frac{1}{7^2}\Big)\Big(1-\frac{1}{11^2}\Big)\Big(1-\frac{1}{19^2}\Big)} = \Big(1+\frac{1}{7}\Big)\Big(1+\frac{1}{11}\Big)\Big(1+\frac{1}{19}\Big)$$

Berndt asks: if this is an isolated result, or are there others? After some poking with Mathematica, it turns out that, together with $p= 2$, we can use the primes of form $4n+1$,

$$\sqrt{2\,\Big(1-\frac{1}{2^6}\Big) \Big(1-\frac{1}{5^2}\Big)\Big(1-\frac{1}{13^2}\Big)\Big(1-\frac{1}{17^2}\Big)} = \Big(1+\frac{1}{5}\Big)\Big(1+\frac{1}{13}\Big)\Big(1+\frac{1}{17}\Big)$$

(Now why did Ramanujan miss this $4n+1$ counterpart?) More generally, given,

$$\sqrt{m\,\Big(1-\frac{1}{n^2}\Big) \Big(1-\frac{1}{a^2}\Big)\Big(1-\frac{1}{b^2}\Big)\Big(1-\frac{1}{c^2}\Big)} = \Big(1+\frac{1}{a}\Big)\Big(1+\frac{1}{b}\Big)\Big(1+\frac{1}{c}\Big)$$

Q: Let $p =a+b+c,\;q = a b + a c + b c,\;r =abc$. For the special case $m = 2$, are there infinitely many integers $1<a<b<c$ such that, $$n =\sqrt{\frac{2(p-q+r-1)}{p-3q+r-3}}$$ and $n$ is an integer? (For general $m$, see T. Andrew's comment below.)

Note: A search with Mathematica reveals numerous solutions, even for prime $a,b,c$. It is highly suggestive there may be in fact parametric solutions.

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    $\begingroup$ General $m$: $$n=\sqrt{\frac{m(r-q+p-1)}{(m-1)r-(m+1)q+(m-1)p-(m+1)}}$$ $\endgroup$ – Thomas Andrews Mar 28 '16 at 20:25
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    $\begingroup$ This question reminds me of a proof I conjectured for Fermat's last theorem out of basic trigonometry. If you can find a parametric solution I think you'll be close to it. I think it's possible to show that every set of two or more prime numbers has a unique relationship with only one equation of the form $a^n+b^n=c^n$ and that this is always a pythagorean triangle which therefore precludes them satisfying any other $n\neq2$. The terms in your equation represent the dimensions of multiple triangles combined. It was just a conjecture that looked quite viable at the time, that this reminds me of $\endgroup$ – user334732 May 19 '16 at 18:01
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    $\begingroup$ @RobertFrost: 1. A very interesting conjecture! Even without solving FLT completely, it might form the basis of a nice elementary solution to Abel's Conjecture [about FLT], which currently has no such proof. 2. Perhaps there is a deep connection here with Bini's recurrence (also applied to FLT, c.f. Ribenboim's Fermat's Last Theorem for Amateurs), which deals with numbers of the form given here for $p,q,r$. $\endgroup$ – Kieren MacMillan Aug 23 '16 at 12:46
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    $\begingroup$ @RobertFrost: No… Abel stated in 1823 that if $x, y, z$ are nonzero relatively prime integers such that $0<x<y<z$ and $x^n+y^n =z^n\ (n>2)$, then none of $x, y, z$ are prime-powers. It's been partly proven using analytical techniques, but not completely proven even with “heavy machinery” [excluding Prof. Wiles' result, of course!]. $\endgroup$ – Kieren MacMillan Aug 23 '16 at 19:35
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    $\begingroup$ @RobertFrost: <books.google.ca/…> $\endgroup$ – Kieren MacMillan Aug 24 '16 at 11:34
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It can be transformed to next equation.

$$\displaystyle m=\frac{n^2}{n^2-1}\frac{a+1}{a-1}\frac{b+1}{b-1}\frac{c+1}{c-1}$$

so, function m decrease monotonously ,when all values are bigger than 2. If whatever m is, this equation has only finite solutions(when all are integer), $2\leq m\leq12(n=2,a=2,b=3,c=5)$.

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