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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    $\begingroup$ @Cameron's point is that those definitions are confusingly written. It would be much more readable with an equals sign for each of $p$, $q$, and $r$. They're not that expensive. $\endgroup$ Mar 28, 2016 at 19:39
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    $\begingroup$ General $m$: $$n=\sqrt{\frac{m(r-q+p-1)}{(m-1)r-(m+1)q+(m-1)p-(m+1)}}$$ $\endgroup$ Mar 28, 2016 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 $\endgroup$ May 19, 2016 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 $p,q,r$. $\endgroup$ Aug 23, 2016 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!]. $\endgroup$ Aug 23, 2016 at 19:35

2 Answers 2


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), $2\leq m\leq12(n=2,a=2,b=3,c=5)$.


I was able to find a general equation for your second relation with the $4n+1$ primes. For the first relation, discovered by Ramanujan, a general formula can be found here where the variable is $a=3$.

Here is my general solution:

$$\sqrt{\cfrac{n}{n-4}\bigg(1-\cfrac{1}{n^2}\bigg)\bigg\{1-\cfrac{1}{(n-3)^2}\bigg\}\bigg\{1-\cfrac{1}{(2n-3)^2}\bigg\}\bigg\{1-\cfrac{1}{(2n+1)^2}\bigg\}}=\bigg(1+\cfrac{1}{n-3}\bigg)\bigg(1+\cfrac{1}{2n-3}\bigg)\bigg(1+\cfrac{1}{2n+1}\bigg)$$ in which the relation you presented is obtained from $n=8$.

Often Ramanujan's formulae are derived from his incredible intuition. Likewise, I used my intuition and let $a=a_0n+a_1$, $b=b_0n+b_1$ and $c=c_0n+c_1$. I then matched these values to your relation to evaluate each variable around $n$ and solving for $m$ through some heavy algebra, lots of things cancelled out and outputted $m=n/(n-4)$.

If you want a nicer result, substitute $n\mapsto n+2$ so $m=(n+2)/(n-2)$, but then your relation would be obtained by letting $n+2=8\Rightarrow n=6$ instead. Also, in this generalisation, $n=3$ would yield prime denominators.


Found some more general solutions through trial and error:

$$\sqrt{\cfrac{3n+2}{3n-4}\bigg(1-\cfrac{1}{n^2}\bigg)\bigg\{1-\cfrac{1}{(2n+1)^2}\bigg\}\bigg\{1-\cfrac{1}{(3n+1)^2}\bigg\}\bigg\{1-\cfrac{1}{(6n-7)^2}\bigg\}}$$ $$=\bigg(1+\cfrac{1}{2n+1}\bigg)\bigg(1+\cfrac{1}{3n+1}\bigg)\bigg(1+\cfrac{1}{6n-7}\bigg)$$

Even when $m$ is a square number! $\;\style{display: inline-block; transform: rotate(90deg)}{\Rsh}$

$$\bigg(1+\cfrac 1{n-1}\bigg)\sqrt{\bigg(1-\cfrac{1}{n^2}\bigg)\bigg\{1-\cfrac{1}{(2n+1)^2}\bigg\}\bigg\{1-\cfrac{1}{(3n-1)^2}\bigg\}\bigg\{1-\cfrac{1}{(6n-5)^2}\bigg\}}$$ $$=\bigg(1+\cfrac{1}{2n+1}\bigg)\bigg(1+\cfrac{1}{3n-1}\bigg)\bigg(1+\cfrac{1}{6n-5}\bigg)$$ and $$\bigg(1+\cfrac 1{n-1}\bigg)\sqrt{\bigg(1-\cfrac{1}{n^2}\bigg)\bigg\{1-\cfrac{1}{(2n+1)^2}\bigg\}\bigg\{1-\cfrac{1}{(3n-2)^2}\bigg\}\bigg\{1-\cfrac{1}{(6n-1)^2}\bigg\}}$$ $$=\bigg(1+\cfrac{1}{2n+1}\bigg)\bigg(1+\cfrac{1}{3n-2}\bigg)\bigg(1+\cfrac{1}{6n-1}\bigg)$$

as well as an outlier.

$$\sqrt{3\bigg(3+\frac{4}{n-1}\bigg)\bigg(1-\cfrac{1}{n^2}\bigg)\bigg\{1-\cfrac{1}{(2n+1)^2}\bigg\}\bigg\{1-\cfrac{1}{(3n)^2}\bigg\}\bigg\{1-\cfrac{1}{(6n-1)^2}\bigg\}}$$ $$=\bigg(3-\frac 1n\bigg)\bigg(1+\cfrac{1}{2n+1}\bigg)\bigg(1+\cfrac{1}{3n}\bigg)\bigg(1+\cfrac{1}{6n-1}\bigg)$$

And last one, most similar to the main one in the link:

$$\sqrt{\cfrac{n+1}{n-1}\bigg(1-\cfrac{1}{n^2}\bigg)\bigg\{1-\cfrac{1}{(2n+1)^2}\bigg\}\bigg\{1-\cfrac{1}{(3n+1)^2}\bigg\}\bigg\{1-\cfrac{1}{(6n+5)^2}\bigg\}}$$ $$=\bigg(1+\cfrac{1}{2n+1}\bigg)\bigg(1+\cfrac{1}{3n+1}\bigg)\bigg(1+\cfrac{1}{6n+5}\bigg)$$

The magical factor common among all of these solutions: $1+\frac 1{2n+1}$


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