Entry 24

There is this curious identity$$\sqrt{2\,\Big(1-\frac{1}{3^2}\Big) \Big(1-\frac{1}{7^2}\Big)\Big(1-\frac{1}{11^2}\Big)\Big(1-\frac{1}{19^2}\Big)} = \Big(1+\frac{1}{7}\Big)\Big(1+\frac{1}{11}\Big)\Big(1+\frac{1}{19}\Big)$$ Berndt, the primary editor of Ramanujan's Notebooks, asks: if this is an isolated result, or are there others? After a quick session with Mathematica, it turns out we can use the rest of the primes within that range $$\sqrt{2\,\Big(1-\frac{1}{2^6}\Big) \Big(1-\frac{1}{5^2}\Big)\Big(1-\frac{1}{13^2}\Big)\Big(1-\frac{1}{17^2}\Big)} = \Big(1+\frac{1}{5}\Big)\Big(1+\frac{1}{13}\Big)\Big(1+\frac{1}{17}\Big)$$ It seems far-fetched that he randomly scribbled fractions with prime denominators and just so happen to observe the equality. But with the description that "Every positive integer is one of Ramanujan's personal friends" (quote by Littlewood), then that might be just how he did it.