A Beautiful Relationship

The Indian Mathematics Genius "Shri Srinivasa Ramanujan" found a relationship in a few primes.

Relationship of Ramanujan: \[\frac{1}{4}+2=(1\frac{1}{2})^2\] \[\frac{1}{4}+2*3=(2\frac{1}{2})^2\] \[\frac{1}{4}+2*3*5=(5\frac{1}{2})^2\] \[\frac{1}{4}+2*3*5*7=(14\frac{1}{2})^2)\] \[\frac{1}{4}+2*3*5*7*11*13*17=(714\frac{1}{2})^2\] Before going to my relationship, I define that a nth Pythagorean Triplet is: \[(2n+1,2(n(n+1)),2(n(n+1))+1)\]

My Relationship: $\frac{1}{2}+2=\frac{5}{2}$ $\frac{1}{2}+2*3=\frac{13}{2}$ $\frac{1}{2}+2*3*5=\frac{61}{2}$ $\frac{1}{2}+2*3*5*7=\frac{421}{2}$ $\frac{1}{2}+2*3*5*7*11*13*17=\frac{1021021}{2}$

Here is the relation:

In the first line, the numerator 5, is the hypotenuse of the 1st Pythagorean Triplet

In the second line, the numerator 13, is the hypotenuse of the 2nd Pythagorean Triplet

In the third line, the numerator 61, is the hypotenuse of the 5th Pythagorean Triplet

In the fourth line, the numerator 421, is the hypotenuse of the 14th Pythagorean Triplet

In the fifth line, the numerator 1021021, is the hypotenuse of the 714th Pythagorean Triplet

Now, taking Ramanujan's relationship,

In the first line, the whole number in the mixed fraction is 1

In the second line, the whole number in the mixed fraction is 2

In the third line, the whole number in the mixed fraction is 5

In the fourth line, the whole number in the mixed fraction is 14

In the fifth line, the whole number in the mixed fraction is 714

Finally, checking mine and Ramanujan's corresponding lines, you can find out the relationship!!!

If not, please feel free to ask in the comments!!!

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Comments

Hi Mohammed. This relationship is indeed fascinating! Just a few thoughts:

First, these Pythagorean triples appear to be what are called almost-isosceles Pythagorean triples. They can almost be generated using the expressions you gave. However, I believe you meant to say $2n + 1$ instead of $2n - 1$ . Other than that, it is correct.

Also I think you meant to say "the 5th Pythagorean Triplet" instead of the "the 3rd Pythagorean Triplet". The 3rd triplet doesn't produce a hypotenuse of 61.

Ramanujan's relationship can be expressed as follows:

$\frac{1}{4} + p_n = {(n + \frac{1}{2})}^2$

Where $p_n$ is the product of prime numbers which is used to get the expression on the right, ${(n + \frac{1}{2})}^2$ . I have labeled it as such because we do not yet know how the product and the expression are related. We just know that they go together.

Your relationship can be expressed as:

$\frac{1}{2} + p_n = \frac{H_n}{2}$ ,

Where $H_n$ is the hypotenuse of the $n$ th almost-isosceles Pythagorean triple.

As it turns out, this is because we can derive the second equation from the first:

$\frac{1}{4} + p_n = n^2 +n+\frac{1}{4}$

$\frac{1}{4} + p_n = n(n+1)+\frac{1}{4}$

$\frac{1}{2} + p_n = n(n+1)+\frac{1}{2}$

$\frac{1}{2} + p_n = \frac{2n(n+1)+1}{2}$

$\frac{1}{2} + p_n = \frac{H_n}{2}$

Very neat that products of prime numbers relate to Pythagorean triples (specifically their hypotenuses) in this way!

I would be interested in knowing the relationship between those products of prime numbers and the numbers Ramanujan came up with for the other side of the equation (the mixed number which was squared). For instance, why did those products work? Would other products work? I noticed they were increasing sequentially by factors of prime numbers (adding on the next prime each time) until the last one. Are we able to continue the sequence, and if not, why?

Subject for research in my spare time perhaps! :)

David Stiff - 1 year, 2 months ago

Hi David Stiff,

Thank you very much for commenting me. the relation in the first line for Ramanujan, he got 1 as whole number in the mixed fraction whereas I got 1st Pythagorean triplet similarly, in the second line for Ramanujan, he got 2 as whole number and I got 2nd Pythagorean triplet. Similarly, it happens for all mine and Ramanujan's corresponding lines.

Thank you for finding my typo.

Bye

Mohammed Imran - 1 year, 2 months ago

Superb David Stiff, I was expecting someone to find the hidden truth behind this fascination. Also, if $p_{n}$ can be expressed in the form $x(x+1)$ , we can include that as well!!!

Thanks Mohammed!

You're welcome!!!

Check out for my new Notes on brilliant. They are "A new proof for 1+2+..+n=n(n+1)/2"and "A question but not a doubt 1"

Wow. I didn't know this.. great job!

Adhiraj Dutta - 1 year, 1 month ago

thank you!

Mohammed Imran - 1 year, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$