New Primitive Pythagorean triplets via Triangular numbers.

This a Pythagorean identity which gives every primitive triplets of the form $a^2$ + $b^2$ $=$ $c^2$ such that g.c.f.(a,b,c)=1 and c-a=1.

$(\,4T)\,^2$ + $(\,8T+1)\,$ $=$ $(\,4T+1)\,^2$

The proof is very, very easy.At first observe that it follows the identity $(\,a+b)\,^2$ $=$ $a^2$ + $2ab$ + $b^2$ ,but clever part is the $(\,8T+1)\,$ because it is an odd square itself.So if you want to check it out pick any Triangular number i.e. nos. of the form $\frac{n(n+1)}{2}$ for n being any natural number. (Here $T$ denotes any Triangular number.)

#NumberTheory

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`2 \times 3`	$2 \times 3$
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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}`	$\boxed{123}$