PS: The equations are rendered using MathJax library. Please allow JS if you can't read equations properly.

While lying semi-awake in the morning, reading the first part of Bill Habbard's "

**Message Contains No Recognisable Symbols**", the thoughts drifted to how in a right-angled triangle we have:

$$ a^2 + b^2 = c^2 $$

That led me to:

$$\begin{align}

a^2 &= c^2 - b^2 \\

&= (c + b) \cdot (c - b)

\end{align}$$

Now, we have triplets like $(3, 4, 5)$; $(5, 12, 13)$ etc. where 2 of the values are consecutive. What if we fix $ c - b = 1 $; that is, the hypotenuse and one of the bases. We get to:

$$\begin{align}

a^2 &= (c + b) \cdot 1 \\

&= ((b + 1) + b) \\

&= 2 \cdot b + 1

\end{align}$$

So, if I get any odd perfect square, I can always get its corresponding values to form a

*co-prime Pythagorean triplet*(or a

**primitive triplet**). To test it out, here is a list of first 10 such triplets:

$$

(3, 4, 5) \\

(5, 12, 13) \\

(7, 24, 25) \\

(9, 40, 41) \\

(11, 60, 61) \\

(13, 84, 85) \\

(15, 112, 113) \\

(17, 144, 145) \\

(19, 180, 181) \\

(21, 220, 221)

$$

Another interesting thing about these triplets is, the other two sides are the medians for our odd-square values. For eg. first odd-square is $9$. The other two sides will be around $\dfrac{9}{2} = 4.50$, which gives the other 2 sides: $4$ and $5$.

From another such triplet $ (x, y, z) $, you can arrive at the immediate next one as follows: $$ \left(x + 2, y + 2 \cdot (x + 1), z + 2 \cdot (x + 1) \right) $$

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