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2017-10-09

Pythagorean triplets

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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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