Saturday, 30 March 2013


Today is the $90^{th}$ day of the year.

$90 = 2 \times 3^2 \times 5$

$90 = 3^2 + 9^2$
$90 = 1^2 + 5^2 + 8^2$
$90 = 4^2 + 5^2 + 7^2$

All primitive Pythagorean triples can be derived from the following:
$$a = k^2 - l^2,  b = 2kl,  c = k^2 + l^2$$
$for  k, l \in \mathbb{N}  with  k > l > 0,  (k,l) = 1  and  k \not\equiv l  mod  2$
What this is saying is that choose any old integers $k$ and $l$ such that
  • $k$ is larger than $l$;
  • $k$ and $l$ are co-prime (have no common factors other than one);
  • If $k$ is odd then $l$ is even or vice-versa.
then $a, b  and  c$  as derived above will form a primitive Pythagorean triple and all primitive Pythagorean triples are of this form.
For proof of this see
Recalling that an odd number squared is odd and an even number squared is even and that one of $k$ and $l$ is odd and the other is even then we know that $k^2 - l^2$ is either an odd minus an even or en even minus an odd. Which ever is the case the result is odd. Similarly, $k^2 + l^2$ is odd. This means that all primitive Pythagorean triples have one odd-length leg and one odd-length hypotenuse. The remaining leg, the one that is $2kl$, is, of course, even. However, we know that one of $k$ and $l$ is even also so that means that leg $b$ is a multiple of $4$.
The corollary of all this is that no number, $n$, that is even but not a multiple of $4$ i.e. where $n \equiv 2  (mod  4)$, can be a member of a primitive Pythagorean triple. which in turn means that $90 \equiv 2  (mod  4)$ is not a member of any primitive Pythagorean triple.

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