SUNDAY, 31 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 www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/pythagtriple.pdf.
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.