<s>[INST] Why there are exactly 100 distinct (not necessarily primitive) Pythagorean triples $(a,b,c)$ with $c<100$? Why there are exactly 100 distinct (not necessarily primitive) Pythagorean triples $(a,b,c)$ with $c<100$?
Using the fact that all primitive Pythagorean triples can be generated by the following:
$a=2uv,  b=u^2-v^2, c=u^2+v^2,$ 
where $u>v, u$ and $v$ are of different parity (i.e., one is even and the other odd), and $u$ and $v$ are relatively prime (i.e., their greatest common divisor is 1), then
let $a^2+b^2=c^2$ and suppose $a,b$ to both be odd. Then $a^2+b^2=1+1=2,$ and 2 is not a square root in mod 4, thus one, either $a$ or $b$ must be even. Since $gcd(a,b)=1$ then if, say $a$ is even then $b$ must be odd, meaning $c$ is also odd. 
So I have solved for the $2uv$ portion but do not know how to get to the $u^2-v^2$ or $u^2+v^2$?
 [/INST] Let us find all these c<100. The following formula give all Pythagorean triples:
3) $a=m^2-n^2, b=2mn, c=m^2+n^2$
$c=m^2+n^2<100$ ⇒ $m<10, n<10$ ⇒ $(m,n)= (9,2), (9, 4),(8,1), (8,3),(8,5),(7,2),(7,4),(7,6),(6,1),(6,5),(5,2),(5,4),(4,1),(4,3),(3,2),(2,1)$
That is there is 16 primitive triples.The primitives less than 50 are:


*

*$5$ which gives 18 other triples less than 100.

*$13$ which gives 6 other triples less than 100.

*$17$ which gives 4 other triples less than 100.

*$25$ which gives 2 other triples less than 100.

*$29$ which gives 2 other triples less than 100.

*$37$ which gives 1 other triple less than 100.

*$41$ which gives 1 other triple less than 100.
That is we have 34 triples other than primitives, so number on triples less than 100 is:
$$2\times(34+16)=100$$
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