How to sensibly use Euclid's formula for Pythagorean triples. I've tried playing around with Euclid's formula ($A=m^2-n^2$, $B=2mn$, $C=m^2+n^2$) but I can't see any pattern in the triples it generates or how to predict what numbers will work other than being mutually prime. Here are examples:
$f(3,2)=(5,12,13), $
$f(4,1)=(15,8,17), $
$f(4,3)=(7,24,25), $
$f(5,2)=(21,20,29), $
$f(5,3)=(16,30,34), $
$f(5,4)=(9,40,41), $
$f(6,5)=(11,60,61), $
I've heard that the formula generates all primitives but I can't even figure out how to get $(3,4,5)$. How should I be using it to get, for example, the series $$(3,4,5), (5,12,13), (13,84,85), (85,3612,3613), (3613,6526884,6526885)$$ that is a natural pairing of side $C$ of one triple with side $A$ of the triple that follows? I can do it easily using other formulas but this is the $standard$ that everyone accepts. 
Update: My problems in understanding are the seeming lack a pattern of the triples produced and the seeming invalidity of the statement: "Primitives will be produced if and only if $m$ and $n$ are co-prime." One comment says f(2,1)=(3,4,5) but $1$ is not prime. Let's accept it anyway, but then we find that f(3,1)=(8,6,10) is not primitive. We also find that f(5,3)=(16,30,34), f(7,3)=(40,42,58), and f(7,5)=(24,70,74) are each twice a primitive with the positions of $A$ and $B$ switched. I get a poor reception whenever I touch on the other functions I mentioned so I'm trying to find out how Euclid's formula can give me the power of prediction I seek.
Can I know the nature of the GCD(A,B,C) for a specific combination of $m$ and $n$? Can I know the difference between subsequent values of $A$ for incremental values of $m$ or $n$? Can I know the difference between $B$ and $C$ for a specific combination of $m$ and $n$? Can I find triples with matching sides, areas, or perimeters? If I can find answers to at least some of these questions, it will allow me to stop further exploration of them for a paper I'm writing. If Euclid's formula provides no such power, may I then assume that my contribution is original and perhaps non-trivial?
 A: All non-zero numbers $m > n$ will produce a triple. That triple will be primitive if and only if $m$ and $n$ are coprime, and one of the two is even.
Suppose $a^2 + b^2 = c^2$ is a primitive Pythagorean triple. It is easy to see that $c$ must be odd, and one of $a$ or $b$ is even. W.l.o.g., $b$ is even. Then we can rearrange: $$b^2 = c^2 - a^2 = (c+a)(c-a)$$
Since $c$ and $a$ are both odd, $c+a$ and $c-a$ are even. Let $r = \frac {c+a}2$ and $s = \frac{c-a}2$. Then $c = r+s$ and $a = r - s$. Now if $p\mid r$ and $p\mid s$, we also have $$p \mid (r+s) = c$$ and $$p \mid r-s = a$$. Since $a$ and $c$ are coprime, $p = 1$. Therefore $r$ and $s$ are also coprime.
But $rs = \frac{c^2 -a^2}4 = \left(\frac b2\right)^2$, which can only be if $r$ and $s$ themselves are perfect squares. Let $m = \sqrt r$ and $n = \sqrt s$. Then $$a = r - s = m^2 - n^2\\b = 2\sqrt{rs} = 2mn\\c = r + s = m^2 + n^2$$
Since every triple is a multiple of some primitive triple, the general formula for triples is $$a = k(m^2 - n^2)\\b = 2kmn\\c=k(m^2 + n^2)$$
A: Conditions on primitive triples include:


*

*coprime m,n

*exactly one of m,n  is even ( because if both are even, or both are odd all three parts are even)

*one number in the triple will always be divisible by 4.

*one number (possibly the same one as above) will be divisible by 5.

*one number will ( possibly the same as either or both of the above) will be divisible by 3.


The last 3 of these show up as follows( applying to all triples):
Mod 3:


*

*if m,n are non-zero then $m^2-n^2\equiv 0\bmod 3$

*if at least one of m,n are 0 mod 3, then $2mn\equiv 0\bmod 3$

*if both m,n are 0 mod 3 ( won't happen primitively), then $m^2+n^2\equiv 2mn\equiv m^2-n^2\equiv 0\bmod 3$
Mod 4:


*

*nearly same as above except Mod 4


Mod 5:


*

*if m,n are the same, or additive inverses mod 5, then $m^2-n^2\equiv 0\bmod 5$

*if at least 1 of m,n are 0 mod 5, then $2mn\equiv 0\bmod 5$

*if m,n are both non-zero mod 5, and their squares aren't the same then $m^2+n^2\equiv 0\bmod 5$
Coprime simply is another name for relatively prime, meaning not sharing a factor other than 1, in common. 
area of a primitive pythagorean triple triangle is always $nm^3-mn^3$
Perimeter is always $2m^2+2mn$
For the even side, side matching, comes down to factoring it into different breakdowns like $24=2(3)(4)=2(6)(2)$ or $12=2(3)(2)=2(6)(1)$ 
$C-B=(m-n)^2$ and  $A=A+2m+1\qquad \text{ if $m=m+1$}, \qquad A=A+2n-1,\qquad \text{ if $n=n-1$}$ etc.
