Formulas for calculating pythagorean triples

I'm looking for formulas or methods to find pythagorean triples. I only know one formula for calculating a pythagorean triple and that is euclid's which is:

\begin{align} &a = m^2-n^2 \\ &b = 2mn\\ &c = m^2+n^2 \end{align}

With numerous parameters.

So are there other formulas/methods?

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What is wrong with this one? It gets all the primitive ones if $m,n$ have no common factor and one is even. –  Ross Millikan May 18 '13 at 14:26
Why don't you like these formulas? –  Boris Novikov May 18 '13 at 14:26
@RossMillikan I was looking for one that could be easier to implement into a program. I am making a program that generates pythagorean triples. I have implemented this formula but I would like to know if there are more or if there are better more advance ones. EDIT: Also, this formula is a little cumbersome to implement, anyway. –  Jeel Shah May 18 '13 at 14:28
@gekkostate: That's the most efficient formula for generating primitive triples.Are you looking for all the triples in order? –  Inceptio May 18 '13 at 14:33
en.wikipedia.org/wiki/… –  lab bhattacharjee May 18 '13 at 14:33

It doesn't seem cumbersome to me. Just loop on $m$ starting at $2$. Loop on $n$ starting at $1$ or $2$ depending on the parity of $m$ and going up to $m-1$. Check the GCD using Euclid's algorithm. If it is $1$ you have a primitive set, so calculate $a,b,c$. If you want all sets up to some $N$, multiply $a,b,c$ by all values up to $\lfloor \frac Nc \rfloor$. Stop the $m$ loop at $\sqrt N$

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The one I was taught in my lectures was:

Assume that $(x, y, z)$ is a Pythagorean triple in which $x$ is odd, so that $y$ is even and $z$ is odd. For similar reasons, we assume that $p$ and $q$ are coprime $(x,y,z,p,q \in \mathbb{Z})$. The theory of Pythagorean triples then tells us that there are nonzero integers $p$ and $q$ (with $p > q$)such that

$$x + iy = (p + iq)^2 \hspace{1.5cm} z = |p + iq|^2 = p^2 + q^2.$$

If $x$ is odd then one of $p$ and $q$ must be even and the other is odd.

In case it isn't obvious, the way it works is this: let's first check with the smallest three triples. Let's pick two small numbers satisfying the criteria for $p$ and $q$, i.e take $p = 2$ and $q = 1$ (one of them is odd, both are co-prime and $p > q$). First, to find our $z$ value, we simply do $p^2 + q^2 = 1^2 + 2^2 = 5$. Now, let's calculate our $x$ and $y$ values:

$$x + iy = (p + iq)^2 = (2 + i)(2 + i) = 4 + 4i - 1 = 3 + 4i.$$

From here, we see that $x = 3$ and $y = 4$, which we know to be true as $3^2 + 4^2 = 5^2$/

This might not give you ALL the Pythagorean triples but its another way of doing it, seeing as you only need to pick $p,q$ and the rest are worked out for you.

The coding I've ever done is Matlab so I can only really think how to put this in terms of Matlab (and even then it might be wrong), so I'm not sure if it will work for your program or whatever you're making, but here is another way anyway.

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Euclid's method is only one of many for producing Pythagorean triples. See the following article for some other methods.

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I can tell formula for any odd number but this formula always works.

take any odd number=(x) . Solve for $(a+a+1 = x^2)$.

$(a)$ and $(a+1)$ will be the two numbers that will satisfy pythagoran triplet with $x$.

$(x)$ should be a natural number.

$(a+1)$ will be hypotenuse, $(a)$ will be the longer side ( if you consider perpendicular is greater, then $(a)$ is the perpendicular) and $(x)$ will be the shortest side.

This is the new trick which is invented by me shreyas.

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