# Derivation of Pythagorean Triple General Solution Starting Point:

I was reading on proof wiki about the derivation of the general solution to the pythagorean triple diophantine equation: $$x^2 + y^2 = z^2,$$ where $x,y,z > 0$ are integers. I came across the following general solution to the primitive function: \begin{align*} x &= 2mn\\ y &= (m^2 - n^2)\\ z &= (m^2 + n^2)\\ \end{align*} for coprime $m,n$.

I looked at the proof of it working (if you square $x$ and $y$ and add it it does indeed equal $z^2$)

My one qualm was, how the hell did they start with that? For example, is there a natural way by starting with the original problem that you end up with the expression above?

I noticed that when attempting to derive the general solution myself, from start to finish,

I would begin by noting I can find all pairs such that $z$ and $y$ differ by a constant $k$... but I cannot make that final leap to end up with the equation above so that for any given $k$ you can find a solution.

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That's not a generating formula for all triples. If we make restriction on $m$ and $n$ (relatively prime, opposite parity, $m \gt n$) we get all primitive triples. To see don't get all triples, we can't get $9,12,15$. – André Nicolas Jun 17 '13 at 23:31
find a rational parameterization of the unit circle. – yoyo Jun 17 '13 at 23:41
@André: That's the result of primitive triplets $\{3,4,5 \}$, right? – Inceptio Jun 18 '13 at 7:46
Yes. The details are nicely done in the answer by robjohn. – André Nicolas Jun 18 '13 at 10:56

There are a number of ways to find those expressions organically. One way is via complex numbers (and this method can be found in the first chapter of Marcus' Number Fields).

Suppose $x^2 + y^2 = z^2$, with $x,y,z\in\Bbb N$, and also suppose that $x$, $y$, and $z$ have no common factor. Looking mod $4$, we can tell that $z$ must be odd: the only squares mod $4$ are $0$ and $1$, so the only possible equations are $0^2 + 0^2\equiv 0^2\mod 4$ and $0^2 + 1^2 \equiv 1^2\mod 4$. But in the first case, everything is divisible by $4$, so we don't have a primitive triple.

Proceeding, we can factor the left hand side and get $(x + iy)(x - iy) = z^2$. So, we now have a problem concerning $\Bbb Z[i] = \{a + bi\mid a,b\in\Bbb Z\}$. One can show that $\Bbb Z[i]$ has unique factorization of elements into primes (for a precise definition of all this, look to any abstract algebra text), so suppose some prime $p$ divides $x + iy$. $p$ clearly divides $z^2$ an even number of times, so we want to show that $p$ does not divide $x - iy$. If it did, we would have $p\mid (x + yi) + (x - yi)$, or $p\mid 2x$. Since $p$ divides $z$ as well, and $x$ and $z$ are relatively prime, we know that we can find $n,m\in\Bbb{Z}$ such that $2xm + zn = 1$ (recall that $z$ is odd). However, this implies that $p$ divides $1$, but the only elements dividing $1$ in $\Bbb{Z}[i]$ are $\pm 1$, $\pm i$, none of which are primes - so we have a contradiction. Therefore, by unique factorization, we must have $x + iy = u\alpha^2$, where $u\in\{\pm1,\pm i\}$. Writing $\alpha = m + in$, we have $\alpha^2 = m^2 - n^2 + i2mn$, so $x = \pm \left(m^2 - n^2\right)$, $y = \pm 2mn$, and solving for $z$, we get $z = \pm\left(m^2 + n^2\right)$.

If $m$ and $n$ aren't coprime, we don't obtain a primitive triple, as each of $x$, $y$, and $z$ will have a factor in common.

There are numerous other ways of deriving these ways of representing $x$, $y$, and $z$ as well, some of which include parameterizing the unit circle or moving $y^2$ to the other side and writing $x^2 = (z + y)(z - y)$.

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This answered my follow up before I asked it, thanks! – frogeyedpeas Jun 18 '13 at 0:27

Consider the intersection of the line $y=t(x+1)$ with the circle $x^2+y^2=1$. We get points on the circle $$(x,y)=((1-t^2)/(1+t^2),2t/(1+t^2)).$$ Let $t=m/n$ be rational, plug into $x^2+y^2$ and clear denominators to get $$(n^2-m^2)^2+(2mn)^2=(n^2+m^2)^2$$

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Here is the way to generate all relatively prime pythagorean triples:

Theorem: Let $m$ and $n$ be positive integers so that \begin{align} &m\gt n\tag{1}\\ &m+n\text{ is odd}\tag{2}\\ &m\text{ and }n\text{ are relatively prime}\tag{3} \end{align} Then, \begin{align} a &= m^2 - n^2\\ b &= 2mn\\ c &= m^2 + n^2 \end{align}\tag{4} gives all positive, relatively prime $a$, $b$, and $c$ so that $$a^2 + b^2 = c^2\tag{5}$$

Proof: $(5)\Rightarrow(4):$

Suppose $a$, $b$, and $c$ are positive, relatively prime, and $a^2 + b^2 = c^2$.

Because $(2k)^2 = 4k^2$ and $(2k+1)^2 = 4(k+1)k + 1$, the square of an even integer must be $0 \bmod{4}$ and the square of an odd integer must be $1 \bmod{4}$.

At least one of $a$ and $b$ must be odd; otherwise $a$, $b$, and $c$ would share a common factor of $2$. If both are odd, then $c^2$ would need to be $2 \bmod{4}$, which is impossible. Thus, one must be even and the other must be odd. This means that $c$ must be odd. Without loss of generality, let $b$ be even.

Let $M = (c+a)/2$ and $N = (c-a)/2$. Then \begin{align} a &= M - N\tag{6}\\ c &= M + N\tag{7}\\ b^2 &= 4MN\tag{8} \end{align} Thus, we have that $M \gt N \gt 0$ and one of $M$ and $N$ must be even and the othermust be odd. Furthermore, $\gcd(M,N)$ divides $a$, $b$, and $c$; thus, $\gcd(M,N) = 1$. Since $b^2 = 4MN$ and $\gcd(M,N) = 1$, both $M$ and $N$ must be perfect squares. Let $M = m^2$ and $N = n^2$, where $m$ and $n$ are positive; then, $(1)$, $(2)$, $(3)$, and $(4)$ are satisfied.

$(4)\Rightarrow(5):$

Suppose $(1)$, $(2)$, $(3)$, and $(4)$ are satisfied. Then $(5)$ is satisfied: \begin{align} a^2 + b^2 &= (m^2 - n^2)^2 + (2mn)^2\\ &= m^4 - 2 m^2 n^2 + n^4 + 4 m^2 n^2\\ &= m^4 + 2 m^2 n^2 + n^4\\ &= (m^2 + n^2)^2\\ &= c^2\tag{9} \end{align} Furthermore, $a$ and $b$ are relatively prime since \begin{align} \gcd(a,b) &= \gcd(m^2-n^2,2mn)\\ &\:\mid\:\gcd(m-n,2) \gcd(m-n,m) \gcd(m-n,n)\\ &\times\gcd(m+n,2) \gcd(m+n,m) \gcd(m+n,n)\\ &=\gcd(m+n,2)^2 \gcd(n,m)^4\\ &= 1\tag{10} \end{align}\\ $\square$

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