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.

Thanks ahead of time!

  • $\begingroup$ 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 $9,12,15$. $\endgroup$ Jun 17, 2013 at 23:31
  • $\begingroup$ find a rational parameterization of the unit circle. $\endgroup$
    – yoyo
    Jun 17, 2013 at 23:41
  • $\begingroup$ @André: That's the result of primitive triplets $\{3,4,5 \}$, right? $\endgroup$
    – Inceptio
    Jun 18, 2013 at 7:46
  • $\begingroup$ Yes. The details are nicely done in the answer by robjohn. $\endgroup$ Jun 18, 2013 at 10:56
  • $\begingroup$ The hardest part for me is the values of the variables equaling. $\endgroup$
    – user645636
    Feb 23, 2019 at 13:50

4 Answers 4


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

  • $\begingroup$ magic step is imagining the line $y=t(x+1)$ $\endgroup$
    – karakfa
    Apr 22, 2017 at 17:43
  • $\begingroup$ @karakfa it's not as esoteric as it seems, as one wants to cancel the negative root of $y = 1 - x^2$ $\endgroup$
    – BoLe
    Aug 3, 2019 at 19:02

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.


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 $$ \begin{align} \gcd(a,b) &= \gcd(m^2-n^2,2mn)\\ &\:\mid\:\gcd(m-n,2) \gcd(m-n,m) \gcd(m-n,n)\\ &\times\gcd(m+n,2) \gcd(m+n,m) \gcd(m+n,n)\\ &=\gcd(m+n,2)^2 \gcd(n,m)^4\\ &= 1\tag{10} \end{align}\\ $$ $\square$


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 $x^2 = (z + y)(z - y)$.

  • $\begingroup$ This answered my follow up before I asked it, thanks! $\endgroup$ Jun 18, 2013 at 0:27

The difficulty seems to be that if you specify integer lengths for the two legs, or integer lengths for the hypotenuse and one of the legs, the remaining side will most often have irrational length. If one starts in this way, it is perhaps not immediately obvious how to do the number-theoretic analysis to determine the conditions under which the quantity under the square-root sign is a perfect square.

Interestingly, there is a slightly modified geometry problem in which specifying rational lengths for two of three parameters automatically leads to the third also being rational. This sort of problem has a long history, going back to ancient Mesopotamia and Egypt. See Poles and walls in Mesopotamia and Egypt by Duncan J. Melville and page 307 of Methods and traditions of Babylonian mathematics: Plimpton 322, Pythagorean triples, and the Babylonian triangle parameter equations by Jöran Friberg.

Here's one version: a vertical cane of unknown length sits against a wall. The cane slips so that the top of the cane moves down the wall a distance $u$ and the bottom of the cane moves out (perpendicularly) from the wall a distance $a$. How long is the cane?

Let the length of the cane be $c$. In its new position, the cane, the wall, and the ground form a right triangle with legs $a$ and $b=c-u$ and hypotenuse $c$. Now $$ c^2=a^2+(c-u)^2=a^2+c^2-2cu+u^2 $$ so that $2cu=a^2+u^2$ and $$ c=\frac{a^2+u^2}{2u}. $$ Notice that we have computed $c$ without taking any square roots. This means that if we choose the lengths $a$ and $u$ to be rational, then so are $c$ and $c-u$. Therefore we get rational Pythagorean triples this way: $$ (a,b,c)=(a,c-u,c)=\left(a,\frac{a^2-u^2}{2u},\frac{a^2+u^2}{2u}\right). $$ If we want integer triples, we can multiply through by $2u$ and require $a$ and $u$ to be integer. Now that we have a form that gives integer triples, we can, if we wish, investigate whether integer triples always have this form and go on to analyze the conditions under which such triples are primitive.

Added in response to question in comments concerning integer solutions: In the cane problem, the only condition on $u$ is $0<u<a$. If you want $a$, $b$, and $c$ to be integers then you have to restrict $a$ and $u$ to integers and you have to require that $a^2+u^2$ be divisible by $2u$. So, for example, you could let $u=1$ and $a=3$. Then $c=\frac{3^2+1^2}{2\cdot1}=5$ and $b=5-1=4$. A second example: let $u=2$, $a=12$. Then $c=\frac{12^2+2^2}{2\cdot2}=37$ and $b=37-2=35$. A third: $u=9$, $a=15$. Then $c=\frac{15^2+9^2}{2\cdot9}=17$ and $b=17-9=8$.

In order for $a^2+u^2$ to be divisible by $2u$, observe that divisibility by $2$ requires that $a$ and $u$ have the same parity, while divisibility by $u$ requires that $u$ divide $a^2$. If $u$ is odd, these necessary conditions are also sufficient. If $u$ is even, there is a slight additional constraint: write $u=2u'$ and $a=2a'$. Then $4(a'^2+u'^2)$ must be divisible by $4u'$. So $u'$ must divide $a'^2$.

  • $\begingroup$ how do you determine the value of $u$ in the equation that gives $c$ in terms of $a$ and $u$? Can $u$ take any arbitrary value? Can you provide an example with integers (not rationals). Thanks. $\endgroup$
    – user25406
    Feb 21, 2021 at 14:13
  • 1
    $\begingroup$ I've added to the post to address your question. $\endgroup$ Feb 21, 2021 at 16:13

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