Pythagorean Triples

Summary: Integer triples $(p,q,r)$ with $p^2+q^2=r^2$; Euler derives the general parametric formula $(2mn,n^2-m^2,n^2+m^2)$ as a byproduct of rationalizing $\sqrt{1+x^2}$.

Sources: chapter-2.0.4

Last updated: 2026-05-05

---

Definition

A Pythagorean triple is a set of three positive integers $p,q,r$ satisfying

$p^2+q^2=r^2.$

Small examples: $(3,4,5)$, $(5,12,13)$, $(8,15,17)$, $(7,24,25)$.

Euler’s Derivation (§43–44)

Euler arrives at the general formula by solving: for which rational $x$ is $1+x^2$ a perfect square? Setting $\sqrt{1+x^2}=x+p$ and writing $p=m/n$, he finds

$x=\frac{n^2-m^2}{2mn}.$

Substituting back:

$1+\frac{(n^2-m^2{)}^2}{(2mn{)}^2}=\frac{(n^2+m^2{)}^2}{(2mn{)}^2},$

which, after multiplying through by $(2mn{)}^2$, gives

$\boxed{(2mn{)}^2+(n^2-m^2{)}^2=(n^2+m^2{)}^2.}$

Taking $p=2mn$, $q=n^2-m^2$, $r=n^2+m^2$ with $m,n$ any integers ($n>m>0$) produces a Pythagorean triple. (source: chapter-2.0.4, §44)

Selected Triples from the Formula

$n$	$m$	$p=2mn$	$q=n^2-m^2$	$r=n^2+m^2$
2	1	4	3	5
3	2	12	5	13
4	1	8	15	17
4	3	24	7	25
5	2	20	21	29

Two Corollaries (§44)

Because $(n^2+m^2{)}^2-(2mn{)}^2=(n^2-m^2{)}^2$, the same formula also solves:

• Sum of two squares is a square: $p=2mn$, $q=n^2-m^2$, $r=n^2+m^2$.
• Difference of two squares is a square: $p=n^2+m^2$, $q=2mn$, $r=n^2-m^2$.

Iterated for Fermat’s $n=4$

The same parametrization, applied twice, drives Euler’s infinite-descent proof in ch2.0.13-impossibility-biquadrate-sums. Starting from a hypothetical $x^4+y^4=z^2$ solution, treat $(x^2,y^2,z)$ as a Pythagorean triple, then re-parametrize the inner Pythagorean structure to produce a strictly smaller fourth-power equation $t^4+u^4=v^2$. See fermats-last-theorem-n4.

Lagrange’s Chord-Method Generalization

The same parametric pattern Euler discovers for $1+x^2=\square$ is, in Lagrange’s hands (Add. V Art. 57), the general chord-method parametrization for any rational quadratic. Given one rational point $(f,g)$ on the conic $a+bx+c{x}^2=y^2$:

$x=\frac{f{m}^2-2gm+b+cf}{m^2-c},m\in \mathbb{Q}.$

For $a+bx+c{x}^2=1+0\cdot x+1\cdot x^2$ with seed $(f,g)=(0,1)$, this reduces to $x=-2m/(m^2-1)$ — equivalent (after $m=n/m$ rescaling) to Euler’s $(2mn,n^2-m^2,n^2+m^2)$. See add5-rational-quadratic-surds for the general form.

Related pages

• ch2.0.4-surd-rationalization
• ch2.0.13-impossibility-biquadrate-sums
• algebraic-identities
• indeterminate-analysis
• square-roots-and-irrational-numbers
• sums-of-two-squares
• fermats-last-theorem-n4
• infinite-descent
• add5-rational-quadratic-surds