Periodic Continued Fractions

Summary: If the partial denominators of a simple continued fraction repeat with finite period, the continued fraction satisfies a polynomial equation of degree at most $2$ in itself — so its value is a quadratic irrational. Euler works through the single-letter case ($x=(\sqrt{a^2+4}-a)/2$, generating $\sqrt{5},\sqrt{2},\sqrt{13},\sqrt{5},\sqrt{29},\dots$ as $a$ runs over $1,2,3,4,\dots$), the two-letter case (which extends the catalog to every square root), and three- and four-letter cases (whose discriminants reduce to the two-letter case).

Sources: chapter18 (§376–§379).

Last updated: 2026-05-11

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Why periodic CFs give quadratic irrationals (§376)

The key observation is that dropping leading periods does not change the value of a periodic continued fraction. If $x$ is purely periodic with period $k$, then $x$ appears as a tail of itself, and the three-term recurrence turns this self-reference into a polynomial equation in $x$ of degree at most $2$.

Euler’s archetypal example: $x=1/(2+1/(2+1/(2+\cdots )))$ satisfies

$x=\frac{1}{2+x},\text{i.e.,}{x}^2+2x=1,\text{so}x=\sqrt{2}-1.$

The convergents $1/0,0/1,1/2,2/5,5/12,12/29,29/70,\dots$ approximate $\sqrt{2}-1=0.41421356\dots$ with $29/70=0.41428571\dots$ — error in the hundred-thousandths place after only six terms.

Single-letter period (§377)

For arbitrary $a\in {\mathbb{Z}}_{>0}$, $x=1/(a+x)$ gives $x^2+ax=1$, so

$x=\frac{\sqrt{a^2+4}-a}{2}.$

Running through $a=1,2,3,4,\dots$:

$a$	partial quotients	$x$	$\sqrt{a^2+4}$
$1$	$1,1,1,1,\dots$	$\frac{\sqrt{5}-1}{2}$	$\sqrt{5}$
$2$	$2,2,2,2,\dots$	$\sqrt{2}-1$	$\sqrt{8}=2\sqrt{2}$
$3$	$3,3,3,3,\dots$	$\frac{\sqrt{13}-3}{2}$	$\sqrt{13}$
$4$	$4,4,4,4,\dots$	$\sqrt{5}-2$	$\sqrt{20}=2\sqrt{5}$
$5$	$5,5,5,5,\dots$	$\frac{\sqrt{29}-5}{2}$	$\sqrt{29}$

Convergence accelerates as $a$ grows: for $a=4$ the sixth convergent $305/1292$ approximates $\sqrt{5}-2$ with error $<1/(1292\cdot 5473)$.

This method handles exactly the square roots of $a^2+4$ — i.e., numbers that are the sum of two squares.

Two-letter period (§378)

To extend the method to all square roots Euler considers $x=1/(a+1/(b+x))$, which gives $a{x}^2+abx=b$ and

$x=\frac{-ab+\sqrt{a^2{b}^2+4ab}}{2a}.$

Choosing $a=2,b=7$: $x=(-7+3\sqrt{7})/2$, and the convergent sequence $1/2,7/15,15/32,112/239,239/510,\dots$ gives $\sqrt{7}\approx 2024/765=2.6457516$ versus the true $\sqrt{7}=2.64575131\dots$, error $<3/10^7$. (Euler’s $\sqrt{7}$ approximation.)

Three- and four-letter periods (§379)

A three-letter period gives a quadratic with discriminant $(abc+a+b+c{)}^2+4$, which Euler observes is again a sum of two squares — so it does not produce square roots inaccessible to the two-letter method. The same goes for four-letter periods. So purely periodic CFs with periods $\ge 2$ do not enlarge the class of irrationals reachable, only re-parametrise it.

Modern statement (Lagrange’s theorem)

The collection of theorems §376–§379 together establish one direction of what is now called Lagrange’s theorem on periodic continued fractions (1770): a real number has a periodic simple continued fraction expansion if and only if it is a quadratic irrational. The converse direction — every quadratic irrational has a periodic CF — was not stated by Euler in the Introductio, but the §381 Euclidean-algorithm method gives a procedure for computing the CF expansion of any real number, and applied to $\sqrt{2}$ in §381 Example II it recovers the $2,2,2,\dots$ pattern of §376.

Related pages

• continued-fraction
• convergents-of-a-continued-fraction
• euclidean-algorithm-continued-fraction — §381 Example II verifies $\sqrt{2}=[1;2,2,2,\dots ]$ by running the algorithm on a decimal
• best-rational-approximations — periodic CFs give fast best-approximations to quadratic irrationals
• chapter-18-on-continued-fractions