ch2.0.7-pell-equation-method

Ch2.0.7 — Of a Particular Method, by which the Formula $a{n}^2+1$ becomes a Square in Integers

Summary: Presents Pell’s iterative method for finding the minimal integer solution $(n,m)$ of $m^2=a{n}^2+1$ for any positive non-square $a$; proves the method always terminates; gives closed-form solutions when $a$ is near a perfect square; and tabulates results for $a=2$ to $100$.

Sources: chapter-2.0.7

Last updated: 2026-05-08

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The Problem (§96–97)

The integer-solution method of ch2.0.6-integer-solutions-quadratic-squares requires solving

$m^2=a{n}^2+1$

in integers. Fractional solutions are trivial (set $m=1+np/q$ and solve for $n$), but integer solutions require a different approach entirely.

The equation is impossible if:

• $a$ is negative (left side would exceed right for large $n$).
• $a$ is a perfect square (then $a{n}^2$ is a square, and no square increases by 1 to give another square in integers).

For every positive non-square $a$, at least one integer solution exists. Euler attributes the method to John Pell, an English mathematician. (source: chapter-2.0.7, §97)

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Pell’s Method (§98–104)

The idea is to reduce the problem by a sequence of substitutions, each yielding a simpler formula of the same type, until a formula $a{t}^2+1$ appears whose trivial solution $t=0$ is available. Back-substituting recovers the value of $n$.

Step 1: Estimate $\lfloor \sqrt{a}\rfloor$ to bound $m$. Write $m=\lfloor \sqrt{a}\rfloor \cdot n+p$ (or $-p$) and square to get a relation between $n$ and $p$.

Step 2: Solve for $n$. The expression involves $\sqrt{(\cdots )p^2+(\cdots )}$. Bound $n$ in terms of $p$ to pick the next substitution $n=k\cdot p+q$.

Step 3: Repeat until a term $\sqrt{a{t}^2+1}$ (same coefficient as the original) appears. Then set $t=0$ and trace back.

Examples

$a=2$ (§98): $\sqrt{2n^2+1}=n+p$ gives $n=p+\sqrt{2p^2-1}$. Setting $p=1$: $\sqrt{2p^2-1}=1$, so $n=2$, $m=3$.

$a=3$ (§99): $\sqrt{3n^2+1}=n+p$ gives $n=(p+\sqrt{3p^2-2})/2$; then $n=p+q$, leading to $p=q+\sqrt{3q^2+1}$. Set $q=0$: $p=1$, $n=1$, $m=2$.

$a=5$ (§100): $\sqrt{5n^2+1}=2n+p$ (since root $>2n$); two substitutions yield $p=2q+\sqrt{5q^2+1}$. Set $q=0$: $p=1$, $n=4$, $m=9$.

$a=6$ (§101): $\sqrt{6n^2+1}=2n+p$; one further substitution yields $p=2q+\sqrt{6q^2+1}$. Set $q=0$: $p=1$, $n=2$, $m=5$.

$a=7$ (§102): Three substitutions yield $r=2s+\sqrt{7s^2+1}$. Set $s=0$: back-substituting gives $n=3$, $m=8$.

$a=8$ (§103): $m=3n-p$ (since $m<3n$); one substitution gives $n=3p+\sqrt{8p^2+1}$, which is already in the base form. Set $p=0$: $n=1$, $m=3$.

Guarantee (§104): For any positive non-square $a$, the process always terminates. At each stage the “active variable” decreases, so the chain of substitutions must reach a $\sqrt{a{t}^2+1}$ term and terminate. No formula exists to predict in advance how many steps are needed.

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Hard Case: $a=13$ (§105)

Requires nine substitution steps. The chain is:

$n\to p\to q\to r\to s\to t\to u\to v\to x\to y\to z=0$

Back-substituting:

Variable	Value
$z$	$0$
$y$	$1$
$x$	$1$
$v$	$2$
$u$	$3$
$t$	$5$
$s$	$33$
$r$	$38$
$q$	$71$
$p$	$109$
$n$	$180$
$m$	$649$

Thus the minimal solution is $n=180$, $m=649$, and $13\times 180^2+1=649^2$. (source: chapter-2.0.7, §105)

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Special Closed-Form Cases (§107–111)

When $a$ is near a perfect square, the first substitution already terminates:

Form of $a$	Minimal $n$	$m$	Verification
$a=e^2-2$	$n=e$	$m=e^2-1$	$(e^2-2)e^2+1=e^4-2e^2+1=(e^2-1{)}^2$
$a=e^2-1$	$n=2e$	$m=2e^2-1$	$(e^2-1)(2e{)}^2+1=4e^4-4e^2+1=(2e^2-1{)}^2$
$a=e^2+1$	$n=2e$	$m=2e^2+1$	$(e^2+1)(2e{)}^2+1=4e^4+4e^2+1=(2e^2+1{)}^2$
$a=e^2+2$	$n=e$	$m=e^2+1$	$(e^2+2)e^2+1=e^4+2e^2+1=(e^2+1{)}^2$

Examples: $a=23=5^2-2$ gives $n=5$, $m=24$; $a=24=5^2-1$ gives $n=10$, $m=49$; $a=17=4^2+1$ gives $n=8$, $m=33$; $a=11=3^2+2$ gives $n=3$, $m=10$.

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Table of Minimal Solutions (§106)

A selection from Euler’s table ($a=2$ to $100$, perfect squares omitted):

$a$	$n$	$m$		$a$	$n$	$m$
2	2	3		13	180	649
3	1	2		29	1820	9801
5	4	9		41	320	2049
6	2	5		46	3588	24335
7	3	8		58	2574	19603
8	1	3		61	226153980	1766319049
10	6	19		73	267000	2281249
11	3	10		85	30996	285769
12	2	7		94	221064	2143295
17	8	33		97	6377352	62809633

Notable: $a=61$ requires an enormous $n=226153980$. This is one of the most famous examples in number theory. (source: chapter-2.0.7, §106)

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Related pages

• pell-equation
• ch2.0.6-integer-solutions-quadratic-squares
• ch2.0.5-impossibility-quadratic-squares
• indeterminate-analysis
• quadratic-residues