ch2.0.6-integer-solutions-quadratic-squares

Ch2.0.6 — Of the Cases in Integer Numbers, in which the Formula $a{x}^2+b$ becomes a Square

Summary: Given a seed integer solution to $a{x}^2+b=y^2$ and integer values $m$, $n$ satisfying the pell-equation $m^2=a{n}^2+1$, derives a recurrence that produces infinitely many further integer solutions; applies the method to triangular, pentagonal, and hexagonal numbers that are simultaneously squares.

Sources: chapter-2.0.6

Last updated: 2026-05-08

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Setup (§79–80)

After reducing by the substitution of ch2.0.5-impossibility-quadratic-squares (§63), the problem is to find all integer $x$ making $a{x}^2+b=y^2$. Two pre-conditions are required:

1. A seed solution: integers $f$, $g$ with $a{f}^2+b=g^2$.
2. A Pell pair: integers $m$, $n$ with $m^2=a{n}^2+1$ (see ch2.0.7-pell-equation-method).

Without a seed, there may be no integer solutions at all; the method cannot create them from nothing.

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Derivation (§81–86)

Long method (§81–83)

Since $a{x}^2+b=y^2$ and $a{f}^2+b=g^2$, subtracting:

$a(x+f)(x-f)=(y+g)(y-g).$

Introducing auxiliary integers $p$, $q$, then setting $ap(x+f)=q(y+g)$ and $q(x-f)=p(y-g)$, and solving for $x$ and $y$ yields fractional expressions. To make them integers, set $m=(a{p}^2+q^2)/(a{p}^2-q^2)$ and $n=2pq/(a{p}^2-q^2)$. Computing $m^2-a{n}^2$ shows it equals $1$, so $m$ and $n$ must satisfy the Pell equation.

Short method (§86)

Since $y^2-a{x}^2=g^2-a{f}^2$ and $m^2-a{n}^2=1$, multiply both equations:

$y^2-a{x}^2=(g^2-a{f}^2)(m^2-a{n}^2).$

Expanding the right side and setting $y=gm+afn$ forces

$x^2=f^2{m}^2+2fgmn+g^2{n}^2=(fm+gn{)}^2.$

Thus $x=fm+gn$ and $y=gm+afn$, an entirely integer result.

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The Core Recurrence (§83–84, §94)

Given a Pell pair $(m,n)$ with $m^2=a{n}^2+1$, every known solution $(f,g)$ yields a new one:

$x_{\text{new}}=mf+ng,y_{\text{new}}=mg+anf.$

Substituting the new solution back in place of $(f,g)$ produces yet another, giving an infinite sequence. The sequence $f,A,B,C,D,\dots$ satisfies the second-order recurrence (§95):

$E=2m\cdot D-C$

(and identically for $y$), so each term is determined by the two preceding terms without needing the $y$-sequence at all.

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Worked Examples

Question 1 (§87): $2x^2-1=y^2$

$a=2$, $b=-1$. Seed: $f=1$, $g=1$ ($2\cdot 1-1=1$). Pell pair: $2n^2+1=m^2$ gives $n=2$, $m=3$.

Recurrences: $x=3f+2g$, $y=3g+4f$. Starting from $(f,g)=(1,1)$:

$x$	$y$
$1$	$1$
$5$	$7$
$29$	$41$
$169$	$239$
$985$	$1393$
$5741$	$8119$

The $x$-values follow $1,5,29,169,985,5741,\dots$ with each term $=6\times (\text{prev})-(\text{prev-prev})$.

Question 2 (§88): Triangular numbers that are squares

The $z$-th triangular number $T_z=z(z+1)/2$ is a square $x^2$ iff $8x^2+1=(2z+1{)}^2$. So $a=8$, $b=1$. Seed: $f=0$, $g=1$. Pell: $8n^2+1=m^2$ gives $n=1$, $m=3$.

$x$	$y=2z+1$	$z$
$0$	$1$	$0$
$1$	$3$	$1$
$6$	$17$	$8$
$35$	$99$	$49$
$204$	$577$	$288$
$1189$	$3363$	$1681$

The triangular squares are $T_0=0$, $T_1=1$, $T_8=36$, $T_{49}=1225$, $T_{288}=41616$, $T_{1681}=1413721$, $\dots$ (source: chapter-2.0.6, §88)

Question 3 (§89): Pentagonal numbers that are squares

The $z$-th pentagon is $(3z^2-z)/2$. Setting equal to $x^2$ leads to $24x^2+1=(6z-1{)}^2$. $a=24$, $b=1$. Seed: $f=0$, $g=1$. Pell: $24n^2+1=m^2$ gives $n=1$, $m=5$.

$x$	$y$	$z=(y+1)/6$
$0$	$1$	$1/3$
$1$	$5$	$1$
$10$	$49$	$25/3$
$99$	$485$	$81$
$980$	$4801$	$2401/3$

Integer pentagons that are squares: $P_1=1$, $P_{81}=9801$, $\dots$ (source: chapter-2.0.6, §89)

Question 4 (§90): $7x^2+2=y^2$

Seed: $f=1$, $g=3$ ($7+2=9$). Pell: $7n^2+1=m^2$ gives $n=3$, $m=8$.

$x$	$y$
$1$	$3$
$17$	$45$
$271$	$717$

Question 5 (§91): Triangular numbers that are pentagons

Equating $z$-th triangle to $q$-th pentagon reduces (via a completing-the-square step) to $3x^2-2=y^2$, with $a=3$, $b=-2$. Seed: $f=1$, $g=1$. Pell: $3n^2+1=m^2$ gives $n=1$, $m=2$.

$x$	$p$ (triangular root)	$q$ (pentagonal root)
$1$	$0$	$1/3$ (not integer)
$3$	$1$	$1$
$11$	$5$	$10/3$ (not integer)
$41$	$20$	$12$

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Extension to Formula with Middle Term (§92–93)

For $a{x}^2+bx+c=y^2$, with known seed $a{f}^2+bf+c=g^2$, subtract and factor:

$(x-f)(ax+af+b)=(y-g)(y+g).$

The same procedure applies, introducing $p^2=a{q}^2+1$. The integer formulas become:

$x=2gpq+f(a{q}^2+p^2)+b{q}^2,y=g(a{q}^2+p^2)+2afpq+bpq.$

Hexagonal squares (§93): $2x^2-x=y^2$ ($a=2$, $b=-1$, $c=0$). Seed: $f=1$, $g=1$. Pell: $p^2=2q^2+1$ gives $q=2$, $p=3$.

$x$	$y$
$1$	$1$
$25$	$35$
$841$	$1189$

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Related pages

• ch2.0.5-impossibility-quadratic-squares
• ch2.0.7-pell-equation-method
• pell-equation
• polygonal-numbers
• indeterminate-analysis
• quadratic-equations
• vieta-formulas