ch2.0.4-surd-rationalization

Ch2.0.4 — On the Method of Rendering Surd Quantities of the Form $\sqrt{a+bx+c{x}^2}$ Rational

Summary: Develops four systematic rules for finding rational values of $x$ that make $a+bx+c{x}^2$ a perfect square; demonstrates infinite solution families and a bootstrap technique for exploiting a known solution; derives Pythagorean triples as a byproduct.

Sources: chapter-2.0.4

Last updated: 2026-05-05

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The Problem (§38–39)

Given constants $a,b,c$, find rational values of $x$ such that

$a+bx+c{x}^2=y^2$

for some rational $y$. When $c=0$ this is trivial: set $a+bx=y^2$ and solve $x=(y^2-a)/b$. The interesting case is the full quadratic, where the solution depends critically on the nature of $a$, $b$, and $c$.

Euler restricts to degree $\le 2$ in $x$; higher-degree formulas “require different methods which will be explained in their proper places.” The goal is rational (not necessarily integer) values of $x$.

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Prototype: Making $\sqrt{1+x^2}$ Rational (§40–45)

The simplest non-trivial case is $\sqrt{1+x^2}$. Three examples confirm that rational solutions exist: $x=3/4\Rightarrow \sqrt{25/16}=5/4$, $x=4/3\Rightarrow 5/3$, $x=5/12\Rightarrow 13/12$.

Method 1 (§42–43)

Set $\sqrt{1+x^2}=x+p$. Squaring: $1+x^2=x^2+2px+p^2$, so $x^2$ cancels and

$x=\frac{1-p^2}{2p}.$

Writing $p=m/n$ and clearing denominators:

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

For every integer pair $(m,n)$ this produces a valid $x$. A table of small values:

$n$	$m$	$x$
2	1	3/4
3	1	4/3
3	2	5/12
4	1	15/8
4	3	7/24
5	1	12/5
5	2	21/20
5	3	8/15
5	4	9/40

Method 2 (§45)

Set $\sqrt{1+x^2}=1+mx/n$. Squaring: $x^2=2mx/n+m^2{x}^2/n^2$; dividing by $x$ and solving:

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

Both methods produce the same underlying identity.

Pythagorean Triple Byproduct (§44)

Setting $p=(n^2-m^2)/(2mn)$, compute $1+x^2$:

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

which gives the general Pythagorean triple formula:

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

As a corollary, one can find two squares whose sum is a square ($p=2mn$, $q=n^2-m^2$, $r=n^2+m^2$) and two squares whose difference is a square ($p=n^2+m^2$, $q=2mn$, $r=n^2-m^2$).

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Four Rules for the General Formula (§46–54)

Rule 1: $c$ is a perfect square (§46–47)

Write the formula as $a+bx+f^2{x}^2$. Set

$\sqrt{a+bx+f^2{x}^2}=fx+\frac{m}{n}.$

The $f^2{x}^2$ terms cancel; solving for $x$:

$x=\frac{m^2-n^{2}a}{n^{2}b-2mnf},\sqrt{a+bx+f^2{x}^2}=\frac{mnb-m^{2}f-n^{2}af}{n^{2}b-2mnf}.$

Setting $p=m^2-n^{2}a$ and $q=n^{2}b-2mnf$, the formula $a{q}^2+bpq+f^2{p}^2$ is also a perfect square, giving infinitely many integer solutions.

Rule 2: $a$ is a perfect square (§48)

Write the formula as $f^2+bx+c{x}^2$. Set

$\sqrt{f^2+bx+c{x}^2}=f+\frac{mx}{n}.$

The $f^2$ terms cancel; solving for $x$:

$x=\frac{2mnf-n^{2}b}{n^{2}c-m^2},\sqrt{f^2+bx+c{x}^2}=\frac{n^{2}cf+m^{2}f-mnb}{n^{2}c-m^2}.$

Special sub-case $a=0$ (§49)

When $a=0$ the formula is $bx+c{x}^2=x(b+cx)$. Set $\sqrt{bx+c{x}^2}=mx/n$, giving

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

Application — triangular numbers that are also squares. The $x$-th triangular number $T_x=x(x+1)/2$ is a perfect square iff $x^2+x=2T_x$ is a perfect square. With $b=1$, $c=1$:

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

Selected values:

$m$	$n$	$x$	$T_x$	$\sqrt{T_x}$
3	2	8	36	6
7	5	49	1225	35
17	12	288	41616	204

(source: chapter-2.0.4, §49)

Rule 3: $b^2-4ac$ is a perfect square (§50–52)

If $b^2-4ac=d^2$ then the formula factors over the rationals. Setting $b^2-4ac=d^2$, the roots of $a+bx+c{x}^2=0$ are $(-b\pm d)/(2c)$, so

$a+bx+c{x}^2=(cx+\frac{b-d}{2})(x+\frac{b+d}{2c}).$

More generally for a product $(f+gx)(h+kx)$, set

$\sqrt{(f+gx)(h+kx)}=\frac{m(f+gx)}{n}.$

Then $h+kx=m^2(f+gx)/n^2$, so

$x=\frac{f{m}^2-h{n}^2}{k{n}^2-g{m}^2}.$

Rule 4: $p^2+qr$ form (§54)

When none of the first three rules applies, write $a+bx+c{x}^2=p^2+qr$ where $p$, $q$, $r$ are all linear in $x$. Set

$\sqrt{p^2+qr}=p+\frac{mq}{n},$

so $qr=2mpq/n+m^2{q}^2/n^2$. Dividing by $q$: $n^{2}r=2mnp+m^{2}q$, a linear equation in $x$.

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Worked Questions

Question 1 (§52): $2x^2-2$ is a square

Factor: $2x^2-2=2(x+1)(x-1)$. Set root $=m(x+1)/n$. Then:

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

With $m=n=1$: $x=3$, and $2(9)-2=16=4^2$. With $m=3$, $n=2$: $x=-17$, so $x=17$, and $2(289)-2=576=24^2$.

Question 2 (§53): $6+13x+6x^2$ is a square

$b^2-4ac=169-144=25$, so Rule 3 applies. Factors: $(2+3x)(3+2x)$. Set root $=m(2+3x)/n$:

$x=\frac{3n^2-2m^2}{3m^2-2n^2},\text{requires }\frac{2}{3}<\frac{m^2}{n^2}<\frac{3}{2}.$

With $m=6$, $n=5$: $m^2/n^2=36/25$ (in range), so $x=3/58$.

Question 3 (§55): $2x^2-1$ is a square

$a=-1,b=0,c=2$: $b^2-4ac=8$ (not a square); neither $a$ nor $c$ is a square. Rule 4: write $2x^2-1=x^2+(x-1)(x+1)$. Set root $=x+m(x+1)/n$:

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

With $m=1$, $n=2$: $x=5/7$, and $2(25/49)-1=1/49=(1/7{)}^2$.

Question 4 (§56): $2x^2+2$ is a square

Write $2x^2+2=4+2(x+1)(x-1)$. Set root $=2+m(x+1)/n$:

$x=\frac{4mn+m^2+2n^2}{2n^2-m^2}.$

With $m=n=1$: $x=7$, and $2(49)+2=100=10^2$.

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Bootstrap: Using a Known Solution (§58–62)

When a particular value $x_0$ is known to make $a+b{x}_0+c{x}_0^2=k^2$, substitute $x=x_0+y$:

$a+b(x_0+y)+c(x_0+y{)}^2=k^2+(b+2c{x}_0)y+c{y}^2.$

This new formula has $a^{\prime }=k^2$ (a perfect square), so Rule 2 applies. Setting root $=k+my/n$ and solving for $y$ gives an infinite family of solutions in $x$.

Example (§59): $2+7x^2$. At $x=1$: formula $=9=3^2$. Set $x=1+y$: formula $=9+14y+7y^2$. Rule 2 with root $=3+my/n$:

$y=\frac{6mn-14n^2}{7n^2-m^2},x=\frac{6mn-7n^2-m^2}{7n^2-m^2}.$

Example (§60): $5x^2+3x+7$. At $x=-1$: formula $=9$. Set $x=y-1$: formula $=5y^2-7y+9$. Root $=3-my/n$:

$x=\frac{2n^2-6mn+m^2}{5n^2-m^2}.$

Example (§61): $7x^2+15x+13$. Trial: $t=3,u=1$ gives $7(9)+45+13=121=11^2$, so $x=3$ works. Set $x=y+3$: formula $=7y^2+57y+121$. Root $=11+my/n$:

$x=\frac{36n^2-22mn+3m^2}{m^2-7n^2}.$

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Impossibility and Chapter V Preview (§62)

The formula $3x^2+2$ cannot be made into a perfect square for any rational $x$. Substituting $x=t/u$ gives $3t^2+2u^2$, which never yields a perfect square. Since such unmanageable formulas are numerous, Euler promises Chapter V will provide “characters by which their impossibility may be perceived.”

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Related pages

• indeterminate-analysis
• pythagorean-triples
• rationalization
• polygonal-numbers
• discriminant
• completing-the-square
• ch2.0.3-compound-indeterminate-equations
• ch1.4.7-extraction-roots-polygonal-numbers