add5-rational-quadratic-surds

Additions Chapter V — Rational Solutions of Quadratic Surds

Summary: Lagrange’s fifth Addition gives the first systematic decision procedure for rationality of $\sqrt{a+bx+c{x}^2}$. Reducing to the canonical equation $A{p}^2+B{q}^2=z^2$, he constructs a strictly decreasing sequence $A,A^{\prime },A^{\prime \prime },\dots$ of integer coefficients via residue substitutions $z=nq-A{q}^{\prime }$ with $|n|\le A/2$ and $A\mid n^2-B$. Termination is guaranteed: either the procedure exhibits a rational solution, or it reaches an impossibility — no trial-and-error required.

Sources: additions-5

Last updated: 2026-05-10

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Position in the Additions (Appendix to Chapter IV)

This chapter occupies Articles 49–60 (pp. 470–477) and is the centerpiece of the Additions: a complete, finite, mechanical algorithm for a problem Euler treated only by isolated tricks in ch2.0.4-surd-rationalization.

It is the Appendix to Chapter IV of the Additions — i.e. it depends on the residue-condition lemma of Add. IV Art. 48.

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Reduction to the Canonical Form (Arts. 49–51)

Step 1 — Eliminate the linear term (Art. 49)

For $\sqrt{a+bx+c{x}^2}=y$, square and rearrange:

$2cx+b=\sqrt{4c{y}^2+b^2-4ac}.$

So $\sqrt{a+bx+c{x}^2}$ is rational iff $\sqrt{4c{y}^2+(b^2-4ac)}$ is rational. Setting $A=4c$, $B=b^2-4ac$:

$\boxed{\sqrt{A{y}^2+B}\text{ rational}}$

with $A,B\in \mathbb{Z}$ (any signs).

Step 2 — Strip square factors (Art. 50)

If $A={\alpha }^2{A}^{\prime }$ and $B={\beta }^2{B}^{\prime }$ where $A^{\prime },B^{\prime }$ are square-free, then

$A{y}^2+B={\alpha }^2{A}^{\prime }{y}^2+{\beta }^2{B}^{\prime }.$

Setting $y^{\prime }=\alpha y/\beta$ reduces the problem to making $A^{\prime }{y}^{\prime 2}+B^{\prime }$ a square (and afterwards $y=(\beta /\alpha )y^{\prime }$).

Henceforth assume $A,B$ square-free.

Step 3 — Clear denominators (Art. 51)

Write $y=p/q$ with $\gcd (p,q)=1$. The condition becomes

$A{p}^2+B{q}^2=z^2$

for integers $p,q,z$.

Coprimality lemma: $\gcd (q,A)=1$ and $\gcd (p,B)=1$.

Proof: If a prime $\ell$ divided both $q$ and $A$, then ${\ell }^2\mid A{p}^2$ (since $A$ is square-free, $\ell \mid A$ to the first power; combined with ${\ell }^2\mid q^2$, we get ${\ell }^2\mid B{q}^2=z^2-A{p}^2$, but $\ell$ divides $A{p}^2$ only to the first power — contradiction). Similarly for $\gcd (p,B)$.

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Lagrange’s Reduction Algorithm (Arts. 52–55)

WLOG $A>B$ in absolute value. The algorithm produces a strictly decreasing sequence $|A|,|A^{\prime }|,|A^{\prime \prime }|,\dots$ until either a square coefficient appears (success) or no valid residue exists (failure).

Single Reduction Step (Arts. 52–53)

Rewrite $A{p}^2=z^2-B{q}^2$. Since $\gcd (q,A)=1$, the Add. IV Art. 48 lemma gives $z=nq-A{q}^{\prime }$ with $n,q^{\prime }\in \mathbb{Z}$ and $|n|\le A/2$. Substituting:

$A{p}^2=(nq-A{q}^{\prime }{)}^2-B{q}^2=(n^2-B)q^2-2nAq{q}^{\prime }+A^2{q}^{\prime 2}.$

For this to be divisible by $A$, the constant term $n^2-B$ must be divisible by $A$. Writing $n^2-B=A{A}^{\prime }$ and dividing through by $A$:

$\boxed{p^2=A^{\prime }{q}^2-2nq{q}^{\prime }+A{q}^{\prime 2}}$

with

$A^{\prime }=\frac{n^2-B}{A}.$

Why $|A^{\prime }|<|A|$

Since $|n|\le A/2$ and $|B|<|A|$:

$|A^{\prime }|=\frac{|n^2-B|}{|A|}\le \frac{(A/2{)}^2+|B|}{|A|}\le \frac{A/4+1}{1}\cdot \text{something}$

— more carefully, $|A^{\prime }|\le \max (|n^2/A|,|B/A|)<|A|$. So the new leading coefficient is strictly smaller.

Three Outcomes (Art. 53)

After one reduction step, examine $A^{\prime }$:

1. $A^{\prime }$ is a square → solved by inspection: take $q^{\prime }=0$, $q=1$, $p=\sqrt{A^{\prime }}$.

2. $|A^{\prime }|<|B|$ (or becomes so after stripping square factors) → multiply the equation by $A^{\prime }$ and use $A{A}^{\prime }-n^2=-B$:

   $A^{\prime }{p}^2=(A^{\prime }q-n{q}^{\prime }{)}^2-B{q}^{\prime 2}$

   so $\sqrt{B{p}^{\prime 2}+A^{\prime }}\cdot \sqrt{?}$… rearranging: $B{y}^{\prime 2}+C$ must be a square (with $C=A^{\prime }$, $y^{\prime }=q^{\prime }/p$). The roles of "$A$" and "$B$" have swapped, with smaller integers. Recurse.

3. $|A^{\prime }|\ge |B|$ → set $q=v{q}^{\prime }+q^{\prime \prime }$ with $v$ chosen so $|n-v{A}^{\prime }|\le A^{\prime }/2$, get

   $p^2=A^{\prime }{q}^{\prime \prime 2}-2n^{\prime }{q}^{\prime \prime }{q}^{\prime }+A^{\prime \prime }{q}^{\prime 2}$

   where $n^{\prime }=n-v{A}^{\prime }$, $A^{\prime \prime }=(n^{\prime 2}-B)/A^{\prime }$, and again $|A^{\prime \prime }|<|A^{\prime }|$. Iterate inside the same problem until reaching outcome 1 or 2.

Termination (Art. 55)

The descent forces a strictly decreasing sequence of positive integers

$|A|>|A^{\prime }|>|A^{\prime \prime }|>\cdots >|B|,$

then (after a swap) $|B|>|C|>|D|>\cdots$, and so on. By infinite descent, the procedure terminates in finitely many steps with one of:

• A square coefficient (problem solvable, root explicitly constructed by back-substitution)
• An impossibility: at some step no valid $n$ with $|n|\le A^{(k)}/2$ and $A^{(k)}\mid n^2-B^{(k)}$ exists.

This is the first complete decision procedure for rationality of $\sqrt{a+bx+c{x}^2}$ — see lagrange-reduction-algorithm.

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Worked Example 1 — $7+15x+13x^2=\square$ (Art. 56)

Setup: $a=7$, $b=15$, $c=13$. Then $4c=52$ and $b^2-4ac=225-364=-139$. Need $52y^2-139=\square$.

Strip square factor: $A=52=4\cdot 13$, so reduce to $A=13$ (and remember to divide final $y$ by 2).

Setup: $13p^2-139q^2=z^2$. Since $|139|>|13|$, swap roles: $-139p^2+13q^2=z^2$, i.e. $-139p^2=z^2-13q^2$.

First reduction: Need $n$ with $|n|\le 139/2<70$ and $139\mid n^2-13$.

• Try $n=41$: $n^2-13=1681-13=1668=139\cdot 12$. ✓
• New equation (divide by $-139$): $p^2=-12q^2+82q{q}^{\prime }-139q^{\prime 2}$.

$A^{\prime }=-12$, not a square; $|-12|<|-13|$, so swap. Multiply by $-12$:

$-12p^2=(-12q+41q^{\prime }{)}^2-13q^{\prime 2}.$

Setting $y^{\prime }=q^{\prime }/p$, need $13y^{\prime 2}-12=\square$. Strip square factor of $12=4\cdot 3$: reduce to $13r^2-3=\square$ (and remember to multiply final $y^{\prime }$ by 2).

Second reduction: $13r^2-3s^2=z^{\prime 2}$. Need $m$ with $|m|\le 13/2<7$ and $13\mid m^2+3$.

• Try $m=6$: $36+3=39=13\cdot 3$. ✓
• New equation: $r^2=3s^2-12s{s}^{\prime }+13s^{\prime 2}$.

$A^{\prime \prime }=3$, not a square. Apply Art. 54: set $s=\mu s^{\prime }+s^{\prime \prime }$ with $\mu$ chosen so that $|6-3\mu |\le 3/2$. Take $\mu =2$: $6-3\cdot 2=0$. New equation:

$r^2=3s^{\prime \prime 2}+s^{\prime 2}.$

The coefficient of $s^{\prime 2}$ is $1=1^2$ — a square! Solved.

Simplest solution: $s^{\prime \prime }=0$, $s^{\prime }=1$, $r=1$, hence $s=\mu =2$, so $y^{\prime }=r/s=1/2$. Doubling for the stripped factor: $y^{\prime }=1$.

Back-substitution: $q^{\prime }/p=1$, so $q^{\prime }=p$. From $-12p^2=(-12q+41p{)}^2-13p^2$ we get $(-12q+41p{)}^2=p^2$, i.e. $-12q+41p=\pm p$.

• Case $+$: $12q=40p$, so $y=q/p=10/3$. Halving (for the original square strip): $y=5/3$.
• Case $-$: $12q=42p$, so $y=q/p=7/2$. Halving: $y=7/4$ — but we need to recompute: actually $y=21/12=7/4$, but the chapter writes $y=21/12$ and then directly substitutes.

Setting $7+15x+13x^2=(5/3{)}^2=25/9$: $13x^2+15x+7-25/9=0$, i.e. $117x^2+135x+38=0$. By the quadratic formula: $x=\frac{-135\pm 21}{234}$, giving $\boxed{x=-19/39}$ or $\boxed{x=-2/3}$.

The Case $-$ root yields $x=-21/52$ or $x=-3/4$.

General Parametrization (Art. 57 Scholium)

Once one rational $x=f$ is known with $a+bf+c{f}^2=g^2$, all rational $x$ making the form a square are given by

$\boxed{x=\frac{f{m}^2-2gm+b+cf}{m^2-c}}$

with $m\in \mathbb{Q}$ a free parameter.

Proof sketch: Set $\sqrt{a+bx+c{x}^2}=g+m(x-f)$ and use $a+bf+c{f}^2=g^2$ to eliminate the constants; solving for $x$ gives the formula. This is a chord-method argument: given one rational point on the conic, intersect with a line of variable rational slope $m$ to sweep out all other rational points.

For Example 1 with $f=-2/3$, $g=5/3$:

$x=\frac{-(2/3)m^2-(10/3)m+15+13\cdot (-2/3)}{m^2-13}=\frac{19-10m-2m^2}{3(m^2-13)}.$

This is the general rational expression for all $x$ making $7+15x+13x^2$ a square.

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Worked Example 2 — $23y^2-5=\square$ Impossible (Art. 58)

$A=23$, $B=-5$, both square-free. So $23p^2-5q^2=z^2$.

First reduction: Need $n$ with $|n|\le 23/2<12$ and $23\mid n^2+5$. Try $n=0,\pm 1,\dots ,\pm 11$:

$n^2+5(\mathrm{mod}23)$: $5,6,9,14,21,7,18,8,0,\dots$ — at $n=8$: $64+5=69=23\cdot 3$. ✓

(Lagrange notes this is the only value.) New equation: $p^2=3q^2-16q{q}^{\prime }+23q^{\prime 2}$.

$A^{\prime }=3$, less than $|B|=5$. Multiply by $3$ and rearrange:

$3p^2=(3q-8q^{\prime }{)}^2+5q^{\prime 2}.$

Setting $y^{\prime }=q^{\prime }/p$, need $-5y^{\prime 2}+3=\square$. Square-free $A_{\text{new}}=5$, $B_{\text{new}}=3$.

Second reduction: $-5r^2+3s^2=z^{\prime 2}$, i.e. $-5r^2=z^{\prime 2}-3s^2$. Need $m$ with $|m|\le 5/2<3$ and $5\mid m^2-3$.

Try $m=0,1,2$: $m^2-3=-3,-2,1$ — none divisible by $5$.

Conclusion: The procedure terminates with no valid residue. The original equation $23y^2-5=\square$ has no rational solution — proved by exhaustive descent.

This is a remarkable result: certifying impossibility in finitely many steps, without ad-hoc tricks. Compare the residue-class arguments of ch2.0.5-impossibility-quadratic-squares which are typically tailored case-by-case.

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Corollary — General Quadratic in Two Unknowns (Art. 59)

If $a+bx+cy+d{x}^2+exy+f{y}^2=0$ should be made to admit rational $(x,y)$, solve for $y$:

$2fy+ex+c=\pm \sqrt{(c-ex{)}^2-4f(a+bx+d{x}^2)}=\pm \sqrt{\alpha +\beta x+\gamma x^2}$

where $\alpha =c^2-4af$, $\beta =2ce-4bf$, $\gamma =e^2-4df$. Apply Lagrange’s algorithm to $\sqrt{\alpha +\beta x+\gamma x^2}$ — every rational quadratic in two unknowns reduces to the canonical problem.

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Forward References (Art. 60 Scholium)

Lagrange points the reader to:

• Mémoires de l’Académie de Berlin, 1767: “Nouvelle Méthode pour résoudre les Problèmes Indéterminés” — the original presentation; first direct method, no trial.
• Mémoires, 1770 and following: theory of the form of divisors of $z^2-B{q}^2$ — a precursor to genus theory and quadratic reciprocity. By inspecting the form of $A$ alone, one can sometimes certify $A{p}^2=z^2-B{q}^2$ unsolvable without running the full descent.

This connects to the composition law developed in Add. II via Pell’s equation and the proto-genus phenomena Euler observed in ch2.0.11-quadratic-form-factorization.

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Related pages

• lagrange-reduction-algorithm — concept page on the descending $A,A^{\prime },A^{\prime \prime },\dots$ procedure
• add4-integer-method-linear-y — Art. 48 lemma that justifies $z=nq-A{q}^{\prime }$
• ch2.0.4-surd-rationalization — Euler’s case-by-case rationalization rules (the problem this chapter systematizes)
• ch2.0.5-impossibility-quadratic-squares — Euler’s residue-class impossibility arguments
• rationalization — concept page on radical elimination
• indeterminate-analysis — overarching framework
• pythagorean-triples — Art. 57 parametrization is the chord-method analog for the unit circle
• pell-equation — Art. 60 forward references
• binary-quadratic-forms — the form-divisor theory mentioned in Art. 60
• infinite-descent — the proof technique underlying termination