add6-double-triple-equalities

Additions Chapter VI — Double and Triple Equalities

Summary: Lagrange’s sixth Addition addresses the Diophantine problem of making two or three formulas simultaneously equal to perfect squares. Linear double equalities and certain triple equalities reduce to a simple equality soluble by Add. V. But two simultaneous quadratics in one variable lead to a quartic equality with no general method — pointing forward to Chapter IX.

Sources: additions-6

Last updated: 2026-05-10

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Position in the Additions

This chapter occupies Articles 61–63 (pp. 478–480). It is short — three articles — and acts as an overview chapter locating the reach and limits of Lagrange’s method against the Diophantine vocabulary inherited from Bachet, Fermat, and Euler.

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Terminology (Art. 61)

In the Diophantine tradition:

• Simple equality: one formula in one or more unknowns to be made a perfect power (square, cube, etc.).
• Double equality: two formulas in the same unknowns, each to be made a perfect power.
• Triple equality: three formulas, each a perfect power.

Lagrange’s Add. V resolves all simple equalities of degree $\le 2$ when the target power is a square. This chapter asks: how far does that machinery extend?

See double-and-triple-equalities for the concept page.

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Linear Double Equality (Art. 62, first form)

Problem: Solve in rational $x$:

$a+bx=\square \text{and}c+dx=\square .$

Reduction: Set $a+bx=t^2$ and $c+dx=u^2$. Eliminating $x$:

$\frac{t^2-a}{b}=\frac{u^2-c}{d}⟹d{t}^2-b{u}^2=ad-bc.$

Multiply through by $d$:

$\boxed{(dt{)}^2=db{u}^2+d(ad-bc).}$

So the difficulty reduces to making $db{u}^2+d(ad-bc)$ a square — a simple equality in $u$, of the form $\sqrt{A{u}^2+B}$ rational with $A=db$, $B=d(ad-bc)$. Lagrange’s algorithm of Add. V applies directly.

After finding $u$, recover $x=(u^2-c)/d$.

Variant — Quadratic terms with no constant (Art. 62, second form)

Problem:

$a{x}^2+bx=\square \text{and}c{x}^2+dx=\square .$

Trick: Substitute $x=1/x^{\prime }$ and multiply each formula by $x^{\prime 2}$:

$a+b{x}^{\prime }=\square ,c+d{x}^{\prime }=\square ,$

reducing back to the first form.

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Hard Case — Two Quadratics with Constants (Art. 62, third form)

Problem:

$a+bx+c{x}^2=\square \text{and}\alpha +\beta x+\gamma x^2=\square .$

Lagrange’s approach: Solve the first equality by his algorithm; if $f$ is one solution with $a+bf+c{f}^2=g^2$, the Art. 57 parametrization gives all rational $x$ as

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

Substituting into the second formula and clearing $(m^2-c{)}^2$:

$\alpha (m^2-c{)}^2+\beta (m^2-c)(f{m}^2-2gm+b+cf)+\gamma (f{m}^2-2gm+b+cf{)}^2=\square .$

This is a quartic in $m$ to be made a square. No general method exists for this — Lagrange remarks: “all we can do is to find successively different solutions, when we already know one.”

Forward reference: Chapter IX of the Additions treats methods for chaining solutions of degree-4 equalities (the Fermat secant-and-tangent method in modern language).

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Triple Equality with Two Unknowns (Art. 63)

Problem: Make all three of

$ax+by,cx+dy,hx+ky$

simultaneously squares.

Setup: $ax+by=t^2$, $cx+dy=u^2$, $hx+ky=s^2$. Eliminate $x$ and $y$ from the three equations using two of them; the third gives a constraint:

$(ak-bh)u^2-(ck-dh)t^2=(ad-cb)s^2.$

Setting $z=u/t$, divide by $t^2$:

$\boxed{\frac{ak-bh}{ad-cb}{z}^2-\frac{ck-dh}{ad-cb}=\square .}$

This is a simple equality in $z$ — apply Add. V’s algorithm.

Having found $z$, set $u=tz$. The first two equations give

$x=\frac{d-b{z}^2}{ad-cb}{t}^2,y=\frac{a{z}^2-c}{ad-cb}{t}^2.$

So the two-unknown triple equality is solvable by Lagrange’s machinery.

Single-unknown reduction (Art. 63, end)

If we restrict to $y=1$ (only $x$ free), the problem becomes “find $x$ making $ax+b$, $cx+d$, $hx+k$ all squares” — but the constraint $y=1$ adds the equation $(a{z}^2-c)/(ad-cb)=1/t^2$, which combined with the previous simple equality leads (after setting $f$ to be one solution and using the Art. 57 parametrization in $z$) to:

$\frac{a(f{m}^2-2gm+ef{)}^2-c(m^2-e{)}^2}{ad-cb}=\square$

— again a quartic in $m$, no general method.

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Summary of the Method’s Reach

Problem	Equality form	Lagrange’s method	Status
Simple, degree 1	$a+bx=\square$	Direct (trivial)	✓
Simple, degree 2	$a+bx+c{x}^2=\square$	Add. V algorithm	✓
Double linear	$a+bx,c+dx$ both $\square$	Reduces to simple	✓
Double linear-after-substitution	$a{x}^2+bx,c{x}^2+dx$ both $\square$	$x\mapsto 1/x^{\prime }$ then simple	✓
Double quadratic	$a+bx+c{x}^2,\alpha +\beta x+\gamma x^2$ both $\square$	Quartic — no general rule	✗
Triple in two unknowns	three linear forms in $x,y$ all $\square$	Reduces to simple	✓
Triple in one unknown	three linear forms in $x$ all $\square$	Quartic — no general rule	✗

The pattern: whenever elimination produces a single quadratic in one variable, Add. V solves it; whenever it produces a quartic, the problem escapes the algorithm.

This boundary is precisely where elliptic curves would later enter — but for Lagrange in 1771, the chapter ends with the honest admission: “we have not yet any general rule.”

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Related pages

• double-and-triple-equalities — concept page on the Diophantine terminology
• add5-rational-quadratic-surds — the simple-equality solver invoked throughout
• lagrange-reduction-algorithm — the descent procedure used for each simple equality
• indeterminate-analysis — overarching framework
• ch2.0.9-quartic-surd-rationalization — Euler’s main-text treatment of related quartic problems
• ch2.0.14-questions-squares — Euler’s showcase Diophantine questions, several of which are double equalities in disguise