Double and Triple Equalities

Summary: A simple equality in the Diophantine tradition is one formula to be made a perfect power; a double equality is two such conditions on the same unknowns simultaneously, and a triple equality three. Lagrange’s Add. VI shows that every linear double equality and every linear-in-two-unknowns triple equality reduces to a simple equality solvable by his Add. V algorithm — but two simultaneous quadratics in one unknown produces a quartic equality with no general method.

Sources: additions-6

Last updated: 2026-05-10

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Terminology

Name	Form	Number of formulas
Simple equality	$f(x,y,\dots )=\square$ (or cube, etc.)	1
Double equality	$f_1=\square$ and $f_2=\square$	2
Triple equality	$f_1=f_2=f_3=\square$	3

The target power is most often a square (sometimes a cube, as in ch2.0.10-cubic-formula-as-cube and ch2.0.15-questions-cubes). Lagrange treats only the square case.

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Reducible Cases

Any of the following double or triple equalities reduces, by elimination, to a simple equality solvable by Lagrange’s reduction algorithm:

Linear double, one unknown (Add. VI Art. 62)

$a+bx=\square ,c+dx=\square ⟹db{u}^2+d(ad-bc)=\square .$

Linear double with no constant, one unknown (Add. VI Art. 62)

$a{x}^2+bx=\square ,c{x}^2+dx=\square \stackrel{x=1/x^{\prime }}{\to }a+b{x}^{\prime }=\square ,c+d{x}^{\prime }=\square ,$

then as in the previous case.

Triple linear, two unknowns (Add. VI Art. 63)

$ax+by=\square ,cx+dy=\square ,hx+ky=\square$

eliminates to a single simple equality

$\frac{ak-bh}{ad-cb}{z}^2-\frac{ck-dh}{ad-cb}=\square \text{where }z=u/t.$

The full $(x,y)$ solution is then read off from the original first two equations.

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Hard Cases

Double quadratic, one unknown

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

Substituting the Art. 57 parametrization for the first equality into the second produces a quartic in the parameter $m$ — no general method.

Triple linear, one unknown

$ax+b=\square ,cx+d=\square ,hx+k=\square$

(equivalently the two-unknown triple equality with $y=1$ adjoined as a constraint) — also leads to a quartic.

The hard cases are exactly those where elimination yields a degree-4 condition. Diophantine quartics escape the Add. V machinery and were the subject of Add. IX (the secant-and-tangent chord method that historically anticipated elliptic-curve arithmetic).

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Connection to the Main Text

Several of Euler’s Questions on Squares are double equalities in disguise:

• ch2.0.14-questions-squares §217–218: “If $a+x$ and $a-x$ are both squares, then $a$ is a sum of two squares” — a double equality with no $x^2$ terms (linear case, solvable).
• ch2.0.14-questions-squares §229–230: ”$p^2+q^2$ and $p^2+3q^2$ are never both squares” — double quadratic, proved impossible by infinite descent rather than by Lagrange’s method.

For cubes:

• ch2.0.15-questions-cubes features problems like “find $x$ with $x+1$ and $x^2+1$ both cubes” — a cube-target double equality, again outside Lagrange’s framework (he treats only square targets).

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Related pages

• add6-double-triple-equalities — the full chapter exposition
• add5-rational-quadratic-surds — the simple-equality solver invoked by reductions
• lagrange-reduction-algorithm — the descent procedure applied to each reduced simple equality
• ch2.0.14-questions-squares — main-text problems that are implicit double equalities
• ch2.0.15-questions-cubes — main-text problems with cube-target equalities
• indeterminate-analysis — overarching framework