add7-integer-quadratic-method

Additions Chapter VII — General Integer Method for $x^2-A{y}^2=B$

Summary: Lagrange’s direct, terminating procedure for finding all integer values of $y$ that make $\sqrt{A{y}^2+B}$ rational, equivalently solving $x^2-A{y}^2=B$ in integers; reduces an arbitrary quadratic Diophantine equation in two unknowns to a Pell equation $t^2-A{u}^2=1$ via continued fractions.

Sources: additions-7 (Articles 64–84).

Last updated: 2026-05-10.

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This chapter is the Appendix to Chapter VI of the Additions and is the first complete published algorithm for the general indeterminate quadratic of two unknowns. Articles 1–5 of Chapter VI had given only rational solutions; here Lagrange gives a method that decides solvability in integers and, when solvable, exhibits all solutions.

The reduction (Arts. 64–65)

Article 64 — square-free reduction. If $B$ has no square factor, then in any integer solution $y$ must be prime to $B$. (Otherwise: from $x^2=A{y}^2+B$, a prime $\alpha$ dividing both $y$ and $B$ would force ${\alpha }^2\mid B$, contradicting square-freeness.) When $B$ has square factors ${\alpha }^2,{\beta }^2,\dots$, one resolves each branch $x^2-A{y}^2=B/{\alpha }^2,B/{\beta }^2,\dots$ separately and multiplies the resulting $x,y$ by $\alpha ,\beta ,\dots$.

Article 65 — substitution to unit RHS. Assuming $y$ prime to $B$, the substitution $x=ny-Bz$ (Chapter IV/Art. 48) yields $(n^2-A)y^2-2nByz+B^2{z}^2=B.$ Since the second and third terms are visibly divisible by $B$, the term $(n^2-A)y^2$ must be too, and since $y$ is prime to $B$, this forces $B\mid n^2-A.$ Setting $C=(n^2-A)/B$ and dividing through gives the auxiliary equation $C{y}^2-2nyz+B{z}^2=1.$ The right side is now unity, much simpler than $B$. To find admissible $n$, scan $|n|\le B/2$ and keep only those with $n^2\equiv A(\mathrm{mod}B)$. If none exist, $x^2-A{y}^2=B$ is unsolvable in integers. Lagrange notes (cf. Mémoires de Berlin 1767, pp. 194 and 274) that a priori methods can find these $n$ in many cases.

Solving the auxiliary $C{y}^2-2nyz+B{z}^2=1$ — first method (Arts. 66–69)

The discriminant of the binary form on the left is $(2n{)}^2-4BC=4(n^2-BC)=4A$, so the trichotomy is governed by the sign and shape of $A$ (after using $n^2-BC=A$).

Case $A<0$ (Art. 67). Reduce $n/C$ to a continued fraction, form the convergents, and try the numerators for $y$ and denominators for $z$. Either a solution is found or the equation is certified unsolvable.

Case $A>0$, non-square (Art. 68). Apply the method of Art. 33ff: with $E=4A$, expand $a=(n\pm \sqrt{A})/C$ via the simultaneous recursions $Q^{(0)}=-n,P^{(0)}=C,Q^{(k+1)}={\mu }^{(k)}{P}^{(k)}+Q^{(k)},P^{(k+1)}=\frac{Q^{(k+1)2}-A}{P^{(k)}},$ with ${\mu }^{(k)}$ chosen so that $|-Q^{(k)}\mp \sqrt{A}/P^{(k)}|$ is bounded. Continue until two pairs repeat; if any $P^{(k)}=+1$, the corresponding convergent gives $(y,z)$. Otherwise the equation has no integer solutions.

Case $A$ a square (Art. 69). With $A=a^2$, the form $C{y}^2-2nyz+B{z}^2$ factors into two rational linear factors $cy\pm fz$ and $by\pm gz$. Their product equals $1$ only if each is $\pm 1$, giving an immediate finite check.

Solving the auxiliary — second method (Arts. 70–71)

This is Lagrange’s reduction algorithm in its mature form. Repeatedly apply substitutions $y=m{y}^{\prime }+y^{\prime \prime }$ to drive the middle coefficient $|2n|$ below the smaller of the two outer coefficients. The new $n^{\prime },n^{\prime \prime },\dots$ form a strictly decreasing sequence of non-negative integers, so the process terminates.

After reduction, the form becomes $L{\xi }^2-2N\xi \psi +M{\psi }^2=1\text{with}{N}^2-LM=A,|2N|\le L,M.$ Multiplying by $M$ and setting $v=M\psi -N\xi$ yields $v^2-A{\xi }^2=M.$ A short case analysis (Art. 71) shows:

• If $A<0$: solvability requires $M=1$, after which $\xi =0,\psi =\pm 1$.
• If $A>0$ non-square: $M<\sqrt{A}$ and the equation falls under the theorem of Art. 38.

Lagrange comments that this method has the advantage of not requiring trial; the calculation depending on $A$ alone (Art. 71) serves all equations with the same $A$.

Generating all solutions via Pell (Arts. 72–75)

Once a single solution $(p,q)$ of the auxiliary $C{y}^2-2nyz+B{z}^2=1$ is known, all others are derived as follows. Pick $r,s$ with $ps-qr=1$ (always possible since $p,q$ are coprime). The substitution $y=pt+ru$, $z=qt+su$ transforms the auxiliary into $P{t}^2+2Qtu+R{u}^2=1$ with $P=C{p}^2-2npq+B{q}^2=1$, $Q=Cpr-n(ps+qr)+Bqs$, $R=C{r}^2-2nrs+B{s}^2$, and $Q^2-PR=A$. Choosing the free integer $m$ in $r=\rho +mp$, $s=\sigma +mq$ to make $Q=0$ reduces it to the pell-equation $t^2-A{u}^2=1.$ Article 73 dispatches the cases $A<0$ (only $u=0,t=1$, hence at most one solution) and $A$ a positive square (only $t=\pm 1,u=0$). The interesting case is $A>0$ non-square, where infinitely many solutions exist (Art. 37).

Article 75 — closed form. With $(t^{\prime },u^{\prime })$ the least solution of $t^2-A{u}^2=1$, all solutions arise from $t\pm u\sqrt{A}=(t^{\prime }\pm u^{\prime }\sqrt{A}{)}^m,m\in {\mathbb{Z}}_{>0},$ giving the rational expansions $t=\frac{(T+V\sqrt{A}{)}^m+(T-V\sqrt{A}{)}^m}{2},u=\frac{(T+V\sqrt{A}{)}^m-(T-V\sqrt{A}{)}^m}{2\sqrt{A}}.$ Lagrange supplies a binomial-theorem expansion of these and proves they exhaust all solutions: any putative intermediate $(\theta ,v)$ between consecutive powers $m,m+1$ would yield a $u<V$, contradicting minimality of $V$.

Generalisation (Arts. 76–80)

The same machinery resolves the most general two-variable quadratic Diophantine equation $a{r}^2+brs+c{s}^2+dr+es+f=0.$ Writing $A=b^2-4ac$, $g=bd-2ae$, $h=d^2-4af$, one needs $A{s}^2+2gs+h$ to be a square, say $y^2$. Setting $B=g^2-Ah$ reduces this to making $\sqrt{A{y}^2+B}$ rational — exactly the case treated above. The values of $r,s$ are then recovered by integrality conditions. Article 77 shows that $r,s$ depend on $t,u$ via expressions $r=(\alpha t+\beta u+\gamma )/\delta$, $s=({\alpha }^{\prime }t+{\beta }^{\prime }u+{\gamma }^{\prime })/{\delta }^{\prime }$, and Articles 78–79 give a finite algorithm for the residues of $t,u$ modulo any divisor $\Delta$. The scholium (Art. 80) further extends to the homogeneous form $a{r}^2+2brs+c{s}^2=f$ and to higher-degree binary forms $a{r}^m+b{r}^{m-1}s+\cdots =f$.

Examples (Arts. 81–83)

Example 1 (Art. 81). Make $\sqrt{30+62s-7s^2}$ rational. After multiplying by $7$ and substituting $x=7s-31$, the equation becomes $x^2+7y^2=1171$. Here $A=-7$, $B=1171$ (prime). Searching $|n|<580$ with $n^2\equiv -7(\mathrm{mod}1171)$ yields $n=\pm 321$; the auxiliary is $88y^2\mp 642yz+1171z^2=1.$ The reduction algorithm of Art. 70 produces successive coefficients $1171,88,11,1,7$; making the leading coefficient $1$ gives the solution. Tracing back: $s=7$ (and the fractional value $s=13/7$, which still rationalises the radical to $11$).

Example 2 (Art. 82). Find all $y$ making $\sqrt{13y^2+101}$ rational. Here $A=13$, $B=101$, $n=\pm 35$ (since $35^2-13=1212=12\cdot 101$). After reduction the auxiliary becomes $y^{\prime \prime \prime 2}-13y^{\prime \prime 2}=1$, the standard $\sqrt{13}$ Pell equation. Using the table of Art. 41: convergents of $\sqrt{13}$ give the primitive values $y=74$ (and $-34$); from $t^2-13u^2=1$ with $(T,V)=(649,180)$, all solutions are $y=\pm 74t\pm 267u,y=\pm 34t\pm 123u,$ where $t,u$ run through powers of $(649+180\sqrt{13}{)}^m$.

Example 3 (Art. 83). Find all $y$ making $\sqrt{79y^2+101}$ rational. Here $A=79$, $B=101$, $n=\pm 33$. The reduction quickly produces $(7y^{\prime }\pm 3y^{\prime \prime }{)}^2-79y^{\prime \prime 2}=-7$. Consulting the $\sqrt{79}$ table (Art. 41), the upper series is $1,15,2$ (cyclically) and never contains $7$; so the equation has no integer solution.

Refutation of Euler’s induction rule (Art. 84 — Scholium)

Euler, in Novi Commentarii Petrop. vol. IX, conjectured by induction that $x^2-A{y}^2=B$ is solvable whenever $B$ is a prime of the form $4An+r^2-A$ for some $r$. Example 3 disproves this: $101$ is prime and equals $4\cdot 79\cdot (-4)+38^2-79$ (so $A=79$, $n=-4$, $r=38$), yet $x^2-79y^2=101$ admits no integer solution. Lagrange even strengthens his counterexample: $733=4\cdot 79\cdot (-2)+101$ is prime and has the form $x^2-79y^2$ (with $x=38,y=3$), yet $101$ does not — refuting also the strengthened rule that requires $b$ itself to be prime.

This scholium marks an early instance of an inductive number-theoretic conjecture being decisively disproved by an algorithmic decision procedure.

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