ch2.0.11-quadratic-form-factorization

Ch2.0.11 — Of the Resolution of the Formula $a{x}^2+bxy+c{y}^2$ into its Factors

Summary: Investigates when the binary quadratic form $a{x}^2+bxy+c{y}^2$ admits multiple integer factorizations. Three cases by the sign and squareness of the discriminant $b^2-4ac$. The middle term is removed by $x=(z-by)/(2a)$, reducing to $a{x}^2+c{y}^2$. Imaginary factorizations $(x+y\sqrt{-c})(x-y\sqrt{-c})$ produce the Brahmagupta-Fibonacci identity $(p^2+c{q}^2)(r^2+c{s}^2)=x^2+c{y}^2$, which generates compound factorizations and explains why some primes split as sums of squares. Concludes with proto-genus theory: the forms $2x^2+3y^2$ and $x^2+6y^2$ multiply by the rule $\mathrm{I}\times \mathrm{I}=\mathrm{II},\mathrm{I}\times \mathrm{II}=\mathrm{I},\mathrm{II}\times \mathrm{II}=\mathrm{II}$.

Sources: chapter-2.0.11

Last updated: 2026-05-09

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The Three Cases (§162–163)

Restrict $x,y$ to integers (any rational solution reduces here by clearing denominators). Three cases govern factorability:

Case	Condition on $b^2-4ac$	Form’s factors
1	positive square	two distinct rational factors $(fx+gy)(hx+ky)$
2	zero	one repeated rational factor $(fx+gy{)}^2$
3a	positive non-square	two simply irrational factors
3b	negative	two complex (imaginary) factors

This is the discriminant criterion familiar from ch1.4.9-nature-quadratic-equations and discriminant, but used here for factorability over $\mathbb{Q}$ rather than for real-vs-imaginary roots.

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Eliminating the Middle Term (§167)

For the irrational/complex case it is convenient to remove $bxy$. Substitute $x=(z-by)/(2a)$:

$a{x}^2+bxy+c{y}^2=\frac{z^2+(4ac-b^2)y^2}{4a}.$

So the analysis reduces to the diagonal form $a{x}^2+c{y}^2$ (after rescaling).

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Case 1: Rational Factors (§164–165)

If $a{x}^2+bxy+c{y}^2=(fx+gy)(hx+ky)$, set each factor to a sub-product: $fx+gy=pq$, $hx+ky=rs$. Then $a{x}^2+bxy+c{y}^2=pqrs$, with up to four factors. Solving the linear system in $x,y$:

$x=\frac{kpq-grs}{fk-hg},y=\frac{frs-hpq}{fk-hg}.$

Integrality requires the numerators be divisible by $fk-hg$ — typically arranged by choosing $p,r$ or $q,s$ as multiples of the denominator.

Worked example — $x^2-y^2=(x+y)(x-y)$. Setting $x+y=pq$ and $x-y=rs$:

$x=\frac{pq+rs}{2},y=\frac{pq-rs}{2},$

so $pq$ and $rs$ must both be odd or both even. Taking $p=7,q=5,r=3,s=1$: $x=19$, $y=16$, hence $19^2-16^2=105=7\cdot 5\cdot 3\cdot 1$.

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Case 2: Repeated Factor (§166)

If $a{x}^2+bxy+c{y}^2=(fx+gy{)}^2$, set $fx+gy=pqr$ to obtain $p^2{q}^2{r}^2$ — arbitrarily many square factors. One of $x,y$ remains free.

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Case 3: Imaginary Factors and the Brahmagupta-Fibonacci Identity (§169–171)

Here the formula factors only over $\mathbb{C}$:

$x^2+c{y}^2=(x+y\sqrt{-c})(x-y\sqrt{-c}).$

If $x^2+c{y}^2$ is a product of two integers each of the same form, then its complex factors must each split correspondingly. Set

$x+y\sqrt{-c}=(p+q\sqrt{-c})(r+s\sqrt{-c}),x-y\sqrt{-c}=(p-q\sqrt{-c})(r-s\sqrt{-c}).$

Expanding and collecting rational/irrational parts:

$\boxed{x=pr-cqs,y=ps+qr,}$

and the product identity

$\boxed{(p^2+c{q}^2)(r^2+c{s}^2)=x^2+c{y}^2.}$

This is Euler’s derivation of the brahmagupta-fibonacci-identity. See also the more general case-$c\to -c$ version below.

Two Representations from the Sign of $q$

Since only $q^2$ appears, $q\to -q$ gives a second pair:

$x=pr+cqs,y=ps-qr.$

Hence a number expressible as a product $(p^2+c{q}^2)(r^2+c{s}^2)$ admits two representations as $x^2+c{y}^2$. For $c=1$, $p=3,q=2,r=2,s=1$: $13\cdot 5=65=4^2+7^2=8^2+1^2$. Multiplying three such factors gives four representations: $5\cdot 13\cdot 17=1105=33^2+4^2=32^2+9^2=31^2+12^2=24^2+23^2$.

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Sums of Two Squares — Simple vs Compound (§172)

Numbers expressible as $x^2+y^2$ with $\gcd (x,y)=1$ split into:

• Compound: products of smaller representable numbers (use the identity above).
• Simple: irreducibly representable; namely $1,2,5,9,13,17,29,37,41,49,\dots$.

Among the simple numbers Euler observes:

• All odd primes in this list are $\equiv 1(\mathrm{mod}4)$ (i.e. of form $4n+1$).
• The squares $9,49,\dots$ in the list have roots $3,7,\dots$ in form $4n-1$ (which themselves are not representable, but their squares are).
• No number of form $4n-1$ is a sum of two squares (the sum of an even and an odd square is $\equiv 1(\mathrm{mod}4)$).

Euler then asserts (footnote 83): every prime of form $4n+1$ is a sum of two squares. “This is undoubtedly true, but it is not easy to demonstrate it” — Fermat’s two-square theorem, first proved by Euler. See sums-of-two-squares.

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The Form $x^2+2y^2$ (§173–174)

Same imaginary-factorization trick with $\sqrt{-2}$:

$(p^2+2q^2)(r^2+2s^2)=x^2+2y^2,x=pr-2qs,y=ps+qr.$

Numbers of form $x^2+2y^2$ up to 50: $1,2,3,4,6,8,9,11,12,16,17,18,19,22,24,25,27,32,33,34,36,38,41,43,44,49,50$. Simple ones (besides squares): $1,2,3,11,17,19,41,43$. All prime simples are $\equiv 1$ or $3(\mathrm{mod}8)$; primes $\equiv 5,7(\mathrm{mod}8)$ are never of this form.

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The General Form $x^2+c{y}^2$ (§175–176)

Same identity with general $c$:

$\boxed{(p^2+c{q}^2)(r^2+c{s}^2)=x^2+c{y}^2,x=pr\mp cqs,y=ps\pm qr.}$

Replacing $c$ by $-c$:

$(p^2-c{q}^2)(r^2-c{s}^2)=x^2-c{y}^2,x=pr\pm cqs,y=ps\pm qr.$

This gives the multiplication law for the Pell-related forms of ch2.0.7-pell-equation-method: solutions of $x^2-c{y}^2=1$ form a multiplicative semigroup under this composition.

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Coefficient on the First Term: $a{x}^2+c{y}^2$ (§177–179)

Now $a{x}^2+c{y}^2=(x\sqrt{a}+y\sqrt{-c})(x\sqrt{a}-y\sqrt{-c})$. The naïve copy of the previous trick fails (irrational $x,y$). The correct ansatz is:

$x\sqrt{a}+y\sqrt{-c}=(p\sqrt{a}+q\sqrt{-c})(r+s\sqrt{-ac}).$

Expanding gives

$\boxed{a{x}^2+c{y}^2=(a{p}^2+c{q}^2)(r^2+ac{s}^2),x=pr-cqs,y=qr+aps.}$

Notice the two factor types:

• Type I: $a{x}^2+c{y}^2$ (the original form).
• Type II: $x^2+ac{y}^2$ (which subsumes the previously studied $x^2+c{y}^2$ family with $c\to ac$).

A product of two Type-I numbers turns out to be Type II:

$(a{p}^2+c{q}^2)(a{r}^2+c{s}^2)=(apr+cqs{)}^2+ac(ps-qr{)}^2.$

So with $x=apr+cqs$, $y=ps-qr$, we land in $x^2+ac{y}^2$, i.e. Type II.

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Proto-Genus Multiplication Table (§179–180)

The composition rule:

$\boxed{\mathrm{I}\times \mathrm{I}=\mathrm{II},\mathrm{I}\times \mathrm{II}=\mathrm{I},\mathrm{II}\times \mathrm{II}=\mathrm{II}.}$

This extends to longer products: $\mathrm{I}\times \mathrm{I}\times \mathrm{I}=\mathrm{I}$, $\mathrm{I}\times \mathrm{II}\times \mathrm{II}=\mathrm{I}$, etc. — Type-I count modulo 2 determines the result, so the two genera form $\mathbb{Z}/2\mathbb{Z}$ under multiplication.

This is a remarkable pre-Gaussian glimpse of genus theory for binary quadratic forms (Gauss, Disquisitiones Arithmeticae 1801).

Worked Example: $a=2,c=3$

Type	Form	Numbers $\le 50$
I	$2x^2+3y^2$	$2,3,5,8,11,12,14,18,20,21,27,29,30,32,35,44,45,48,50$
II	$x^2+6y^2$	$1,4,6,7,9,10,15,16,22,24,25,28,31,33,36,40,42,49$

Check the rule: $35(\mathrm{I})\times 31(\mathrm{II})=1085$ should be Type I. Indeed $1085=2x^2+3y^2$ has four representations: $(x,y)=(23,3),(19,11),(17,13),(1,19)$.

Simple numbers: Type I: $2,3,5,11,29$; Type II: $1,7,31$.

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Why This Matters

This chapter is the methodological heart of Euler’s number-theoretic chapters. The Brahmagupta-Fibonacci identity unifies:

• Sums of two squares (Fermat’s two-square theorem).
• Pell’s equation composition (cf. pell-equation).
• Cube and higher-power transformations of $a{x}^2+c{y}^2$ (next chapter, ch2.0.12-quadratic-form-as-power).
• Proto-genus theory for binary forms.

The use of imaginary numbers $\sqrt{-c}$ — long after they were dismissed as “impossible” in ch1.1.13-imaginary-quantities — to extract integer identities is, as Euler will note in §191, “remarkable, as we are brought to solutions, which absolutely required numbers rational and integer, by means of irrational, and even imaginary quantities.”

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Related pages

• ch2.0.12-quadratic-form-as-power
• ch2.0.7-pell-equation-method
• ch2.0.5-impossibility-quadratic-squares
• brahmagupta-fibonacci-identity
• sums-of-two-squares
• imaginary-numbers
• pell-equation
• quadratic-residues
• indeterminate-analysis
• discriminant