ch2.0.12-quadratic-form-as-power

Ch2.0.12 — Of the Transformation of the Formula $a{x}^2+c{y}^2$ into Squares, and higher Powers

Summary: Asks when $a{x}^2+c{y}^2$ can be made a perfect $n$-th power for chosen integers $x,y$. Even powers require $a=1$ (or knowledge of one representation case). Odd powers always work for any $a,c$. The technique sets $x\sqrt{a}+y\sqrt{-c}=(p\sqrt{a}+q\sqrt{-c}{)}^n$ and reads off $x,y$ from rational and irrational parts. Yields explicit polynomial parametrizations for squares, cubes, biquadrates, and fifth powers, with worked examples like $11^2+2^2=125=5^3$, $5^2+2=27=3^3$, $2^2+1^2=125^{?}$, and the subtle Pell-related case $2\cdot 4^2-5=27=3^3$ that escapes the basic parametrization.

Sources: chapter-2.0.12

Last updated: 2026-05-09

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Setup and a Reduction (§181)

If $a{x}^2+c{y}^2$ is to be a square (or higher even power), the formula can typically be rewritten with leading coefficient $1$. Example: $2p^2-q^2=(2p+q{)}^2-2(p+q{)}^2$, so by setting $x=2p+q$, $y=p+q$ it becomes $x^2-2y^2$ with $a=1$, $c=-2$. Henceforth focus on $x^2+c{y}^2$.

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Squares: $x^2+c{y}^2=\square$ (§182–185)

Set $(x+y\sqrt{-c})=m(p+q\sqrt{-c}{)}^2$ and conjugate. Multiplying:

$x^2+c{y}^2=m^2(p^2+c{q}^2{)}^2.$

Reading parts:

$\boxed{x=m(p^2-c{q}^2),y=2mpq.}$

For $\gcd (x,y)=1$, take $m=1$:

$x=p^2-c{q}^2,y=2pq,x^2+c{y}^2=(p^2+c{q}^2{)}^2.$

This recovers the parametrizations Euler had derived in ch1.4.5-pure-quadratic-equations and ch2.0.4-surd-rationalization by direct substitution $\sqrt{x^2+c{y}^2}=x+py/q$. Either method gives the same $x/y=(c{q}^2-p^2)/(2pq)$.

Determinate cases (§185)

$c$	Parametrization	Identity
$+1$	$x=p^2-q^2$, $y=2pq$	$x^2+y^2=(p^2+q^2{)}^2$
$-1$	$x=p^2+q^2$, $y=2pq$	$x^2-y^2=(p^2-q^2{)}^2$
$+2$	$x=p^2-2q^2$ or $2p^2-q^2$, $y=2pq$	$x^2+2y^2=(p^2+q^2{)}^2$ etc.
$-2$	$x=p^2+2q^2$, $y=2pq$	$x^2-2y^2=(p^2-2q^2{)}^2$
$+6$	$(a,c)=(1,6)$ or $(2,3)$	$x^2+6y^2=(p^2+6q^2{)}^2=(2p^2+3q^2{)}^2$

When $c$ factors as $c=ab$, each factorization gives a parametrization — directly tied to the Type-I/Type-II decomposition from ch2.0.11-quadratic-form-factorization.

Need a Seed When $a\ne 1$ (§186)

For $a{x}^2+c{y}^2=\square$, one must know a starting case $a{f}^2+c{g}^2=h^2$. Then the substitution

$t=\frac{afx+cgy}{h},u=\frac{gx-fy}{h}$

converts $a{x}^2+c{y}^2$ into $t^2+ac{u}^2$ — exactly the Type II form, where the $a$-coefficient has vanished. Then the general parametrization above applies with $c\to ac$.

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Cubes: $a{x}^2+c{y}^2={\square }^3$ (§187–193)

Now set $x\sqrt{a}+y\sqrt{-c}=(p\sqrt{a}+q\sqrt{-c}{)}^3$ and conjugate. Cubing:

$x\sqrt{a}+y\sqrt{-c}=a{p}^3\sqrt{a}+3a{p}^{2}q\sqrt{-c}-3cp{q}^2\sqrt{a}-c{q}^3\sqrt{-c}.$

Reading parts:

$\boxed{x=a{p}^3-3cp{q}^2,y=3a{p}^{2}q-c{q}^3,}$

with $a{x}^2+c{y}^2=(a{p}^2+c{q}^2{)}^3$.

Crucially, this works for any $a,c$ (no seed needed). Odd powers always succeed; even powers do not, in general — see §200 below.

Sums of Two Squares as Cubes (§188)

With $a=c=1$: $x=p^3-3p{q}^2$, $y=3p^{2}q-q^3$. Example: $p=2,q=1$ gives $x=2,y=11$: $2^2+11^2=125=5^3$.

$x^2+3y^2$ as a Cube (§189)

With $c=3$: $x=p^3-9p{q}^2$, $y=3p^{2}q-3q^3$. Euler’s table:

$p$	$q$	$x$	$y$	$x^2+3y^2$
1	1	8	0	$64=4^3$
2	1	10	9	$343=7^3$
1	2	35	18	$2197=13^3$
3	1	0	24	$1728=12^3$
1	3	80	72	$21952=28^3$
3	2	81	30	$9261=21^3$
2	3	154	45	$29791=31^3$

Question 1 (§192): $x^2+4={\square }^3$

Solve $y^2=4$ in $y=3p^{2}q-q^3=q(3p^2-q^2)=\pm 2$. Trying $q=1$: $3p^2-1=2$ gives $p=1$, $x=p^3-3p{q}^2=-2$, so $x^2=4$. Trying $q=2$: $6p^2-8=-2$ gives $p=1$, $x=p^3-6p\cdot 4=1-24=...$ more carefully $x=p^3-3p{q}^2=1-12=-11$, so $x^2=121$. The only solutions are $x^2=4$ ($\Rightarrow 8=2^3$) and $x^2=121$ ($\Rightarrow 125=5^3$).

Question 2 (§193): $x^2+2={\square }^3$

Use $a=1,c=2$: $x=p^3-6p{q}^2$, $y=3p^{2}q-2q^3$. Solve $y=\pm 1$, requiring $q\mid 1$, so $q=1$, $3p^2-2=\pm 1$, giving $p=1$, $x=1-6=-5$. Hence $5^2+2=27=3^3$ — the unique solution in integers.

Question 4 (§195–196): $2x^2-5={\square }^3$ — A Subtle Failure

Setting $a=2,c=-5$: $x=2p^3+15p{q}^2$, $y=6p^{2}q+5q^3$. Forcing $y=\pm 1$ gives no integer (or even rational) solution. And yet $x=4$ works: $2\cdot 16-5=27=3^3$.

The issue: when $c<0$, the auxiliary form $r^2-ac{s}^2=r^2-(-10)s^2=r^2+10s^2$ becomes $r^2-10s^2$, which can equal $\pm 1$ infinitely often (Pell, pell-equation: $r=3,s=1$; $r=19,s=6$; …). So the cube can be hidden as a Pell-multiple:

$2x^2-5y^2=(f^2-10g^2)\cdot (2p^2-5q^2{)}^3.$

Splitting and reading parts gives a richer family. With $f=3,g=1$ we get the equation $4p^3+18p^{2}q+30p{q}^2+15q^3=\pm 1$, which has many more solutions.

This explains the $x=4$ case and signals that for negative $c$, the basic parametrization is incomplete. (§197 emphasizes this caveat.)

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Biquadrates (Fourth Powers) (§198–200)

Set $x+y\sqrt{-c}=(p+q\sqrt{-c}{)}^4$. Expanding:

$\boxed{x=p^4-6c{p}^2{q}^2+c^2{q}^4,y=4p^{3}q-4cp{q}^3.}$

For $c=1$ ($x^2+y^2={\square }^4$): $x=p^4-6p^2{q}^2+q^4$, $y=4p^{3}q-4p{q}^3$.

$p$	$q$	$x$	$y$	$x^2+y^2$
2	1	7	24	$625=5^4$
3	2	119	120	$28561=13^4$

Even powers always require an existing square representation (so essentially $a=1$); see §200.

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Fifth Powers (§201)

Set $x\sqrt{a}+y\sqrt{-c}=(p\sqrt{a}+q\sqrt{-c}{)}^5$. Expanding:

$\boxed{x=a^2{p}^5-10ac{p}^3{q}^2+5c^{2}p{q}^4,y=5a^2{p}^{4}q-10ac{p}^2{q}^3+c^2{q}^5.}$

For $a=c=1$ ($x^2+y^2={\square }^5$): $x=p^5-10p^3{q}^2+5p{q}^4$, $y=5p^{4}q-10p^2{q}^3+q^5$.

Example: $p=2,q=1$: $x=32-80+10=-38$, $y=80-40+1=41$: $38^2+41^2=1444+1681=3125=5^5$.

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Odd vs Even: The Asymmetry (§200–201)

Stated cleanly:

• Even powers $2k$: $a{x}^2+c{y}^2$ becomes a $2k$-th power only if it is already reducible to a square; equivalently $a=1$ after reduction. A seed solution is needed when $a\ne 1$.
• Odd powers $2k+1$: Always achievable for any $a,c$ via the $n$-th-power-of-conjugate construction.

The reason: odd-power expansion never produces a constant term that must vanish independently; even-power expansion does, forcing seeded constraints.

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Why the Method Is “Remarkable” (§191)

Euler dwells on the meta-point: the construction proves integer identities by manipulating imaginary quantities. He notes a caveat: the principle “if a product is a power, each factor is a power up to a unit” requires no nontrivial common divisors among the factors. For genuine imaginaries $p\pm q\sqrt{-c}$, this is automatic when $\gcd (x,y)=1$. But when $c=-1$, the factors $x+y$ and $x-y$ are real and can share a divisor (e.g., when both $x,y$ are odd), so the method needs the modified ansatz $x+y=2p^3$, $x-y=4q^3$, giving $x^2-y^2=8p^3{q}^3=(2pq{)}^3$.

This is an early instance of the failure of unique factorization — the same gap that subtly affects Euler’s $n=3$ proof of Fermat’s Last Theorem (see sum-of-two-cubes).

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Method Summary Table

Power	Ansatz	$x$	$y$
2	$(p\sqrt{a}+q\sqrt{-c}{)}^2$, scaled by $m$	$m(a{p}^2-c{q}^2)$	$2mpq$ (only when $a=1$ unseeded)
3	$(p\sqrt{a}+q\sqrt{-c}{)}^3$	$a{p}^3-3cp{q}^2$	$3a{p}^{2}q-c{q}^3$
4	$(p+q\sqrt{-c}{)}^4$, requires $a=1$	$p^4-6c{p}^2{q}^2+c^2{q}^4$	$4p^{3}q-4cp{q}^3$
5	$(p\sqrt{a}+q\sqrt{-c}{)}^5$	$a^2{p}^5-10ac{p}^3{q}^2+5c^{2}p{q}^4$	$5a^2{p}^{4}q-10ac{p}^2{q}^3+c^2{q}^5$

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Related pages

• ch2.0.11-quadratic-form-factorization
• ch2.0.13-impossibility-biquadrate-sums
• ch2.0.7-pell-equation-method
• ch2.0.4-surd-rationalization
• ch2.0.10-cubic-formula-as-cube
• brahmagupta-fibonacci-identity
• sum-of-two-cubes
• imaginary-numbers
• pell-equation
• indeterminate-analysis