ch2.0.15-questions-cubes

Chapter XV — Solutions of some Questions, in which Cubes are Required

Summary: Four worked questions transforming formulas into cubes, plus the proof of Fermat’s Last Theorem at $n=3$ (Euler’s celebrated result) and a companion theorem that no sum or difference of two cubes can equal twice a cube. Closes Euler’s treatise on indeterminate analysis (§250).

Sources: chapter-2.0.15

Last updated: 2026-05-09

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Orientation (§241)

Having dispatched square-formula questions in the previous chapter, Euler now addresses formulas that must become cubes. The four problems in this chapter occasion two impossibility theorems (FLT for $n=3$, and $x^3\pm y^3\ne 2z^3$) and a parametric solution for $x^3+y^3+z^3=v^3$.

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Q1 — $x^3+y^3$ a cube (§242), and the §243 Theorem

Set $z=x/y$, so $z^3+1$ must be a cube. Multiply by $y^3$ and use the cube ansatz $z-u$ for the cube root: extensive manipulation produces

$z=\frac{u+1}{2}\pm \sqrt{\frac{-u^3+3u^2-3u-3}{12(u-1)}}$

reducing the problem to making $-3u^4+12u^3-18u^2+9$ a square, with seed $u=1$. Bootstrap $u=1+t$ produces the formula $-12t-3t^4$, which can never be a square unless $t<0$. Substituting $t=-s$ gives $12s-3s^4$, with seed $s=1$. But at $s=1$, $u=0$, and we draw no conclusion. No nontrivial seed exists, hinting at an impossibility proof.

§243 Theorem: $x^3\pm y^3\ne z^3$ — Fermat’s Last Theorem at $n=3$

Proof structure (full details — Euler’s most famous result in number theory):

1. WLOG $\gcd (x,y)=1$. One of $x,y$ even, the other odd.
2. WLOG both $x,y$ odd (handling the sum case; the difference case is analogous).
3. Set $p=(x+y)/2$, $q=(x-y)/2$, so $x=p+q$, $y=p-q$, with $p,q$ of opposite parity. Then

$x^3+y^3=2p^3+6p{q}^2=2p(p^2+3q^2).$

So we must show $2p(p^2+3q^2)$ cannot be a cube.

4. If it were, divisibility by 8 forces $p^2+3q^2$ odd, so $p$ even, and $p/4$ a whole cube.
5. The factors $p$ and $p^2+3q^2$ are coprime unless $3\mid p$. Two cases.

Case 1: $3\nmid p$ — factors coprime, so each separately a cube

To represent $p^2+3q^2$ as a cube via the brahmagupta-fibonacci-identity for $\mathbb{Z}[\sqrt{-3}]$:

$p+q\sqrt{-3}=(t+u\sqrt{-3}{)}^3,$

which gives

$p=t^3-9t{u}^2=t(t-3u)(t+3u),q=3t^{2}u-3u^3=3u(t^2-u^2).$

Then $p/4$ a cube requires $2t(t+3u)(t-3u)\cdot 1/4$ a cube. The three factors $2t,t+3u,t-3u$ are pairwise coprime, so each must separately be a cube:

$t+3u=f^3,t-3u=g^3,2t=f^3+g^3.$

But this gives two smaller cubes $f^3,g^3$ whose sum is a cube (namely $2t$, which must be a cube). Infinite descent: a smaller solution exists, contradicting minimality.

Case 2: $3\mid p$, so $3\nmid q$

Set $p=3r$, formula becomes $\frac{9}{4}r(3r^2+q^2)$. Now $3r^2+q^2$ and $r$ are coprime, $r$ even. Each factor independently a cube; the same Brahmagupta-form argument on $q^2+3r^2$ gives

$q=t(t^2-9u^2),r=3u(t^2-u^2).$

The constraint that $9r/4$ be a cube reduces to $2u(t-u)(t+u)$ a cube, three coprime factors each a cube: $t+u=f^3$, $t-u=g^3$, $2u=f^3-g^3$. Again two smaller cubes whose difference is a cube — descent.

6. Hence neither case admits any solution, completing the proof.

This is Euler’s celebrated proof. The use of $(t+u\sqrt{-3}{)}^3$ implicitly assumes unique factorization in $\mathbb{Z}[\sqrt{-3}]$, which fails (the ring is not a UFD); the proof can be repaired by passing to $\mathbb{Z}[\omega ]$ where $\omega =(-1+\sqrt{-3})/2$. The descent skeleton stays correct.

See fermats-last-theorem-n3 for the proof distilled.

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Q2 — $a^3+b^3+x^3$ a cube (§245-247)

Given two cubes $a^3,b^3$, find $x^3$ making $a^3+b^3+x^3$ a cube. Substitute $x=y-a$: must transform $y^3-3a{y}^2+3a^{2}y+b^3$ into a cube. Take root $fy+b$:

$f^3{y}^3+3b{f}^2{y}^2+3b^{2}fy+b^3=y^3-3a{y}^2+3a^{2}y+b^3.$

Set $f=a^2/b^2$ to kill the linear-in-$y$ mismatch in third-order terms; solve for $y$:

$y=\frac{3a{b}^3(a^3+b^3)}{b^6-a^6}=\frac{3a{b}^3}{b^3-a^3}.$

Hence

$x=y-a=\frac{2a{b}^3+a^4}{b^3-a^3}=a\cdot \frac{2b^3+a^3}{b^3-a^3}.$

Examples:

$(a,b)$	$x$	Identity
$(1,2)$	$17/7$	$9+(17/7{)}^3=(20/7{)}^3$
$(2,3)$	$124/19$	$35+(124/19{)}^3=$ cube
$(3,4)$	$465/37$	$91+(465/37{)}^3=$ cube

§246 Remarkable case $a=b$ — Theorem §247: $x^3\pm y^3\ne 2z^3$

When $a=b$, the formula gives $x=3a^4/0=\infty$: no solution. Euler concludes that

$2a^3+x^3=y^3$

is impossible (the equation $a^3+a^3+x^3=y^3$). Equivalently, no two cubes can sum to twice a cube.

Proof sketch (parallel to §243): WLOG $\gcd (x,y)=1$, both odd. Set $p=(x+y)/2$, $q=(x-y)/2$. Then $x^3+y^3=2p(p^2+3q^2)$, and we want this a $2\cdot$cube, i.e., $p(p^2+3q^2)$ a cube. Two cases on whether $3\mid p$:

• Case 1: $3\nmid p$. Setting $p^2+3q^2=(t^2+3u^2{)}^3$ and $p=t(t^2-9u^2)/(\text{cube})$ leads to the requirement that $2t(t+3u)(t-3u)$ is a cube — a chain of three coprime factors each a cube. This yields $t+3u=f^3$, $t-3u=g^3$, $2t=f^3+g^3$, hence smaller cubes summing to a cube → descent.
• Case 2: $3\mid p$. Parallel argument via $u(t^2-u^2)$ giving $f^3-g^3=2u$, smaller difference of cubes equal to twice a cube → descent.

Hence no integer solution exists.

The exception $y=x$ trivially gives $2a^3+x^3=(a{)}^3+(a{)}^3+x^3=$ trivial when $x=a$, which Euler explicitly excludes.

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Q3 — General $x^3+y^3+z^3=v^3$ (§248)

Now neither $a$ nor $b$ is fixed — find a parametric infinite family. Setting $x=p+q$, $y=p-q$, $z=r-s$, $v=r+s$ converts the equation into

$2p(p^2+3q^2)=2s(s^2+3r^2).$

Both sides have the form $\alpha ({\alpha }^2+3{\beta }^2)$, which by the structure of $\mathbb{Z}[\sqrt{-3}]$ (Article 176) must share a common divisor $t^2+3u^2$. Setting

$p^2+3q^2=(f^2+3g^2)(t^2+3u^2),s^2+3r^2=(h^2+3k^2)(t^2+3u^2)$

yields explicit parametrizations $p=ft+3gu$, $q=gt-fu$, $s=ht+3ku$, $r=kt-hu$. Eliminating gives

$t=3k(h^2+3k^2)-3g(f^2+3g^2),u=f(f^2+3g^2)-h(h^2+3k^2).$

(Free parameters: $f,g,h,k$.)

Worked examples:

Choice	Result
$f=-1,g=1,h=0,k=1$	$-20^3+14^3+17^3=-7^3$, i.e., $14^3+17^3+7^3=20^3$
$f=2,g=1,h=0,k=1$	$30^3=22^3+58^3-54^3$, i.e., $58^3=30^3+54^3+22^3$; dividing by 2: $29^3=15^3+27^3+11^3$
$f=3,g=1,h=1,k=1$	$-12^3+18^3-16^3=2^3$, i.e., $18^3=16^3+12^3+2^3$; dividing by 2: $9^3=8^3+6^3+1^3$
$f=h=1,g=0,k=h$	$5^3,3^3,4^3,6^3$ — the classic $3^3+4^3+5^3=6^3$

The progression $3,4,5$ (consecutive integers!) suggests Q4.

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Q4 — Three numbers in arithmetic progression with common difference 1 whose cubes sum to a cube (§249)

Let middle number $x$, so least is $x-1$, greatest $x+1$. Sum:

$3x^3+6x=3x(x^2+2).$

This must be a cube. Trial gives $x=4$ as a seed (corresponding to $3,4,5\to 6$). Bootstrap with $x=4+y$: $216+150y+36y^2+3y^3$ must be a cube. Root $6+fy$ kills the constant; $f=25/18$ kills the linear term. The remaining quadratic in $y$ gives

$y=-7452/1871,x=32/1871.$

Reducing the question to squares is cleaner: setting $3x(x^2+2)=x^3{y}^3$ gives $x=6/(y^3-3)=36/(6y^3-18)$. Hence the denominator $6y^3-18=18z^3-18$ (with $y=3z$, $z=1+v$) must be a square; bootstrap yields $z=17/32$, $y=51/32$.

After full computation, the second AP solution:

$x-1=\frac{149}{107},x=\frac{256}{107},x+1=\frac{363}{107},$

with cube-sum $(408/107{)}^3$. So besides $(3,4,5;6)$ the next AP-of-difference-1 cube triple is $(149/107,256/107,363/107;408/107)$.

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§250 — Closing the Treatise

“We shall here finish this Treatise on the Indeterminate Analysis, having had sufficient occasion, in the questions which we have resolved, to explain the chief artifices that have hitherto been devised in this branch of Algebra.”

Part II of Elements of Algebra ends here. The “Questions for Practice” follow as fourteen exercises consolidating the material.

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Concept Pages

• fermats-last-theorem-n3 — proved in §243
• sum-of-two-cubes — restates §243 in modern form
• infinite-descent — proof method for both §243 and §247
• brahmagupta-fibonacci-identity — used implicitly in $(t+u\sqrt{-3}{)}^3$ expansion
• three-cubes-as-cube — §248 parametrization, generating $3^3+4^3+5^3=6^3$ etc.

Related pages

• ch2.0.10-cubic-formula-as-cube
• ch2.0.13-impossibility-biquadrate-sums
• ch2.0.14-questions-squares
• fermats-last-theorem-n4
• cubic-equations
• indeterminate-analysis