Three Cubes Summing to a Cube
Summary: The Diophantine equation admits an infinite parametric family of integer solutions, in stark contrast to (Fermat’s Last Theorem at , impossible). Euler’s parametrization in ch2.0.15-questions-cubes §248 generates classics like , , and from four free integers .
Sources: chapter-2.0.15
Last updated: 2026-05-09
The Parametrization
Setting , , , converts into
By the multiplicative theory of (Article 176 / brahmagupta-fibonacci-identity for ), both sides factor through a common term . Setting
p &= ft + 3gu, & q &= gt - fu, \\ r &= kt - hu, & s &= ht + 3ku, \end{aligned}$$ and equating, one obtains a linear equation in $t, u$: $$t = 3k(h^2 + 3k^2) - 3g(f^2 + 3g^2), \qquad u = f(f^2+3g^2) - h(h^2+3k^2).$$ **Free parameters**: $f, g, h, k$ — any four integers (or rationals). The recipe: 1. Choose $f, g, h, k$. 2. Compute $t, u$ via the formulas above. 3. Compute $p, q, r, s$ from $f, g, h, k, t, u$. 4. Set $x = p+q$, $y = p-q$, $z = r-s$, $v = r+s$. The resulting $(x, y, z, v)$ satisfies $x^3 + y^3 + z^3 = v^3$ identically — though some entries may be negative or rational, in which case rearrange/scale. --- ## Worked Examples (§248) ### Family 1: $k = 0, h = 1$ Yields $$x = -3g - (f^2+3g^2)^2 + f, \quad y = -3g + (f^2+3g^2)^2 - f, \quad z = (3g - f)(f^2+3g^2) + 1, \quad v = -(3g-f)(f^2+3g^2) + 1.$$ With $f = -1, g = 1$: $(x, y, z, v) = (-20, 14, 17, -7)$, i.e., $$\boxed{14^3 + 17^3 + 7^3 = 20^3.}$$ Check: $2744 + 4913 + 343 = 8000 = 20^3$. ✓ ### Family 2: $f = 2, g = 1, h = 0, k = 1$ Computes $f^2 + 3g^2 = 7$, $h^2 + 3k^2 = 3$. Then $t = -12$, $u = 14$. After substitution: $$p = -10 \cdot ?, \quad q = -22, \quad r = -12, \quad s = 42$$ (after some sign normalization). Result: $58^3 = 30^3 + 54^3 + 22^3$, i.e., $$\boxed{30^3 + 54^3 + 22^3 = 58^3.}$$ Dividing all by 2: $\boxed{29^3 = 15^3 + 27^3 + 11^3.}$ ### Family 3: $f = 3, g = 1, h = 1, k = 1$ $f^2+3g^2 = 12$, $h^2+3k^2 = 4$. After computation, $(x, y, z, v) = (-12, 18, -16, 2)$, giving $18^3 = 16^3 + 12^3 + 2^3$. Dividing by 2: $$\boxed{9^3 = 8^3 + 6^3 + 1^3.}$$ Check: $512 + 216 + 1 = 729 = 9^3$. ✓ ### Family 4: $g = 0, k = h$, free $f, h$ Gives $$\begin{aligned} x &= 16fh^3 - f^4, \\ y &= 8fh^3 + f^4, \\ z &= 16h^4 - 4hf^3, \\ v &= 16h^4 + 2hf^3. \end{aligned}$$ With $f = h = 1$: $(x, y, z, v) = (15, 9, 12, 18)$. Dividing by 3: $$\boxed{3^3 + 4^3 + 5^3 = 6^3.}$$ Check: $27 + 64 + 125 = 216 = 6^3$. ✓ --- ## The "$3, 4, 5$" Phenomenon — Q4 (§249) The identity $3^3 + 4^3 + 5^3 = 6^3$ has **three consecutive integers** as the cubed terms. Euler's Q4 asks for the next such identity — three integers in AP (common difference 1) whose cubes sum to a cube. The next solution turns out to be **rational**: $$\left(\frac{149}{107}\right)^3 + \left(\frac{256}{107}\right)^3 + \left(\frac{363}{107}\right)^3 = \left(\frac{408}{107}\right)^3.$$ Here $x = 256/107$ is the middle term, with $x \pm 1$ the others. There is no integer-valued analog beyond $(3, 4, 5; 6)$. --- ## Other Notable Examples (Beyond Euler) The general theorem on three-cubes-as-cube has produced many famous identities: - $3^3 + 4^3 + 5^3 = 6^3$ (Euler) - $1^3 + 6^3 + 8^3 = 9^3$ (Euler) - $7^3 + 14^3 + 17^3 = 20^3$ (Euler) - $11^3 + 15^3 + 27^3 = 29^3$ (Euler) - $9^3 + 10^3 = 12^3 + 1^3 = 1729$ (the **Hardy-Ramanujan number**, *two* cubes summing to two-cubes-sum) - $3^3 + 4^3 + 5^3 + 8^3 = 10^3$, etc. Euler's parametrization from §248 is **complete** in that any rational solution can be parametrized this way, "since in the whole calculation we have admitted no arbitrary limitation." --- ## Contrast with $x^3 + y^3 = z^3$ | Equation | Solutions | |----------|-----------| | $x^3 + y^3 = z^3$ | **No** nontrivial integer solutions ([[fermats-last-theorem-n3]]) | | $x^3 + y^3 + z^3 = v^3$ | **Infinitely many** integer solutions, four-parameter family | | $x^3 + y^3 + z^3 = w^3 + 0$ (3-cube sum to cube) | Same family | | $x^3 + y^3 = 2z^3$ | **No** nontrivial solutions ([[ch2.0.15-questions-cubes]] §247) | | $x^3 + y^3 + z^3 = n$ (single integer) | Open in general; e.g., $33$ and $42$ resolved only in 2019 | The jump from "two cubes" to "three cubes" turns impossibility into a four-parameter abundance — a structural shift mirrored at every odd prime $n$ in FLT's history. --- ## Related pages - [[ch2.0.15-questions-cubes]] - [[fermats-last-theorem-n3]] - [[sum-of-two-cubes]] - [[brahmagupta-fibonacci-identity]] - [[indeterminate-analysis]] - [[cubic-equations]]