Ch2.0.9 — Of the Method of rendering Rational the incommensurable Formula
Summary: Treats the radicand-quartic case in three subclasses — first term square, last term square, both squares — yielding up to six new values from a single application; develops the bootstrap from a known solution and the Möbius-style substitution ; closes by acknowledging that the techniques fail at degree 5 and beyond, marking the frontier of the subject.
Sources: chapter-2.0.9
Last updated: 2026-05-09
The Three Subclasses (§128)
For the quartic radicand
three structural cases admit dedicated methods:
| Case | Form | Methods |
|---|---|---|
| 1 | (first term square) | Three-term root |
| 2 | (last term square) | Reverse three-term root |
| 3 | (both squares) | Combination of 1 and 2 plus two extra ansätze; up to six new values |
Cases not in this list require a known solution before any method applies.
Case 1: First Term is a Square (§129–131)
Suppose root . Squaring and matching:
b &= 2fp &&\Rightarrow\; p = b/(2f),\\ c &= 2fq+p^2 &&\Rightarrow\; q = (c-p^2)/(2f). \end{aligned}$$ After dividing the remainder by $x^3$: $$\boxed{\,x = \frac{d-2pq}{q^2-e} = \frac{2pq-d}{e-q^2}.\,}$$ **Failure modes**: when $b=c=0$ (formula $f^2+dx^3+ex^4$), forcing $p=q=0$ collapses to $dx^3+ex^4=0$, hence $x=-d/e$, which simply factors out a square; and when $b=d=0$ (formula $f^2+cx^2+ex^4$), $p=0$ gives $x=0$ trivially. (§131) --- ## Case 2: Last Term is a Square (§132–133) Substituting $x=1/y$ in $a+bx+cx^2+dx^3+g^2x^4$ inverts the formula into Case 1 form $g^2+dy+cy^2+by^3+ay^4$, so the technique is symmetric. Directly: suppose root $= q+px+gx^2$, choose $p$ to kill the fourth term and $q$ to kill the third: $$p = d/(2g), \qquad q = (c-p^2)/(2g).$$ Then $$\boxed{\,x = \frac{a-q^2}{2pq-b} = \frac{q^2-a}{b-2pq}.\,}$$ **Failure mode** (§133): formula $a+cx^2+g^2x^4$ has $b=d=0$, giving $p=0$ and an infinite/undefined $x$. --- ## Case 3: Both Ends are Squares (§134–136) Formula $f^2+bx+cx^2+dx^3+g^2x^4$. All four ansätze below are admissible, **and each admits a sign flip** on $f$ or $g$ (since both appear only squared in the formula), giving six new values in total. ### Ansatz A: Root $= f+px+gx^2$, kill the second terms $2fp = b$ gives $p = b/(2f)$, and the constant and quartic terms vanish automatically. Dividing by $x^2$: $$x = \frac{c-2fg-p^2}{2gp-d} = \frac{p^2+2fg-c}{d-2gp}.$$ Sign-flipping $g \to -g$ gives a second value: $$x = \frac{c+2fg-p^2}{-2gp-d} = \frac{p^2-2fg-c}{2gp+d}.$$ ### Ansatz B: Root $= f+px+gx^2$, kill the fourth terms Choose $d=2gp$, so $p=d/(2g)$. The first and last terms also cancel; dividing the rest by $x$: $$x = \frac{b-2fp}{2fg+p^2-c}.$$ Sign-flipping $f \to -f$: $$x = \frac{b+2fp}{p^2-2fg-c}.$$ So this method together with Ansatz A yields **four** new values; combining with the two from Cases 1 and 2 makes **six** total. **Failure mode** (§136): formula $f^2+cx^2+g^2x^4$ has $b=d=0$, giving $p=0$ and either an undefined or trivial $x$ in every method. --- ## Bootstrap from a Known Solution (§137–138) When neither end is a square, find any $x=h$ with $a+bh+ch^2+dh^3+eh^4 = k^2$, then substitute $x=h+y$ to get a Case 1 formula with leading term $k^2$: $$k^2 + (b+2ch+3dh^2+4eh^3)y + (c+3dh+6eh^2)y^2 + (d+4eh)y^3 + ey^4.$$ ### Worked Reduction for $a+ex^4$ (§138) Even after $x=h+y$ this formula has both middle terms nonzero. Take root $=k+py+qy^2$: $$p = \frac{2eh^3}{k}, \qquad q = \frac{eh^2(k^2+2a)}{k^3}\quad\text{(using }eh^4 = k^2-a\text{)}.$$ After algebraic simplification $$\boxed{\,y = \frac{4hk^2(2a-k^2)}{3k^4-4a^2}, \qquad x = h+y = \frac{h(k^4-8ak^2+4a^2)}{4a^2-3k^4}.\,}$$ This is one of the cleaner closed-form bootstraps in the chapter. --- ## The $(1+y)/(1-y)$ Substitution (§139–140) Since $x=\pm h$ both serve as known cases when only $x^4$ appears, treat the formula as Case 3 by the substitution $$x = \frac{h(1+y)}{1-y}.$$ This sends $y=0 \to x=h$ and $y=1 \to x=\infty$, while making both the new constant term and the leading term square. The Case 3 ansatz $\text{root} = (k+py-ky^2)/(1-y)^2$ recovers the same $x$ as §138 — confirming the consistency of the methods. (§139) ### Numerical Example: $2x^4-1$ (§140) Here $a=-1$, $e=2$, seed $h=1$ (since $2-1=1$), so $k=1$. - First iteration: $x = (1+8+4)/(3-4) = -13$, hence $|x|=13$ gives $2(13)^4-1 = 57121 = 239^2$. - Second iteration with $h=13$, $k=239$: $$x = \frac{13(239^4 + 8 \cdot 239^2 + 4)}{3 \cdot 239^4 - 4} = \frac{42\,422\,452\,969}{9\,788\,425\,919}.$$ The fractions blow up quickly — characteristic of these bootstrap chains. --- ## Even-Powers-Only Variant: $a+cx^2+ex^4$ (§141–144) Substituting $x=h+y$ where $a+ch^2+eh^4=k^2$ produces the Case 1 form $$k^2 + (2ch+4eh^3)y + (c+6eh^2)y^2 + 4ehy^3 + ey^4.$$ With $p=(ch+2eh^3)/k$ and $q=(c+6eh^2-p^2)/(2k)$: $$y = \frac{4eh-2pq}{q^2-e}.$$ **Examples that yield no new value** (because the proposed formula is impossible beyond a known small set): - $1-x^2+x^4$ at $h=k=1$: $p=q=$ values forcing $y=0$. Euler notes the formula is square only at $x=0,\pm 1$. (§142) - $2-3x^2+2x^4$ at $h=k=1$: again $y=0$. (§143) **A productive example — $1+8x^2+x^4$ (§144)**: at $h=2$, $k=7$. The Case 1 method gives $p=32/7$, $q=272/343$, $y=-5880/2911$, $x=-58/2911$ (sign immaterial). The same formula also admits Case 3 treatment (last term $x^4$ is already a square). Setting $x=2+y$ yields $49+64y+32y^2+8y^3+y^4$. Trying root $=7+py+y^2$ recovers the known $x=\pm 2$; instead killing the second term with $p=32/7$ gives $y=-71/28$, hence $x=15/28$, with the formula equal to $(1441/784)^2$. The $g \to -g$ flip (root $=7+py-y^2$) reproduces the same value. --- ## The Frontier: Degree 5 and Beyond (§146) For $k^2+bx+cx^2+dx^3+ex^4+fx^5$ with first term square: - **Three-term root** $k+px+qx^2$: kills second and third terms; remainder, divided by $x^3$, is **quadratic** in $x$ — so $x$ depends on a new square root, defeating the purpose. - **Four-term root** $k+px+qx^2+rx^3$: square has degree 6; killing terms 2, 3, 4 still leaves degree 4, 5, 6 — divided by $x^4$, again quadratic in $x$. Euler concludes: > "We have really exhausted the subject of transforming formulas into squares: we may now proceed to quantities affected by the sign of the cube root." This marks the boundary of 18th-century technique: degree 5 and higher remained intractable. Compare the analogous limit for cube extraction at degree 4 in [[ch2.0.10-cubic-formula-as-cube]] §147. --- ## Summary of Method Yields | Subclass | Methods | New $x$ values per pass | |----------|---------|--------------------------| | Case 1 ($a=f^2$) | Three-term root | 1 | | Case 2 ($e=g^2$) | Reverse three-term root | 1 | | Case 3 (both squares) | Cases 1+2 plus four extra ansätze (with $\pm f$, $\pm g$) | up to 6 | | Neither square | Find a seed $h$, then reduce to Case 1 via $x=h+y$ | depends on iteration | --- ## Related pages - [[ch2.0.8-cubic-surd-rationalization]] - [[ch2.0.10-cubic-formula-as-cube]] - [[ch2.0.4-surd-rationalization]] - [[rationalization]] - [[indeterminate-analysis]] - [[quartic-equations]]