Does $2764976 + 3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} = 0$?

Question

In the process of my numerical computations I have found a very special identity:

• $\;\;1264483 + 1707789 \,\sqrt[3]{7} - 1238313\,\sqrt[3]{7^2} = 9.313225746154785 \times 10^{-10}$

• $ -1500493 - 2026256\,\sqrt[3]{7} + 1469290\,\sqrt[3]{7^2} = 9.313225746154785 \times 10^{-10}$

Therefore if we subtract these two equations one should find the difference is zero. On my computer I found:

$$ 2764976 + 3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} \stackrel{?}{=} \left\{ \begin{array}{cl} 0 & \text{by hand} \\ -1.862645149230957 \times 10^{-9} & \text{by computer} \end{array}\right. $$

Subtracting these two numbers - which might be the same - we have gotten twice the number!

Answer 1

To expand on the comments of Fabio and Jyrki:

You're claiming that $\sqrt[3]7$ is a root of the quadratic equation

$$2764976 + 3734045x -2707603x^2 = 0$$

but that leads to a contradiction.

The solutions of a quadratic with integer coefficients can be written as

$$x = \frac{p \pm \sqrt q}{r} $$

for some integers $p, q, r$.

If $x^3 = 7$ then

$$\begin{align} \left(\frac{p \pm \sqrt q}{r}\right)^3 = 7\\ \frac{p^3 \pm 3p^2\sqrt q + 3pq +q\sqrt q}{r^3} = 7\\ \frac{(p^3 + 3pq) \pm \sqrt q(3p^2 +q)}{r^3} = 7\\ \end{align}$$

Now the only way the expression on the LHS can be rational (let alone integer) is if $\sqrt q$ is rational. But $q$ is an integer, so its square root can only be rational if $q$ is a perfect square. But that would make $x = \frac{p \pm \sqrt q}{r}$ rational, and we know that $x = \sqrt[3]7$ is not rational.

Answer 2

$$ 276497+3734045x−2707603x^2 $$ is a quadratic with two roots: $$ x = \frac{-3734045\pm \sqrt{3734045^2 + 4 \cdot 2764976 \cdot 2707603}}{2 \cdot (−2707603)} $$ neither of which is a cube root of $7$, which can be verified by cubing them (or by knowing about the degrees of algebraic numbers, per @JackD'Aurizio).

Answer 3

No. Mathematica tells us that its value is about $-2.0876013027695663896 \times 10^{-9}$.

$$1264483 + 1707789 \times 7^{1/3} - 1238313 \times 7^{2/3} = -3.1767789172657775703*10^{-10}$$ according to Mathematica.

$$-1500493 - 2026256 \times 7^{1/3} + 1469290 \times 7^{2/3} = 1.7699234110429886326 \times 10^{-9}$$ similarly.

These are calculated with Mathematica's arbitrary-precision arithmetic, so it is "guaranteed" not to have the rounding errors that floating-point arithmetic can often introduce.

Answer 4

Assume $$ \tag 12764976 + 3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} = 0$$ Multiply by $\sqrt[3]7$ to obtain $$ \tag 22764976\,\sqrt[3]7 + 3734045\,\sqrt[3]{7^2} -18953221 = 0.$$ The system of linear equations $$\begin{align} 3734045 x-2707603 y&=-12764976\\ 22764976 x+3734045 y&=18953221 \end{align} $$ has rational solutions, i.e., we are forced to conclude that $$\sqrt[3]7=\frac{-12764976\cdot 3734045+2707603\cdot 18953221}{3734045\cdot 3734045+2707603\cdot 22764976} =\frac{3652803231343}{75581609374553}$$ is rational, which is absurd! (Not to mention that this rational number is $\approx 0.05$).

Answer 5

If $\sqrt[3]{7}$ and $\left(\sqrt[3]{7}\right)^2$ were linearly dependent over $\mathbb{Q}$, $\;\alpha=\sqrt[3]{7}$ would be an algebraic number over $\mathbb{Q}$ with degree $\leq 2$, hence $x^3-7$ would be a reducible polynomial over $\mathbb{Q}$. However, $x^3-7$ is an irreducible polynomial over $\mathbb{F}_{13}$, hence $x^3-7$ is an irreducible polynomial over $\mathbb{Q}$ and no linear combination of $1,\sqrt[3]{7},\sqrt[3]{49}$ with rational coefficients $\neq(0,0,0)$ can be zero.

However, $\sqrt[3]{7}$ is pretty close to $$ \frac{-25642+4 \sqrt{708733687}}{42263} $$ (I get that through the Mathematica command RootApproximant), hence, for instance, $$ 252756- 51284\sqrt[3]{7}+42263 \sqrt[3]{49}$$ is pretty close to zero, without actually being equal to zero (namely it is $\approx 2.91\cdot 10^{-11}$) and you may derive more accurate approximations by replacing the continued fraction of $\sqrt[3]{7}$ with a definitely periodic continued fraction, that is always associated to an algebraic number over $\mathbb{Q}$ with degree $2$. For instance, from: $$ \sqrt[3]{7}\approx [1; 1, 10, 2, 16, 2,1,4,2,1,21,1,3,5,\overline{1,2,1}] $$ we get:

$$ 646825196228418 - 676266038270144 \sqrt[3]{7} + 176761726809763 \sqrt[3]{49} \approx 3.9\cdot 10^{-13} $$

and

$$ \sqrt[3]{7}\approx \frac{324048498507980-\sqrt{10}}{169398931559230}-1.37\cdot 10^{-17}. $$

Answer 6

Since you tagged it elementary number theory, let's give a proof at that level that no such equations can exist. If $\,w=\sqrt[3]7\,$ and $\,w^2 = a w + b\,$ and $\,w^2 = cw + d\,$ for rationals $\,a,b,c,d,\, a\ne c\,$ then subtracting we get $\, (a\!-\!c)w + b\!-\!d = 0\,$ so $\,w = (b\!-\!d)/(a\!-\!c)\,$ is rational, contra $\,w=\sqrt[2]7 $ is irrational, by $\,x^3-7\,$ has no rational roots via Rational Root Test.

Answer 7

The scale of integer factors in your expressions is $10^6$ while the roots themselves contain floating point errors in the region of $2^{-52}\approx 4·10^{-15}$ which gives an overall numerical uncertainty of a small multiple of $10^{-9}$.

The accumulated error of the sums is in the same scale as are the actual results reported in the answer of Patrick Stevens, so that your observations are a random result of error cancellations and the gaps between with 64bit floats representable numbers.