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I read on a poster today that Fibonacci showed that $x^3+2x^2+10x=20$ has no solution expressible in radicals, way back when.

I couldn't find the proof anywhere. Does anyone know where I can find it?

Leonardo da Vinci
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    That must be wrong; the Cardano-Tartaglia formula shows that every cubic equation with integer coefficients has a solution expressible in radicals. – MJD May 02 '14 at 22:27
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    See http://www.math.vt.edu/people/brown/doc/fibo_number.pdf for a discussion of what Fibonacci did or did not do. – Barry Cipra May 02 '14 at 22:29
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    The key point in the article Barry Cipra linked: “[Fibonacci] proves that [the real root] is neither an integer, nor rational, nor any of the forms from Book X of Euclid's Elements. He continues, ‘And because it was not possible to solve this equation in any of the above ways, I worked to reduce the solution to an approximation.’” The approximation was (in base-60 notation) $1;22,07,42,33,04,40$, or around $1.36880810785322359396$. – MJD May 02 '14 at 22:40

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He proved that the solution cannot be one of Euclid's irrationals. All Euclid's irrationals are strictly contained in the set of numbers of the form $$ \sqrt[4]{p}\pm\sqrt[4]{q}, \qquad p,q\in\mathbb{Q}. $$ The proof would be similar to (but of course more complicated than) how you prove $\sqrt2$ is not rational.

timur
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Your claim is wrong. Here is the one real root the polynomial has. Define $\alpha = \sqrt[3]{176+3\sqrt{3930}}$. Then the real root is $$\dfrac13\left(-2 - \dfrac{13 \cdot 2^{2/3}}{\alpha} + \alpha\sqrt[3]2\right)$$

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    I believe that the OP meant "with a formula involving only the four basic operations and the extraction of square roots" – mau May 03 '14 at 07:23