Geometers have been much occupied with the general solution of algebraic equations, and several of them have sought to prove its impossibility. But, if I am not mistaken, nobody has succeeded so far.

— Niels Henrik Abel, Memoir on algebraic equations (1824)

Everyone who has been to school knows one formula for solving equations: the quadratic formula. The equation $ax^2+bx+c=0$ has the roots

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

The recipe uses only the four operations and a square root. The Babylonians could already solve problems of this kind almost four thousand years ago, and Euclid solved them with squares and rectangles. The natural question is: are there recipes like this for equations of higher degree?

Degrees 3 and 4

For degree 3 the answer came in Renaissance Italy. Scipione del Ferro, Niccolò Tartaglia and Gerolamo Cardano found the formula, which Cardano published in 1545 in the book Ars magna. For the equation $x^3+px+q=0$, one root is

$$x=\sqrt[3]{-\frac q2+\sqrt{\frac{q^2}4+\frac{p^3}{27}}}+\sqrt[3]{-\frac q2-\sqrt{\frac{q^2}4+\frac{p^3}{27}}}.$$

Let us check it on $x^3+6x-20=0$, that is, $p=6$ and $q=-20$. First the square root:

$$\frac{q^2}4+\frac{p^3}{27}=100+8=108,\qquad\sqrt{108}=6\sqrt3.$$

The formula gives

$$x=\sqrt[3]{10+6\sqrt3}+\sqrt[3]{10-6\sqrt3}.$$

It looks dreadful, but $(1+\sqrt3)^3=1+3\sqrt3+9+3\sqrt3=10+6\sqrt3$, and in the same way $(1-\sqrt3)^3=10-6\sqrt3$. So

$$x=(1+\sqrt3)+(1-\sqrt3)=2.$$

Indeed, $2^3+6\cdot2-20=0$. Soon afterwards Lodovico Ferrari, Cardano’s pupil, found the formula for degree 4, published in the same book. Both use only the four operations and roots: we say these equations can be solved by radicals.

Two hundred and fifty years of attempts

Degree 5 remained. For more than two centuries the greatest mathematicians tried, and every trick that worked up to degree 4 got stuck. In 1771 Lagrange explained why the tricks worked: they depended on how certain expressions in the roots change when the roots are shuffled. For degree 5 the count did not close. In 1799 the Italian Paolo Ruffini published a proof that there was no formula, but it had gaps and almost nobody accepted it.

Abel’s theorem

In 1824 Abel gave the complete proof:

$$\boxed{\text{There is no formula with the four operations and roots for the general equation of degree }n\geq5.}$$

Today the result is called the Abel-Ruffini theorem. The idea of the proof is to imagine that such a formula exists and to see how it would have to be built: in layers, each layer taking a root of expressions made from the previous layers. Abel showed that each of these intermediate roots would itself have to be an expression in the five roots of the equation. This is where the argument of Lagrange and Ruffini comes in: there are $5!=120$ ways of shuffling five roots, and the expressions a formula would need would have to change in a way that no expression in five letters can change. The assumption leads to a contradiction.

It is important to understand exactly what the theorem says. It does not say that no fifth-degree equation can be solved. The equation $x^5-32=0$ has the root $x=\sqrt[5]{32}=2$. What does not exist is a general formula that works for every equation of degree 5 from its coefficients. And there are specific equations, such as

$$x^5-x-1=0,$$

whose real root, $x\approx1.1673$, exists and can be computed to as many decimal places as desired, but cannot be written with radicals in any way. Which equations can be solved and which cannot was the question that Galois answered a few years later.

Exercise: apply Cardano's formula to x³ − 15x − 4 = 0, as Rafael Bombelli did in 1572.

Here $p=-15$ and $q=-4$, and $\frac{q^2}4+\frac{p^3}{27}=4-125=-121$. The formula asks for $\sqrt{-121}=11i$, the root of a negative number: $x=\sqrt[3]{2+11i}+\sqrt[3]{2-11i}$. Bombelli noticed that $(2+i)^3=8+12i-6-i=2+11i$ and that $(2-i)^3=2-11i$. So $x=(2+i)+(2-i)=4$, and indeed $64-60-4=0$. Imaginary numbers entered mathematics through this door: to reach a real answer, the cubic formula sometimes has to pass through them.

Exercise: can the equation x⁵ − 3 = 0 be solved by radicals? Does this contradict Abel's theorem?

Yes, $x=\sqrt[5]3\approx1.2457$ is a root, written with a radical. There is no contradiction: the theorem is about the general equation $ax^5+bx^4+cx^3+dx^2+ex+f=0$, with arbitrary coefficients. Some particular equations of degree 5 can be solved by radicals; what does not exist is a single formula that serves for all of them.

Sources: Niels Henrik Abel, Mémoire sur les équations algébriques (Christiania, 1824) and “Beweis der Unmöglichkeit, algebraische Gleichungen von höheren Graden als dem vierten allgemein aufzulösen”, Journal für die reine und angewandte Mathematik 1 (1826) · Peter Pesic, Abel’s Proof (MIT Press, 2003) · Abel biography, MacTutor.

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