RePoSS #11: The Mathematics of Niels Henrik Abel: Continuation ...

6.6. Combination into an impossibility proof 119
link between root extractions and permutations ensured him that the function v would
take on m values under all permutations of x1, . . . , x5. Since m was a prime and had
to divide 5! by LAGRANGE’S theorem, ABEL argued, the only possibilities were that
m equaled 2, 3, or 5. And since no function of five quantities could have three values
under permutations by the CAUCHY-RUFFINI theorem, ABEL ruled out this possibility.
The two remaining cases were subsequently both brought to contradictions.
The innermost root extraction could not be a fifth root. In the simplest case, corre-
sponding to m = 5, the function v had to have the form of a fourth degree polynomial,
as ABEL had demonstrated:
v = 5√ R =
4
∑ rkx k=0
k 1 .
Through a process of inversion of polynomials in which the quintic equation (6.15) was
used to lower the degree, ABEL found that
x1 =
4
∑ skR k=0
k 5 ,
where s0, . . . , s4 were symmetric functions of x1, . . . , x5. Furthermore, by use of basic
properties of primitive roots of unity, he obtained
s1R 1 5 = 1
�
x1 + α
5
4 x2 + α 3 x3 + α 2 �
x4 + αx5 ,
where α was a primitive fifth root of unity. The left hand side of the equation was a so-
lution to a fifth degree equation, and thus had (at most) five different values, whereas
the right hand side was formally altered by any permutation of x1, . . . , x5 and thus
had 5! = 120 different values. This ruled out the case m = 5, and the innermost root
extraction could not be a fifth root.
The innermost root extraction could not be a square root, neither. ABEL brought
the case m = 2 to a contradiction in a similar way, although it involved studying
expressions of the second order as well. He knew that the root would have to be of
the form
√ R = p + qs,
and the other value under permutations would be
− √ R = p − qs.
Subtracting these two, ABEL concluded that √ R was of the form
√ R = qs,