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

16.3. The division problem 315
Based on a direct application of the addition formulae, ABEL obtained an expression
� �
2µ ¯ωi
φ1 β ± = Rµ ± R
2n + 1
′ �
µ (1 − c2x2 ) (1 + e2x2 )
in which Rµ and R ′ µ were rational functions of x = φ (β). Consequently, ABEL found
that both the functions
ψ (β) ψ1 (β) = λ (β) and ψ (β) 2n+1 + ψ1 (β) 2n+1 = λ1 (β) (16.9)
were rational in x. By direct calculation, he also found that both functions (16.9) were
invariant if another root β + kω+k′ ¯ωi
2n+1
of the equation
φ ((2n + 1) β) = P2n+1
Q2n+1
(16.10)
was substituted for β. Thus, ABEL knew that λ (β) and λ1 (β) were rational in the
coefficients of (16.10), in particular in the quantity φ ((2n + 1) β). When he solved the
system of equations (16.9), ABEL found
ψ (β) = 2n+1
�
�
�
�λ1 (β)
2 +
�
λ1 (β) 2
4
− λ (β). (16.11)
From these, ABEL obtained φ1 (β) and then φ (β) by similar arguments.
However, as ABEL observed, the formula for φ (β) which he had obtained also
contained the quantities
� �
ω
φ
2n + 1
� �
¯ωi
and φ .
2n + 1
Thus, in order to completely solve the problem, these two quantities should also be
determined and ABEL demonstrated how the equation P2n+1 = 0 which determined
these could be reduced to equations of lower degrees, one of degree 2n + 2 and 2n + 2
equations of degree n. Furthermore, ABEL also proved that the equations of degree
n were always solvable by radicals. In the process, ABEL employed tools similar to
those described above as well as some knowledge of primitive roots of an integer. Im-
portantly, ABEL knew qualitatively how the roots were interrelated (by 16.8) and used
this knowledge to investigate the system of roots and prove its reduction to equations
of lower degree, some of which were proved to be solvable by radicals.
16.3.1 Division of the lemniscate
The culmination of ABEL’S research into the division problem was his application of
the theory to the case of the lemniscate. The symmetry of ABEL’S representation of
the elliptic integrals became evident when he chose e = c = 1 to obtain the lemniscate
integral
φ (α) = x, α =
� x
0
dx
√ 1 − x 4 .