CHEZ FERMAT A.D. 1637' Erkka Maula and Eero Kasanen Abstract ...

144 ERKKA MAULA & EERO KASANEN
means [7:87],
p (a,k).2a/(a+1) : p(a,k+1/2) = p(a,k+l/2) : p(a,k).(a+1)/2
where p(a,k).2a/(a+1) < p(a,k+1/2) < p(a,k).(a+l)/2,
p(b,k).2b/(b+l) : p(b,k+l/2) = p(b,k+l/2) : p(b,k).(b+l)/2
where p(b,k).2b/(b+l) < p(b,k+1/2) < p(b,k).(b+l)/2,
p (e,k).2e/(e+l) : p(e,k+l/2) = p(c,k+1/2) : p(c,k).(c+1)/2
where p(c,k).2e/(c+l) < p(c,k+1/2) < p(c,k).(c+l)/2,
for aIl aeute-angled scalene integer-side triangles (a,b,e), a > b
> c > o.
Taking now new harmonie and arithmetical means (of second
degree) of the previous harmonie and arithmetical means of
p(a,k+l), p(a,k); p(b,k+1), p(b,k); p(c,k+l), p(c,k), and sa on, ad
infinitum, one will ob tain ever better approximations from below
and from above, ta the geometric means p(a,k+l/2), p(b,k+l/2),
p(c,k+1/2). As the geometrie means are, by counter-assumption,
sides of a right-angled triangle, aU harmonie means of the same
degree must be sides of acute-angled triangles and an arithmetical
means of the same degree sides of obtuse-angled triangles.
In arder ta establish a contradiction it is sufficient ta prove
that any of the triangles with harmonic means as aides is either
right-angled or obtuse-angled, or any of the triangles with
arithmetical means as sides is either right-angled or aeuteangled.
As Prop. 4 already proves FLT for a11 even exponents, it
suffices in facl to prove that any of the triangles with harmonic
means (of the same degree N) as sides, say (p(a,o"'î), p(b,0"'2),
p(c,0"'3» where 1 ~ k < o"'î < oA2 < 0"'3 < k+l/2, is either
right-angled or obtuse-angled. By Defs. 2-3 that is the case
when p(a,20"'1) ~ p(b,20"'2) + p(c,20"'3). We shall claim that in our
praof reconstruction of Prop. 5 (below). Note that the claim
implies p(a,20"'1) > p(b,20Al) + p(c,20"'1), because p(b,u""2) >
p(b,o"'l) and p(c,0"'3) > p(c,o"'l). This additional implication is
vital.
Given a,b,c and the degree N, the exponents u .... î, u ..... 2, 0 ..... 3 are
computable, of course, but actually the harmonic means of degree
N will suffice. For N = 1,2,3 the harmonic means between p(r,k)
and p(r,k+1) are (where H = r+1 and G = p(H,2)+4r):
2p(r,k+1)/H < 4p(r,k+l)H/G < 8p(r,k+l)HG/(p(G,2)+rp(4H,2» <
••• < m"'(N,harm)(p(r,k+l),p(r,k» < m .... geom(p(r,k+l),p(r,k»; r > 1,
1 are integers, N > 3 indicates the degree of the harmonic
k ~
mean, and increases ad infinitum (notation is for brevity).
Perhaps we must mention in passing rusa the Golden Section.
Note that if we rewrite p(a,2k+1) = p(b,2k+1) + p(c,2k+1) as
p(a,2k+1)/p(b,2k+1) - p(c,2k+1)/p(b,2k+l) = 1, three mutually
exclusive, necessary conditions for.a non-trivial solution are