  A suggestion is put forward regarding a partial proof of FLT(case1), which is
elegant and simple enough to have caused Fermat's enthusiastic remark in the
margin of his Bachet edition of Diophantus' "Arithmetica". It is based on an
extension of Fermat's Small Theorem (FST) to mod p^k for any k>0, and the cubic
roots of 1 mod p^k for primes p=1 mod 6. For this solution in residues the
exponent p distributes over a sum, which blocks extension to equality for
integers, providing a partial proof of FLT case1 for all p=1 mod 6. This simple
solution begs the question why it was not found earlier. Some mathematical,
historical and psychological reasons are presented. . . . . In a companion
paper, on the triplet structure of Arithmetic mod p^k, this cubic root solution
is extended to the general rootform of FLT (mod p^k) (case1), called "triplet".
While the cubic root solution (a^3=1 mod p^k) involves one inverse pair:
a+a^{-1} = -1 mod p^k, a triplet has three inverse pairs in a 3-loop: a+b^{-1}
= b+c^{-1} = c+a^{-1} = -1 (mod p^k) where abc = 1 (mod p^k), which reduces to
the cubic root form if a=b=c (\neq 1) mod p^k. The triplet structure is not
restricted to p-th power residues (for some p \geq 59) but applies to all
residues in the group G_k(.) of units in the semigroup of multiplication mod
p^k.
