Barlow's Observation

Some interesting characteristics of any putative counter-example of 
Fermat's Last Theorem can be deduced quite simply, and can be used
to place very strong constraints on the set of possible counter-
examples.  One of these was noted by Barlow in 1810 (and probably
by others earlier).  Assume non-zero integers x,y,z such that

                 x^p + y^p + z^p  =  0                           (1)

Without loss of generality we consider only primitive solutions, i.e.,
those for which x,y,z are pairwise coprime.  Now we wish to show that
the quantity (x+y+z)^p is of the form pABCD where A,B,C,D are pairwise
coprime integers with

               A = x+y      B = x+z      C = y+z

To prove this, notice that the binomial expansion allows us to write

 [(x+y) + z]^p  =  (x+y)^p  +  p (x+y)^(p-1) z

                              +  [p(p-1)/2] (x+y)^(p-2) z^2


                                +  p (x+y) z^(p-1)

                                 +  z^p                           (2)

Also, since

   -z^p   =   x^p + y^p

          =  (x+y) [x^(p-1) - x^(p-2) y + ... + y^(p-1)]          (3)

we know the last term is divisible by (x+y), and so the whole quantity
[(x+y)+z]^p is divisible by (x+y).  By symmetrical arguments, we can
show that this quantity is also divisible by (x+z) and (y+z).  Further,
it's clear that (x+y), (x+z), and (y+z) must be pairwise coprime,
because (for example) any prime that divides (x+y) also divides
(x+y+z) as we've just seen, and so it must divide z.  It follows that
it a given prime divides TWO of the numbers (x+y), (x+z), (y+z), then
it divides two of the numbers z,y,x, contradicting the pairwise
co-primeness of these these basic numbers.

Equation (3) also shows that the integer (z^p)/(x+y) is coprime to
(x+y), except for a possible factor of p.  To see this, suppose the
contrary, i.e., that the right hand factor of (3) is congruent to zero
modulo some prime divisor q of (x+y).  This means x=-y (mod q), so we
can substitute -x for y into the right hand factor of (3) to give the
putative congruence

                p x^(p-1)  =  0   (mod q)

Obviously since q divides x+y it cannot be a divisor of x (recallng
that x and y are mutually coprime), so this congruence is impossible
unless the prime q happens to equal the exponent p.

Therefore, once we have divided equation (2) by (x+y), including taking
a factor of (x+y) out of the last term z^p, each of the remaining terms
still has a factor of (x+y) or z (which shares every prime divisor of
(x+y)), with the exception of the last term, z^p/(x+y), which we have
just seen is coprime to (x+y), with the possible exception of a factor
of p.  We can also see from the binomial coefficients of (2), excluding
the terms x^p + y^p + z^p which vanish, every one is a multiple of p,
so the entire quantity (x+y+z)^p is a multiple of p, as well as being
a multiple of the mutually coprime quantities A=(x+y), B=(x+z), and
C=(y+z), at most one of which is itself a multiple of p.  Thus, we
                     (x+y+z)^p  =  pABCD

where A,B,C,D are pairwise coprime integers.  Since the left-hand
side is a pth power, and the right-hand factors are coprime, it
follows from unique factorization that each of the right-hand factors
is a pth power power, except for whichever one (if any) of them is
a multiple of p, which must be p^(p-1) times a pth power.  Hence if
none of the integers x,y,z is divisible by p (i.e., in Case 1 of
Fermat's Last Theorem) we have integers a,b,c,d such that

     x+y = a^p      x+z = b^p      y+z = c^p     pD = d^p

On the other hand, if one of the base variables, say, z, is divisible 
by p (i.e., Case 2), then we have

    p(x+y) = a^p      x+z = b^p      y+z = c^p      D = d^p

Return to MathPages Main Menu