Is there a probabilistic argument that Fermat's Last Theorem is true?

In an amazing convergence of recent topics, I have found a
probabilistic proof of Fermat's Last Theorem by that great
mathematician Richard Feynman.  It is from _Review of Modern
Physics_, volume 58, no. 2, April 1986, page 502.

He argues that the spacing between (N)^(1/n) and (N+1)^(1/n)
is (for large N) (1/n)*N^(1/n-1).  So the probability that
x^n + y^n is a perfect nth power is
(x^n+y^n)^(1/n) / (n(x^n+y^n)) [Editor: closing parenthesis added].

So the total probability of any x^n+y^n being perfect nth
power for x > x0 and y > y0 is equal to

int_x0^infty { int_x0^infty 1/n(x^n+y^n)^(-1+1/n) dx dy }
                        = 2/n * cn * x0^(3-n)

where 2cn =
  int_0^infty int_0^infty (u^n+v^n)^(1/n-1) du dv

Feynman notes that c2 converges badly, corresponding to the
infinitude of solutions of x^2 + y^2 = z^2 [Editor: "= x^2" in the
original].  c3 diverges logarithmically so we can't use probability,
must use analytic techniques.

For large n, he goes on to show that the chance of finding a
counterexample is less than 10^-200, so "for my money Fermat's
Theorem is true."

Feynman also points out, however, that a similar argument
would show that something like x^7 + y^13 = z^11 (prime
powers) would have no solutions when in fact it has infinitely
many.

-- David Fry <fry@huma1.harvard.EDU>