Note that the probabilistic argument fails for n=3 for Fermat’s last theorem; call this (3,2) (power=3, number of summands is 2).
So we know (3,2) is impossible; Euler’s conjecture is the equivalent of saying that (n+1,n) is also impossible for all n. However, the probabilistic argument fails for (n+1,n) the same way as it fails for (3,2). So we’d expect Euler’s conjecture to fail, on probabilistic grounds.
In fact, the surprising thing on probabilistic grounds is that Fermat’s last theorem is true for n=3.
Note that the probabilistic argument fails for n=3 for Fermat’s last theorem; call this (3,2) (power=3, number of summands is 2).
So we know (3,2) is impossible; Euler’s conjecture is the equivalent of saying that (n+1,n) is also impossible for all n. However, the probabilistic argument fails for (n+1,n) the same way as it fails for (3,2). So we’d expect Euler’s conjecture to fail, on probabilistic grounds.
In fact, the surprising thing on probabilistic grounds is that Fermat’s last theorem is true for n=3.