There’s a difficulty here involving the fact that every finite set of possible counterexamples has measure zero in the set {(a, b, c, n) | a, b, c, n in N} equipped with the probability measure that assigns each possible counterexample an equal probability.
So all the usual biases and cognitive gotchas involving probability and infinity come into play (even when the people doing the thinking are mathematicians!), and all bets are off.
My hypothesis is that the commonly used prior for counterexample distributions is exponential. As the lower bound K >= a, b, c, n on possible counterexamples increases, the exponential is updated into something close to uniform on the rest of {(a, b, c, n) | a, b, c, n in N}.
This does somewhat dodge the question, but it does make a difference that an infinite set of counterexamples can be associated with each counterexample. That is, if (a,b,c,n) is not a solution to the Fermat equation, then (ka,kb,kc,n) isn’t either for any positive integer k.
There’s a difficulty here involving the fact that every finite set of possible counterexamples has measure zero in the set {(a, b, c, n) | a, b, c, n in N} equipped with the probability measure that assigns each possible counterexample an equal probability.
So all the usual biases and cognitive gotchas involving probability and infinity come into play (even when the people doing the thinking are mathematicians!), and all bets are off.
My hypothesis is that the commonly used prior for counterexample distributions is exponential. As the lower bound K >= a, b, c, n on possible counterexamples increases, the exponential is updated into something close to uniform on the rest of {(a, b, c, n) | a, b, c, n in N}.
This does somewhat dodge the question, but it does make a difference that an infinite set of counterexamples can be associated with each counterexample. That is, if (a,b,c,n) is not a solution to the Fermat equation, then (ka,kb,kc,n) isn’t either for any positive integer k.