Bill Gosper <billgosper@gmail.com> writes:
Wow. Do you still think all the apparently legal squarefrees will eventually show up? —rwg
Yes; the usual heuristics suggest this, and also that they appear so rarely that we it would have been somewhat surprising if a two-digit one appeared in the new data. The numbers x with x^2 = a^4 + b^4 + c^4 seem to be growing as expected: the n-th one is roughly (Cn)^2 for some constant C. That's "as expected" because there are about N^3 candidates a^4+b^4+c^4 of size N^4, and we'd expect about N of them to be squares (size N^2). The constant multiplier of that N^4 is positive but small, both because of the symmetry (we may assume 0 <= a <= b <= c) and because of congruence conditions (notably, that two of a,b,c must be multiples of 5). This makes C somewhat large -- numerically about 60 from the data I have. Now if d is a "legal squarefree" then a large number x is d*square with probability about 4/sqrt(d*x) (I get the factor 4 as 8/2, with the bonus of 8 because only one in eight d's is legal, and the denominator 2 from d(u^2)/du). So, we expect that the n-th solution will work with probability about 4/sqrt(d*(Cn)^2), which is a small constant times 1/n -- around 1 / (15*sqrt(d)) if I did this right. This means that we expect infinitely many solutions, but very sparse because of the logarithmic growth of sum(1/n) and the small coefficient: we'd have to try exp(15*sqrt(d)) solutions x to expect to find one with squarefree part d. Already for d=17 that's more than 10^26 solutions. So the fact that I've not found one among the first 10^4 or so does little do change my expectation that there are infinitely many examples of d=17 "out there". (Especially since I already found the first few thousand of solutions in the previous pass, so the sum of 1/n over the new ones is not log(10^4) ~ 9 but less than 2.) NDE