Six days ago I wrote: On Mon, Dec 24, 2018 at 5:42 PM Elkies, Noam <elkies@math.harvard.edu> wrote: < 5441 arises for {a,b,c} = {1960, 2183, 2360} (with square factor 39^2). < An overnight search roughly tripled the number of examples up to 10^6; < the new ones start 201, 889, 5441, 5601, 6969, 7761, 8049. < No two-digit solutions turned up, but it may be reasonable to guess that < every squarefree 8k+1 arises eventually, though it may take a long time. < Indeed I might venture to call this guess a conjecture once it's known that < there's no "Brauer-Manin obstructuion". < Trying all a,b,c up to N takes time about N^3; that's basically what I did, < though with some tricks [ . . . ] Update: I coded this more carefully to make sure that every primitive solution of a^4+b^4+c^4 = x^2 appears as long as (a,b,c) is in the closed ball of radius R about the origin (i.e. a^2+b^2+c^2 <= R^2), and extended the search to R = 2^20 (so a bit over one million). This took about two weeks of CPU time, which came to about one day on 19 processors. That's probably about as far as it makes sense to take this method, because the run time is proportional to R^3, and the R^2 techniques, albeit with a much larger constant, are surely faster by now. For R = 2^20, there are 15838 primitive solutions with 0 <= a <= b <= c, and as expected the count grows about linearly in R -- e.g. 7981 of the 15838 solutions are in the R=2^19 ball. Also as expected, the squarefree part 1 (re)appeared for (a,b,c) = (95800, 217519, 414560). It was not quite the first duplicate, though: x = 53781801801 works for both (a,b,c) = (38960, 123140, 227107) and (a,b,c) = (6365, 87224, 230740). This x is squarefree (3*7*13*7417*26561); the next duplicate is not: 313049121489 = 3^4 13 269 1105177 for both (a,b,c) = (49600, 91319, 559400) and (a,b,c) = (91319, 426200, 504800). There were no further duplicates even when each x was reduced to its squarefree part. No two-digit squarefree parts turned up; indeed the smallest new squarefree part is 24401 = 13*1877, for (a,b,c) = (52360, 133145, 139384) with x = 1041^2 * 24401. Happy 1^4 + 2^4 + 3^4 + 5^4 + 6^4, --NDE

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