Bill Gosper <billgosper@gmail.com> writes:
_Very_ probably spotty list:
Yes, see below.
the squarefree part of integers of the form sqrt(a^4+b^4+c^4).
Nice question!
Like the congruent numbers, you can't rule them out with finite searching.
Unlike congruent numbers, if there's no solution then we don't know in general how to prove it.
1 doesn't show up until a=414560, b=217519, c=95800 (https://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture).
Unless you allow 0,0,1 . . .
Are there really none with four digits? Two?
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 to avoid having to test directly whether each candidate a^4+b^4+c^4 is a square. To go much beyond that, one would probably want an algorithm that takes much less than N^3 time; one approach is to factor all the candidates a^4+b^4, and then check, for each solution of a^4+b^4 = x^2-y^2 in the desired range, whether y is a square. You can assume that a and b are multiples of 5, with a <= b and a,b not both odd. Factoring should take subexponential time but is still expensive to do N^2 times. If you want to search for solutions of a^4+b^4+c^4 = r^2*d^4 for one or a few target r such as 17 or 33, you can do it in about N^2 time (not N^2 factorizations) and N space using the technique described in Dan Bernstein's "sorted sums" page <http://cr.yp.to/sortedsums.html>. The number of primitive solutions of a^4+b^4+c^4 = x^2 with a,b,c < N should grow more-or-less linearly in N, but I don't think I know a way to find all or even a substantial fraction of them in time significantly less than N^2. Happy New Year's Eve Eve Eve Eve Eve Eve Eve Eve (see today's XKCD <xkcd.com/2089>), --Noam D. Elkies 1: [0, 0, 1] 1 201: [80, 1192, 1535] 117 409: [120, 136, 255] 13 481: [12, 15, 20] 1 889: [7500, 13299, 17960] 647 5441: [1960, 2183, 2360] 39 5601: [4967, 6640, 12320] 169 6969: [7460, 12335, 30748] 371 7761: [3083, 3320, 3740] 51 8049: [7360, 11512, 12025] 159 17089: [59540, 63320, 111653] 891 18161: [18275, 21160, 25852] 219 24961: [60, 65, 156] 1 27641: [9389, 102880, 194300] 1191 28721: [60, 135, 148] 1 29921: [260, 775, 1532] 9 36849: [25883, 42220, 88360] 467 53209: [640, 6085, 6676] 33 59289: [6319, 10460, 23860] 99 65089: [468, 9060, 10865] 47 65441: [72, 175, 240] 1 66889: [800, 1372, 2245] 9 79161: [1936, 7220, 7435] 31 86601: [16372, 50425, 107900] 371 110441: [1640, 1852, 2795] 9 113241: [155, 260, 296] 1 121609: [265, 700, 988] 3 131001: [185, 1072, 7600] 21 178481: [66780, 81036, 86335] 247 183641: [1752, 27920, 36135] 91 212729: [164940, 240840, 279397] 689 272649: [28681, 156800, 176080] 381 345761: [60, 431, 540] 1 362401: [168, 175, 600] 1 364009: [38252, 75225, 87780] 163 384161: [408, 465, 520] 1 407001: [217400, 295157, 380860] 657 428129: [220, 35475, 38112] 67 489689: [8955, 17960, 18252] 31 530881: [420, 580, 609] 1 543121: [2000, 4849, 6080] 9 562489: [1241, 5520, 7800] 11 564689: [2160, 2431, 5160] 7 620321: [120, 264, 785] 1 688441: [61, 1880, 7460] 9 720921: [4220, 14885, 25616] 31 763609: [340, 559, 2620] 3 825929: [78020, 108692, 109145] 147 854401: [65, 180, 924] 1 882889: [355, 780, 792] 1 909321: [80, 764, 835] 1 937561: [2659, 30740, 32680] 39