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A365688 Primitive solutions k to k^2 = u^4 + v^4 + w^4, with u, v, w > 0.

%I A365688 #39 Sep 29 2023 08:26:19
%S A365688 481,24961,28721,65441,69121,113241,345761,362401,384161,530881,
%T A365688 620321,854401,882889,909321,1094481,1163249,1305281,1697761,1855841,
%U A365688 2074281,2294681,2423601,2568369,2576641,2619281,2665721,2696161,2751489,2997761,3151281
%N A365688 Primitive solutions k to k^2 = u^4 + v^4 + w^4, with u, v, w > 0.
%C A365688 A solution is primitive if gcd(u,v,w) = 1.
%C A365688 Multiplying k by a positive square gives the terms in A365657.
%C A365688 A term in this sequence is a square iff its square root is in A003828. The smallest term in A003828 is 422481, so the smallest square in this sequence is 422481^2 = 178490195361. - _Jon E. Schoenfield_, Sep 24 2023
%C A365688 From _David A. Corneth_, Sep 26 2023: (Start)
%C A365688 If k^2 = u^4 + v^4 + w^4 then k^2 - u^4 = (k - u^2)(k + u^2) = v^4 + w^4. Hence to find terms we can iterate over v and w to find values v^4 + w^4 which we then factor into pairs (d, t) such that d*t = (k - u^2)(k + u^2).
%C A365688 It follows that d and t must be even and one of (k - u^2) and (k + u^2) is NOT divisible by 4. Dividing both by 2 gives one of them odd so we only care about odd divisors of (w^4 + v^4)/4. (End)
%C A365688 From _Jon E. Schoenfield_, Sep 28 2023: (Start)
%C A365688 For every integer j, j^4 mod 16 = 0 if j is even, 1 if j is odd. Consequently, if k were even (which would make k^2 divisible by 4), then u,v,w would all have to be even as well, so the solution (k,u,v,w) would not be primitive. Thus every term k is odd, so k^2 mod 8 = 1, so exactly one of u,v,w is odd, and since (u^4 + v^4 + w^4) mod 16 = 1, k^2 mod 16 = 1, so k mod 8 is either 1 or 7 (not 3 or 5, because those would give k^2 mod 16 = 9).
%C A365688 Similarly, for every integer j, j^4 mod 5 = 0 if 5 divides j, 1 otherwise, so if k were divisible by 5 (and k^2 were thus also divisible by 5), u,v,w would all likewise have to be divisible by 5, so the solution (k,u,v,w) would not be primitive. Thus no term k is divisible by 5, so k^2 mod 5 is never 0. This leaves the only possible values of k^2 mod 5 as 1 (when k mod 5 is 1 or 4) and 4 (when k mod 5 is 2 or 3). But k^2 mod 5 must equal (u^4 + v^4 * w^4) mod 5, so k^2 mod 5 cannot be 4; it must be 1, so k mod 5 must be 1 or 4, and exactly one of u,v,w is not divisible by 5.
%C A365688 Thus k mod 40 = 1, 9, 31, or 39; exactly two of u,v,w are even; and exactly two of u,v,w are divisible by 5.
%C A365688 Conjectures:
%C A365688 (1) k mod 8 = 1 (hence k mod 40 is 1 or 9).
%C A365688 (2) Of u,v,w, the two even numbers are divisible by 4. (End)
%H A365688 Jud McCranie, <a href="/A365688/b365688.txt">Table of n, a(n) for n = 1..1093</a>
%e A365688 481^2 = 231361 = 12^4 + 15^4 + 20^4.
%Y A365688 Cf. A003828, A365657.
%K A365688 nonn
%O A365688 1,1
%A A365688 _Jud McCranie_, Sep 16 2023