cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

481, 24961, 28721, 65441, 69121, 113241, 345761, 362401, 384161, 530881, 620321, 854401, 882889, 909321, 1094481, 1163249, 1305281, 1697761, 1855841, 2074281, 2294681, 2423601, 2568369, 2576641, 2619281, 2665721, 2696161, 2751489, 2997761, 3151281
Offset: 1

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Author

Jud McCranie, Sep 16 2023

Keywords

Comments

A solution is primitive if gcd(u,v,w) = 1.
Multiplying k by a positive square gives the terms in A365657.
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
From David A. Corneth, Sep 26 2023: (Start)
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).
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)
From Jon E. Schoenfield, Sep 28 2023: (Start)
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).
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.
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.
Conjectures:
(1) k mod 8 = 1 (hence k mod 40 is 1 or 9).
(2) Of u,v,w, the two even numbers are divisible by 4. (End)

Examples

			481^2 = 231361 = 12^4 + 15^4 + 20^4.
		

Crossrefs

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