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COMMENTS
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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
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)
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)
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