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A383359 Integers m such that m^4 is the sum of squares of two or more consecutive positive integers.

%I A383359 #80 May 12 2025 22:32:43
%S A383359 13,295,330,364,1085,5005,6305,15516,415151,1990368,34011252,42016497,
%T A383359 79565281,139107722,254801664,418093065,667378972,1214995500,
%U A383359 3609736702,4353556896
%N A383359 Integers m such that m^4 is the sum of squares of two or more consecutive positive integers.
%C A383359 a(21) > 10^10. - _Xianwen Wang_, May 08 2025
%C A383359 Terms of A383653 such that the consecutive integers are all positive.
%C A383359 From _David A. Corneth_, May 04 2025: (Start)
%C A383359 The sum of the first m positive squares is f(m) = m*(m + 1)*(2*m + 1) / 6.
%C A383359 The sum of consecutive squares m^2 + (m+1)^2 + ... + t^2 where 0 < m <= t may be written as f(t) - f(m-1) for some t and m.
%C A383359 From there we can factor out t - m - 1 and solve the system of equations going over divisors of 6*m^4.
%C A383359 To get divisors of 6*m^4 we need to factor 6*m^4 which can be done using the factors of 6 and the factors of m. Doing so makes we need to factorize smaller numbers. (End)
%H A383359 Zhining Yang, <a href="https://bbs.emath.ac.cn/forum.php?mod=viewthread&amp;tid=39808">Can be expressed as the fourth power of the sum of squares of consecutive positive integers</a>, Chinese BBS.
%e A383359 295 is a term because 295^4 = 6453^2 + 6454^2 + ... + 6628^2 + 6629^2.
%t A383359 d[m_] := Select[Divisors[6 m^4], 1 < # < Floor@ CubeRoot[3 m^4] &&
%t A383359     IntegerQ[1/6 (-3 (1 + #) + Sqrt[3 (12 m^4/# + 1 - #^2)])] &];
%t A383359 Do[If[Length@d[m] > 0, Print[m]], {m, 1, 10000}]
%Y A383359 Cf. A000330, A097812, A189173, A383367, A383653.
%K A383359 nonn,more
%O A383359 1,1
%A A383359 _Zhining Yang_, May 01 2025
%E A383359 a(11)-a(18) from _Xianwen Wang_, May 04 2025
%E A383359 a(19)-a(20) from _Xianwen Wang_, May 08 2025