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A383653 Integers m such that m^4 is the sum of squares of two or more consecutive integers, positive or negative.

%I A383653 #29 May 13 2025 22:44:50
%S A383653 1,13,26,33,295,330,364,1085,5005,5546,5682,6305,6538,15516,415151,
%T A383653 1990368,3538366,34011252,42016497,79565281,139107722,175761059,
%U A383653 254801664,418093065,667378972,1214995500,3609736702,4353556896
%N A383653 Integers m such that m^4 is the sum of squares of two or more consecutive integers, positive or negative.
%C A383653 a(29) > 10^10.
%C A383653 From _David A. Corneth_, May 04 2025: (Start)
%C A383653 The sum of the first m positive squares is f(m) = m*(m + 1)*(2*m + 1) / 6.
%C A383653 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 A383653 From there we can factor out t - m - 1 and solve the system of equations going over divisors of 6*m^4.
%C A383653 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 A383653 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 A383653 5546 is a term because 5546^4 = (-22205)^2 + (-22204)^2 + ... + 141400^2 + 141401^2.
%t A383653 lst={};Monitor[Do[mm=6 m^4;div=TakeWhile[Divisors[mm][[2;;-2]],2mm/#+1>#^2&];
%t A383653 ans=Select[div,IntegerQ[Sqrt[(2mm/#+1-#^2)/3]]&&Mod[#-Sqrt[(2mm/#+1-#^2)/3],2]==1&];
%t A383653 If[Length[ans]>0,tmp={m,{#,q=Sqrt[(2mm/#+1-#^2)/3],p=(q+1-#)/2}&/@ans};Print[tmp];
%t A383653 AppendTo[lst,tmp]],{m,1,10^4}],m];lst
%Y A383653 Cf. A000330, A097812, A189173, A383359, A383367, A383654.
%K A383653 nonn,more
%O A383653 1,2
%A A383653 _Xianwen Wang_, May 04 2025