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

1, 13, 26, 33, 295, 330, 364, 1085, 5005, 5546, 5682, 6305, 6538, 15516, 415151, 1990368, 3538366, 34011252, 42016497, 79565281, 139107722, 175761059, 254801664, 418093065, 667378972, 1214995500, 3609736702, 4353556896

Offset: 1

Author

Xianwen Wang, May 04 2025

Keywords

• nonn
• more

Comments

a(29) > 10^10.

From David A. Corneth, May 04 2025: (Start)

The sum of the first m positive squares is f(m) = m*(m + 1)*(2*m + 1) / 6.

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.

From there we can factor out t - m - 1 and solve the system of equations going over divisors of 6*m^4.

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)

Examples

			5546 is a term because 5546^4 = (-22205)^2 + (-22204)^2 + ... + 141400^2 + 141401^2.
		

Links

• Zhining Yang, Can be expressed as the fourth power of the sum of squares of consecutive positive integers, Chinese BBS.

Crossrefs

Cf. A000330, A097812, A189173, A383359, A383367, A383654.

Programs

Mathematica

lst={};Monitor[Do[mm=6 m^4;div=TakeWhile[Divisors[mm][[2;;-2]],2mm/#+1>#^2&];
ans=Select[div,IntegerQ[Sqrt[(2mm/#+1-#^2)/3]]&&Mod[#-Sqrt[(2mm/#+1-#^2)/3],2]==1&];
If[Length[ans]>0,tmp={m,{#,q=Sqrt[(2mm/#+1-#^2)/3],p=(q+1-#)/2}&/@ans};Print[tmp];
AppendTo[lst,tmp]],{m,1,10^4}],m];lst