1.Introduction

Problem of sum of consecutive squares equal to a 4th power.
y^4 = (x+1)^2 + (x+2)^2 +....+ (x+n)^2.

2. Method

We consider below diophantine equation.

y^4 = nx^2+n(n+1)x+1/6n(n+1)(2n+1)...............................(1)

Since discriminant for x must be perfect square number, then we obtain

v^2 = -3n^4+3n^2+36y^4n..........................................(2)

We consider equation (2) as quartic equation of n and searched the integer solutions of equation (2).



3.Numerical solutions

Search results of the integer points for equation (1) by brute force.

y<1000000

[  n    x    y ]

[2, 118, 13]
[177, 6452, 295]
[352, 5627, 330]
[1536, 2583, 364]
[2401, 22814, 1085]
[40898, 102854, 5005]
[60625, 130188, 6305]
[185761, 463115, 15516]
[19512097, 28852808, 415151]