Problem of sum of consecutive squares equal to a 4th power. y^4 = (x+1)^2 + (x+2)^2 +....+ (x+n)^2.1.IntroductionWe 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).2. MethodSearch 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]3.Numerical solutions

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