1. The problem statement, all variables and given/known data The sum of two consecutive squares can be a square: for instance, 32 + 42 = 52 (a) Prove that the sum of m consecutive squares cannot be a square for the cases m = 3; 4; 5; 6. (b) Find an example of eleven consecutive squares whose sum is a square. 3. The attempt at a solution The only thing i can think of using is the formula for the sum of square numbers e.g. for part one for the case where m = 3, it simplifies to 3n2 -6n + 5 = a2. I dont know how to prove this cannot be a square. (or e.g. in part b) - (n)(n+1)(2n+1)/6 - (n-11)(n-10)(2n-21)/6 = a2 which simplifies to 11(n2 -10n + 385) = a2 This would suggest to me that a is divisible by 11, however i'm not sure how to generate a number such that it is equal to a square - I've thought of maybe trying to substitute a value for n which will allow me to factor the l.h.s. into a square, and then i'd be done, however I don't have a clue. Is this even the right direction to be going in? Or is there some easy way to solve this with modular arithmetic? Thanks in advance!