## Number Theory Problem (sums of consecutive squares)

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!

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 Recognitions: Gold Member This might help: The term in parenthesis must have a factor of 11. The 385 is already divisible by 11, so the n2-10n must be divisible by 11. The 385 divided by 11 is equal to 35. Therefore, n2-10n divided by 11, when added to 35 must be a perfect square. The factors of n2-10n are n and n-10. Therefore, n is either 11k, or 11k+10, where k is an integer. Find the value of k that works.
 For the case considering m = 3, you got 3n^2 -6n + 5 = a^2. So now, all integers can be represented as 3k, 3k+1, or 3k+2. Try checking what remainders the squares of these general numbers give, when divided by 3, and apply the same idea to your equation too. Also, a good idea of choosing the numbers would be (n-1), (n), (n+1) since its easier to calculate :)

Recognitions:
Gold Member

## Number Theory Problem (sums of consecutive squares)

This refers to my post #2. The value of k that works is k=1, and n = 11. With n = 11, a = 6x11=66.

 Recognitions: Gold Member I checked your algebra in obtaining the equation 11(n2 -10n + 385) = a2, and got a different result: (66n2-661n+2310)/6 = a2

 Tags consecutive squares, maths, modular arithmetic, number theory, problem