Sum of squares

This is the Cannonball problem . The only solutions $(a,N)$ are (1,1) and (24,70).

My solution. Use the fact that the sum of consecutive squares from $1$ to $n$ is $\dfrac{n(n+1)(2n+1)}{6}$ . Now this must be a square, $x^2$ . Rewrite it as $n(n+1)(2n+1)=6x^2$ Now multiplying the two parenthesis we obtain $n(2n^2+3n+1)=6x^2$ . The $gcd(n,2n^2+3n+1)=1$ so let's say that $6\mid n \implies n=6k$ . By substitution in the original equation we obtain $k(6k+1)(12k+1)=x^2$ now all three terms must be perfect squares because they are all coprimes (proved easily thanks to Euclidean algorithm). The smallest value that makes this true is $k=4 \implies n=6\times4=24$ so the smalles solution is $n=24$ and thus $x=70$ .

Python 3.5.1:

def issquare(n):
    if n**0.5==round(n**0.5):
        return True
count=2
while not issquare(sum(i**2 for i in range(1,count+1))):
    count+=1

print (count)

Is there any approach to this question using pure number theory? Please comment.!