Only $1$ and $4900$ are squares as $1+4+9+\ldots+ n^2$

Question

I encountered this fact yesterday: $1$ and $4900$ are the only squares as the sum of $1+4+9+\ldots +n^2$. I was trying to solve this problem using my knowledge of elementary number theory. I reduce it to the point:

Show that $(a,b,c)=(2,5,7)$ is the only positive integer solution to $$ \left\lbrace \begin{array}{} 6\times a^2+1=b^2 \\ 12\times a^2+1=c^2 \end{array} \right. $$ (then let $n=6\times a^2$, you get $4900= 24\times25\times49/6$)

I recognize these as Pell's equations, but I don't know how to proceed.

Answer 1

You are exploring what has been called the Cannonball problem. Édouard Lucas formulated the cannonball problem as a Diophantine equation $$\sum_{n = 1}^N n^2 = M^2$$ or $$\frac{1}{6} N(N + 1)(2N + 1) = \frac{2N^3 + 3N^2 + N}{6} = M^2.$$

Lucas conjectured that the only solutions are $N=1, M=1$, and $N=24, M=70$, using either $1$ or $4900$ cannon balls (as it seems you have observed). It was not until 1918 that G. N. Watson proved it, using a method that is far beyond on the scope of elementary number theory. However it was proved in $1990$ in just over $4$ pages, if you are interested in seeing a simpler proof.

Answer 2

(More of a long comment.) This is an old question but a Pell equation in it caught my eye. The post is about,

$$1^2+2^2+3^2+\dots+n^2=y^2$$

with stated solution $n=24$ for $n>1$. The OP arrived at a pair of Pell equations, one of which is,

$$6\times a^2+1=b^2$$

Or more accurately since $a=2q$ is even,

$$\; \color{blue}{24}\,q^2+1 =p^2$$

If we relax the requirement that the initial square is $1$, but retain the condition the LHS contains $\color{blue}{24}$ squares, then the same Pell equation appears. Define,

$$F(x) = x^2+(x+1)^2+(x+2)^2+\dots+(x+23)^2=y^2$$

Equivalently,

$$F(x) = 24(x-1)(x+24)+70^2=y^2$$

From the structure of $F(x)$, one can clearly that $F(1) = F(-24) = 70^2$. But since we relax the conditions on $x$, then there are infinitely many sets of $24$ consecutive squares such that their sum is a square, given by,

$$x = (p+24q)^2+4\times23\,pq\quad\\[4pt] \quad y = (p+24q)^2+3\times23\,(p+4q)^2$$

where $(p,q)$ solve the Pell equation,

$$p^2-24q^2=1$$

$p_n = 1,5,49,485,\dots$ (A001079)

$q_n =\, 0,1,10,99,\,\dots\,$ (A004189)

The three smallest cases yield,

$$\sum_{k=0}^{23} (1+k)^2 =70^2$$

$$\sum_{k=0}^{23} (1301+k)^2=6430^2$$

$$\sum_{k=0}^{23} (128601+k)^2=630070^2$$

and so on.

P.S. It seems the initial square conveniently has an ending digit of $1$, and is $x \equiv 1\, (\text{mod}\, 100)$. I've tested it for the first 20 $(p,q)$, but I'm not sure it is true for all.