# Is Summation a perfect square?

$\sum_{i=1}^{n^2} (i) = a^2$

$(n,a)\in N$ , $N$ denotes natural number.

Does there exist only one such $a$ to satisfy the above conditions? Note by Akash Shukla
3 years, 7 months ago

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No, there are actually infinitely many $a$.
The Pell Equation, $x^2 - 2y^2 = -1$ has infinite solutions, eg, $(1,1), (7,5), (41, 29)$.
So, we can have, $\sum_{i = 1}^{49} i = \frac{49\cdot 50}{2} = (35)^2$

- 3 years, 7 months ago

Where $x=n$ and $y=\frac{a}{n}$

- 3 years, 7 months ago

Yes, I wanted him to see how the two were related :P

- 3 years, 7 months ago

Oh. Too bad I spoilt it then. :(

- 3 years, 7 months ago

Yes I got the above expression, but couldn't find the other one.

- 3 years, 7 months ago

Other one?

- 3 years, 7 months ago

As there are infinite $a$, so I can't find other value of $a$.

- 3 years, 7 months ago

If you look at Deeparaj's comment or after some simple manipulations, you'll see $41 \cdot 29 = 1189$ is another value of $a$.

- 3 years, 7 months ago

Yes I got this. Thank you so much. it has wonderful connection with pell equation. How do you know that pell equation and my question are related

- 3 years, 7 months ago

$\frac{n^2 + 1}{2}$ had to be a perfect square.

- 3 years, 7 months ago

OH!, Yes. You mean $\dfrac{x^2 + 1}{2} = y^2 \implies$ a perfect square

- 3 years, 7 months ago

Yes, I changed the variables, as shown in the other comments.

- 3 years, 7 months ago