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# 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
11 months, 3 weeks 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, $\sum_{i = 1}^{49} i = \frac{49\cdot 50}{2} = (35)^2$ · 11 months, 3 weeks ago

Yes I got the above expression, but couldn't find the other one. · 11 months, 3 weeks ago

Other one? · 11 months, 3 weeks ago

As there are infinite $$a$$, so I can't find other value of $$a$$. · 11 months, 3 weeks ago

If you look at Deeparaj's comment or after some simple manipulations, you'll see $$41 \cdot 29 = 1189$$ is another value of $$a$$. · 11 months, 3 weeks 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 · 11 months, 3 weeks ago

$$\frac{n^2 + 1}{2}$$ had to be a perfect square. · 11 months, 3 weeks ago

OH!, Yes. You mean $$\dfrac{x^2 + 1}{2} = y^2 \implies$$ a perfect square · 11 months, 3 weeks ago

Yes, I changed the variables, as shown in the other comments. · 11 months, 3 weeks ago

Where $$x=n$$ and $$y=\frac{a}{n}$$ · 11 months, 3 weeks ago

Yes, I wanted him to see how the two were related :P · 11 months, 3 weeks ago