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

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

Does there exist only one such \(a\) to satisfy the above conditions?

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

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

Does there exist only one such \(a\) to satisfy the above conditions?

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## Comments

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TopNewestNo, 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, \[ \sum_{i = 1}^{49} i = \frac{49\cdot 50}{2} = (35)^2 \] – Ameya Daigavane · 9 months, 3 weeks ago

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– Akash Shukla · 9 months, 3 weeks ago

Yes I got the above expression, but couldn't find the other one.Log in to reply

– Ameya Daigavane · 9 months, 3 weeks ago

Other one?Log in to reply

– Akash Shukla · 9 months, 3 weeks ago

As there are infinite \(a\), so I can't find other value of \(a\).Log in to reply

– Ameya Daigavane · 9 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 \( a \).Log in to reply

– Akash Shukla · 9 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 relatedLog in to reply

– Ameya Daigavane · 9 months, 3 weeks ago

\(\frac{n^2 + 1}{2} \) had to be a perfect square.Log in to reply

– Akash Shukla · 9 months, 3 weeks ago

OH!, Yes. You mean \(\dfrac{x^2 + 1}{2} = y^2 \implies\) a perfect squareLog in to reply

– Ameya Daigavane · 9 months, 3 weeks ago

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

– Deeparaj Bhat · 9 months, 3 weeks ago

Where \(x=n\) and \(y=\frac{a}{n} \)Log in to reply

– Ameya Daigavane · 9 months, 3 weeks ago

Yes, I wanted him to see how the two were related :PLog in to reply

– Deeparaj Bhat · 9 months, 3 weeks ago

Oh. Too bad I spoilt it then. :(Log in to reply