Let $m$ and $n$ be positive integers, $m$ odd, $n$ even, and $1+2+...+m$ and $1+2+...+n$ are perfect squares. Find $m+n$ for smallest $m$ and $n$.

Asked May 23 '24 at 09:36

Active May 23 '24 at 10:03

Viewed 69 times

My approach was purely of hit and trial method, since I could see that the question is solved with it with relative ease:

$1+2+3+...+(m-1)+m = k^2$ [Let]

Putting $m=1$

$1=1^2$

Also,

$1+2+3+...+(n-1)+n=l^2$ [Let]

Putting $n=8$

$36=6^2$

Hence $m+n=9$

Since this question was relatively simple and straightforward, it could be done with simple hit and trial. However, I want to learn an algebraic approach to it such that it could be applied to more complex questions as well.

edited May 23 '24 at 10:03

van der Wolf

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asked May 23 '24 at 09:36

Neel Sharma

https://en.wikipedia.org/wiki/Vieta_jumping This could be useful to you with such exercise involving square and minimal quantity. (infinite descente e.g.) – EDX May 23 '24 at 09:46
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See https://en.wikipedia.org/wiki/Square_triangular_number – lhf May 23 '24 at 09:48
@lhf I'm sorry, I am a little new to this stuff. Can you please explain the context in which square triangular numbers can be used here – Neel Sharma May 23 '24 at 09:57
I think the suggestion was that since 1+2+...+m is a triangular number combined with two consecutive triangular numbers summing to a square number you can conclude the result. I could be wrong but that is how I read it. – Red Five May 23 '24 at 10:03
See this post. – Sahaj May 23 '24 at 10:31

Let $m$ and $n$ be positive integers, $m$ odd, $n$ even, and $1+2+...+m$ and $1+2+...+n$ are perfect squares. Find $m+n$ for smallest $m$ and $n$.

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