A Special Number

Problem:

Show that for each positive integer n equal to twice a triangular number, the corresponding expression \sqrt{n+\sqrt{n+\sqrt{n+ \sqrt{n+\ldots}}}} represents an integer.

Solution:

Let n be such an integer, then there exists a positive integer m such that n=(m-1)m=m^{2}-m. We then have n+m=m^{2} so that we have successively

\sqrt{n+m}=m; \sqrt{n + \sqrt{n+m}}=m; \sqrt{n+\sqrt{n+\sqrt{n+m}}}=m and so on. It follows that

\sqrt{n+\sqrt{n+\sqrt{n+ \sqrt{n+\ldots}}}}=m, as required.

Comment: you have to be a bit aware of properties of triangular numbers.

Reference:

1001 Problems in Classical Number Theory by Jean-Marie De Koninck and Armel Mercier, AMS (American Mathematical Society), Indian Edition:

Amazon India link:

https://www.amazon.in/1001-Problems-Classical-Number-Theory/dp/0821868888/ref=sr_1_1?s=books&ie=UTF8&qid=1508634309&sr=1-1&keywords=1001+problems+in+classical+number+theory

Cheers,

Nalin Pithwa.

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