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#1 h Yesterday at 9:23 AM

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Find all positive integers such that $n=\lfloor \sqrt{n}\rfloor^2+\lfloor \sqrt{n}\rfloor$

This post has been edited 1 time. Last edited by m4thbl3nd3r, Yesterday at 9:53 AM

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#2 h Yesterday at 9:45 AM

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isn't this just zero?

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#3 h Yesterday at 9:46 AM

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m4thbl3nd3r wrote:

Find all positive integers such that $n=\lfloor n\rfloor^2+\lfloor n\rfloor$

Since RHS is integer, so is LHS and so $\lfloor n\rfloor=n$ and equation is $\lfloor n\rfloor^2=0$

And unique solution $\boxed{n=0}$

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#4 h Yesterday at 9:53 AM

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Oops! Edited the statement

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#5 h Yesterday at 10:41 AM

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Let $a^2 \leq n \leq a^2+2a$
Then $\lfloor \sqrt{n}\rfloor = a$ and so $n=a^2+a$

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#6 h Yesterday at 10:42 AM

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m4thbl3nd3r wrote:

Find all positive integers such that $n=\lfloor \sqrt{n}\rfloor^2+\lfloor \sqrt{n}\rfloor$

So $n\ge 0$ and we can write $n=m^2+k$ with $k\in\{0,1,...,2m\}$

Equation becomes $m^2+k=m^2+m$

And solutions $\boxed{n\in\{m(m+1)\quad\forall m\in\mathbb Z_{\ge 0}\}}$

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