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juantheron
Poster

Post Number: 12
 Posted on Monday, 04 July, 2011 - 09:07 am:

find all $n\in\mathbb{N}$ which satisfy $\displaystyle \lfloor \sqrt{n}\rfloor = \lfloor \sqrt{n+2011}\rfloor$

* $\lfloor x \rfloor=$ floor function
Clare Fanthorpe
Poster

Post Number: 20
 Posted on Monday, 04 July, 2011 - 09:52 am:

Think about what exactly |_$\sqrt{n}$_|=|_$\sqrt{n+2011}$_| means - can you express it using inequalities?
cypher
Prolific poster

Post Number: 211
 Posted on Monday, 04 July, 2011 - 01:34 pm:

You could also think about it graphically. I believe it makes it easy to solve but I'm not sure what sort of written solution you need to be aiming for.

By the way Clare, if you right click on juantheron's code and select show source you will see "\lfloor and \rfloor".
Clare Fanthorpe
Poster

Post Number: 21
 Posted on Monday, 04 July, 2011 - 02:38 pm:

Thanks! I've just thought of a nice way - what if you write n as a square and a "remainder", the same way you might write a number which is not divisible by 3 as 3m+k where k=1 or 2 (or k=0 if the n is divisible by 3)?