# RMO 2015 Problem 6 Solution

Solution 6

We can write $$a=[a]+{a},$$ where $$[a]$$ denotes the integral part of $$a.$$

Now, we can say that $$0<{a}<1,$$ as $$a\not\in\mathbb{Z}.$$

Let $$a=[a]+\dfrac{1}{2},$$ where $$[a]$$ is odd. Then $${a}=\dfrac{1}{2}.$$

All such integers, must satisfy the property $$2k+1<a<2k+2,$$ where $$k$$ is a non-negative integer.

Then $$a\left(3-{a}\right)=\dfrac{\left(2[a]+1\right)}{2}~\cdot~\left([a]+\dfrac{1}{2}-\dfrac{3}{2}\right)=\dfrac{(2[a]+1)([a]-1)}{2}~.$$

Now, $$[a]=2k+1.$$ Means, $$[a]-1$$ is even.

So $$2|[a]-1.$$

Or, $$=a(3-{a})=\dfrac{(2[a]+1)([a]-1)}{2}$$ is an integer.

Hence, $$a(3-{a})$$ is an integer for all positive reals $$a$$ satisfying $$-$$

$$(I)~2k+1<a<2k+2,$$ for some non-negative integer $$k.$$

$$(II)~{a}=\dfrac{1}{2}~.$$

As $$k$$ takes infinitely many values, number of such positive real numbers $$a$$ is also infinite.

This completes the proof.