Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.2 AIME, harder Prove it Romania

For every positive, odd integer $n$ , prove that
$\left[ \frac{1}{2} + \sqrt{n + \frac{1}{2}} \right] = \left[ \frac{1}{2} + \sqrt{n + \frac{1}{2020}} \right],$
where $[a]$ denotes the integer part of the real number $a$ .

Solution

We will prove by contradiction that there is no integer between $\frac{1}{2} + \sqrt{n + \frac{1}{2020}}$ and $\frac{1}{2} + \sqrt{n + \frac{1}{2}}$ .

Suppose there exists $k \ge 1$ such that $\frac{1}{2} + \sqrt{n + \frac{1}{2}} \ge k > \frac{1}{2} + \sqrt{n + \frac{1}{2020}}$ . We obtain $n \ge k^2 - k - \frac{1}{4} = x_k$ and $n < k^2 - k + \frac{126}{505} = y_k$ .

Since the only integer between $x_k$ and $y_k$ is $k^2 - k$ , we must have $n = k^2 - k = k(k - 1)$ which contradicts $n$ odd.

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