Korea 1995

[tanmoy]

For any positive integer $m$,show that there exist integers $a,b$ satisfying
$\left | a \right |\leq m, \left | b \right |\leq m, 0< a+b\sqrt{2}\leq \frac{1+\sqrt{2}}{m+2}$

[Nirjhor]

Hint: PHP...

[Nayeemul Islam Swad]

Full Solution:

Let $M=\frac{1+\sqrt{2}}{m+2} \text{ and } N(a,b)=a+b\sqrt{2}, 0\leq a,b \leq m.$ So in total there are $(m+1)^2$ $N's.$
Now the highest value among the $N's$ is $m+m\sqrt{2}.$ So each $N$ will be in the interval $[0,m(1+\sqrt{2})]$. Divide this interval into $m(m+2)$ equal parts. Then each part equals $M$.
As there are $m(m+2)=(m+1)^2-1$ parts and $(m+1)^2$ $N's$, some two $N's$, say $N(a_1,b_1),N(a_2,b_2)$ will be in the same part. Then their difference $|N(a_1-a_2,b_1-b_2)|$ can't be more than $M.$