Fix a prime number $p$ and consider integers $0<a,b<p$. Prove that from remainders of $a,2a,3a,\cdots, ba \mod p$ you can choose $\lfloor{\sqrt{b}}\rfloor$ numbers that when put in increasing order form arithmetic progression.
Here is my attempt:
The remainders will be of the form
$a,2a,\cdots,ka,$
$(k+1)a-p,\cdots,k_1a-p$
$(k_1+1)a-2p,\cdots,k_2a-2p$
and so on. Each row will have either $k$ or $k+1$ elements so if $k+1\geq\lfloor{\sqrt{b}}\rfloor$ then we can just choose that row as our arithmetic progression. So suppose that $k+1<\lfloor{\sqrt{b}}\rfloor$ then we will have more then $\lfloor{\sqrt{b}}\rfloor$ rows. Now we could try to choose a arithmetic progression starting in the top row and setting the difference to either $ka-p$ or $(k+1)a-p$. This would correspond to going down or left/right and down in the table, depending on the length of the row. I couldn't really get this approach to work any further.
I would really appreciate solution or any hints.
Best Answer
Notation: Let $ \{ x \} $ denote the remainder when $x$ is divided by $p$.
Let $ \lfloor \sqrt{ b} \rfloor = c$.
Claim:
Proof:
By contradiction. Suppose that none of them satisfy the condition. Then by Pigeonhole Principle applied to these distinct $c$ values as pigeons and the $c-2$ ranges
$$ ( \frac{p}{c}, \frac{2p}{c}), ( \frac{2p}{c}, \frac{3p}{c}), \ldots , ( \frac{(c-2)p}{c}, \frac{(c-1)p}{c}) $$
as holes, then at least 2 of these values lie in the same hole.
Let $n_1 > n_2$ and $ \{ n_1 a \}, \{ n_2 a \}$ lie in the same hole.
Then $ | \{n_1a\} - \{n_2 a\} | < \frac{ p}{c}$, and so $\{ (n_1 - n_2) a \}$ is either $< \frac{p}{c}$ or $> \frac{(c-1)p}{c}$, as desired.
Note that we have to take the "larger index minus the smaller index" to ensure that $ n_ 1 - n_2 > 0 $ is one of the $c$ values.
Corollary:
Notice that $ cn \leq c^2 \leq b$, so we do indeed have $c$ terms.
Notes: