Solution e8fdc983-74cb-45ef-a454-71edfd9af4e4

Points: 1 pt

Lemma 1: Finding a divisor $d_1$ always results in another divisor $d_2$.

If $d_1\mid p$, then
\[
p=d_1d_2
\]
for some integer $d_2$. However, we see that
\[
d_2\mid p,
\]
hence we have 2 divisors.

Lemma 2: $d_1\leq \sqrt{p}$ or $d_2\leq \sqrt{p}$.

Suppose
\[
d_1,d_2>\sqrt{p}.
\]
Then
\[
p=d_1d_2>p,
\]
which is a contradiction because $p\not>p$.

Final answer:
\text{Not finding a divisor }d\leq\sqrt{p}\text{ means there is no divisor }d>\sqrt{p}