Let $n$ be a natural number such that for all prime numbers $p\leq \sqrt[4]{n}$, $p\nmid n$. Find how many different prime factors $n$ can have, and bring an example in every case.
I think that $3$ different prime factors are the most $n$ can have, but I can not justify it.
On
Given that: For all prime numbers $p$ ($\leq \sqrt[4]{n}$), $p\nmid n$.Thus, $n\ge p^4$ and must be a odd number because $2\nmid n$ (smallest $p$). Therefore, $n$ should be $\ge 81$ (for the next prime number $3$). If $n$ is not prime, then smallest $n$ should be $85$ since $85=5\times17$ and $5^4=625\gt 85$. Next possible $n$ is $91$ since $91=7\times13$ and $7^4=2401\gt 91$ so does $95$ ($5\times19$). Therefore, I think, $n$ should have some other limits, which should be included in the question.
Assume that $n$ has at least $4$ prime factors, then $$n=p_1p_2p_3p_4k$$ where $p_1\leq p_2\leq p_3\leq p_4$. Since $p_1$ is the smallest prime factor, we must have $p_1^4\leq n$, which means that $p_1\leq\sqrt[4]{n}$. But this implies that $p\not\mid n$, which is a contradiction.
Therefore $n$ cannot have at least $4$ prime factors.