## Task A-3.1.

How many natural numbers $n$ are there for which there exists a triangle with sides of length

$$
3, \quad \log _{2} n \quad \text{and} \quad \log _{4} n \quad ?
$$
## Solution.

Since $n$ is a natural number, we observe that all values are well-defined. Since $\log _{2} n>0$ and $\log _{4} n>0$ must hold, it follows that $n \neq 1$, or that $n$ is greater than 1.

We notice that all expressions can be written using logarithms with base 2:

$$
\log _{4} n=\log _{2^{2}} n=\frac{1}{2} \log _{2} n, \quad 3=\log _{2} 8
$$

The lengths of the sides satisfy the triangle inequality, that is,

$$
\log _{2} n+\log _{4} n>3, \quad \log _{2} n+3>\log _{4} n \quad \text{and} \quad \log _{4} n+3>\log _{2} n
$$

From the first inequality, we have $\log _{2} n+\frac{1}{2} \log _{2} n>3$, or $\log _{2} n>2$, from which it follows that $n>4$.

From the second inequality, we have $\log _{2} n+\log _{2} 8>\frac{1}{2} \log _{2} n$, or $\log _{2} 8 n>\log _{2} \sqrt{n}$. It follows that $8 n>\sqrt{n}$, from which it follows that $n>\frac{1}{64}$.

From the third inequality, we have $\frac{1}{2} \log _{2} n+\log _{2} 8>\log _{2} n$, or $\frac{1}{2} \log _{2} n<\log _{2} 8$. It follows that $n<64$. Considering all three inequalities, it must hold that $4<n<64$, so we conclude that there are a total of 59 such natural numbers.