We are asked to find $\varphi(n)$ for $n \le 310$ and we are given $n = 4$. It is presumed we have to find $\varphi(4)$.

1. Problem Description

We are asked to find $\varphi(n)$ for $n \le 310$ and we are given $n = 4$. It is presumed we have to find $\varphi(4)$.

2. Solution Steps

The Euler's totient function $\varphi(n)$ counts the positive integers up to $n$ that are relatively prime to $n$.

If $n$ is a prime number $p$, then $\varphi(p) = p - 1$.

If $n = p^k$, where $p$ is a prime number, then $\varphi(p^k) = p^k - p^{k-1}$.

For $n = 4$, we can write it as $n = 2^2$.

Using the formula $\varphi(p^k) = p^k - p^{k-1}$, where $p = 2$ and $k = 2$, we get

$\varphi(4) = 2^2 - 2^{2-1} = 2^2 - 2^1 = 4 - 2 = 2$.

Alternatively, we can list the positive integers up to 4: 1, 2, 3,

4. The integers that are relatively prime to 4 are 1 and

3. Therefore, $\varphi(4) = 2$.

3. Final Answer

$\varphi(4) = 2$