. The following remark will facilitate our task: the point with coordinates $(x, y)$ is visible if and only if $x$ and $y$ are coprime. Let's now count the visible points by separately counting those (strictly) below the diagonal $x=y$, those above, and those on the diagonal. Of course, by symmetry, there are as many visible points below as above. Furthermore, on the diagonal, there is a unique visible point which is $(1,1)$. For a fixed abscissa $x$, there are, by the first remark of this solution, as many visible points below the diagonal as there are integers in $\{0,1, \ldots, x-1\}$ that are coprime with $x$. By definition, this is $\varphi(x)$ where $\varphi$ is the Euler's totient function. In the end, the number we are looking for is therefore:

$$
1+2[\varphi(2)+\varphi(3)+\varphi(4)+\cdots+\varphi(25)]=399
$$

(To quickly compute the first values of $\varphi$, one can use the formulas $\varphi\left(p^{k}\right)=p^{k}-p^{k-1}$ when $p$ is a prime number and $\varphi(a b)=\varphi(a) \varphi(b)$ when $a$ and $b$ are coprime.)
. A point $A$ in the plane with integer coordinates is said to be visible from the origin $O$ if the open segment $] O A[$ does not contain any point with integer coordinates. How many points are visible in $[0,25]^{2} \backslash\{(0,0)\} ?$