Question: We have a regular triangle $ABC$, where side is $15$, point $N$ belongs to $AB$ such that $AN = 5$. Point $M$ belongs to $AC$ such that $AM = 3$. Prove that $BM$ is perpendicular to $CN$.

I tried to make similar triangles carrying the $BK$ segment, where $K$ is the midpoint of $MC$. So, as $NB:AN = MK:AM$ and angle $A$ is common, $ANM$ and $ABK$ triangles are similar, right? Follows that
$BK \| NM$.

I have multiple versions of solving, e.g. if we prove that $MK=KC=OK$, where $O$ is the intersection point of $BM$ and $CN$, then $MOC$ triangle will become right, or we can show that $COM$ and $BMH$, where $H$ is the base of height from $B$, triangles are similar.

So, I can't conclude what to do.

Answer: 
I've tried to write some hints of important properties to assemble to get the proof :

Hint1 : a cyclic group of cardinal $p-1$ has a generator, which is of order exactly $p-1$  

Hint2 : a polynomial of degree $3$ is irreducible on a field if and only if it has no roots  

Hint3 : use arithmetic ! what does it mean for $3$ to divide/not divide $p-1$ ? note that $3$ is prime

The full proof if you want to see it :

Assume $3$ divides $p-1$. If for every $a \in \mathbb F\_p^{\times}$ the polynomial $x^3 - a$ is reducible, then every element of $\mathbb F\_p^{\times}$ has a third-root (a polynomial of degree $3$ is irreducible on a field if and only if it has no roots). Since $\mathbb F\_p^{\times}$ is cyclic of order $p-1$, it has a generator $\alpha$. Let $\beta$ be a third-root of $\alpha$.  

Since $3$ divides $p-1$, there exists an integer $d$ such that $3d = p-1$. We get $\alpha^d = (\beta^3)^d = \beta^{p-1} = 1$, which is absurd since $\alpha$ has order exactly $p-1$. Hence, there exists $a \in \mathbb F\_p^{\times}$ such that $x^3 - a$ is irreducible over $\mathbb F\_p$. Thus, $[\mathbb F\_p(\sqrt[3]{a}): \mathbb F\_p] = 3$.

Now, assume $3$ doesn't divide $p-1$. Let $a \in \mathbb F\_p^{\times}$. Again, let $\alpha$ be a generator of $\mathbb F\_p^{\times}$. There exists an integer $k$ such that $a = \alpha^k$. Since $3$ is prime and doesn't divide $p-1$, $3$ and $p-1$ are coprime. Bézout's identity gives us integers $u, v$ such that $3u + (p-1)v = k$. So $(\alpha^u)^3 = \alpha^k = a$ and $a$ has a third root in $\mathbb F\_p^{\times}$, that is $\alpha^u$. Thus, $[\mathbb F\_p(\sqrt[3]{a}): \mathbb F\_p] = 1$.
$\newcommand{\Fp}{\mathbb{F}\_p}$Note that you want to find exactly those $p$ for which there exists a non-cube-root $a \in \Fp$. Since $0$ is always a cube, this is equivalent to the existence of a non-cube-root $a \in \Fp^{\times}$.

Note that $\Fp^{\times}$ is a cyclic group of order $p - 1$. Let $g$ be a generator of this group.

**Case 1.** $3$ divides $p - 1$.  

In this case, $g$ cannot be a cube. Indeed, if $h^3 = g$, then $g^{(p - 1)/3} = h^{p-1} = 1$, showing that $g$ does not have the correct order.

**Case 2.** $3$ does not divide $p - 1$.  

We will show that $g$ itself is a cube. In this case, every element is then a cube as well.  

Write $1 = 3a + (p - 1)b$ for some integers $a, b \in \Bbb Z$.  

Then, $g = g^{3a} \cdot (g^{p - 1})^b = g^{3a}$. Thus, $h = g^{a}$ is a cube root.