I'm trying to prove that $5^{k/2}\leq R_{k}(3)=min\{n\in\mathbb{N}, \forall c:e(K_{n})\to [k], \exists ab, bc, ca\in e(K_{n}) \wedge c(ab)=c(bc)=c(ac)\} $
My first attempt was to induction: If it's true for $k=r$, then it's true for $k+2$. The base cases are $k=1,2$, which are obvious. For the even numbers the proof is straightforward, but for the odd numbers I can't finish the inductive step, since $5^{1/2}$ it's not an integer.
For $k$ even, I will show that the inequality is strict. For $k=2$ is true, since $5<R_{2}(3)=6$, so we call $T_{k}$ the graph with $5^{k/2}$ vertices, with no triangles and using $k$ colors. Define $T_{k+2}$ to be a disjoint union of five copies of $T_{k}$ using k colors, and for the remaining vertices draw a $K_{5}$ with the two new colors, so that every vertex is one of the copies of $T_{k}$, and this $K_{5}$ does not contain any triangles.
The last solution does not seems to work for $k$ odd since if we have that $5^{r/2}\leq R_{k}(3)$, all I can assure is that there exists a graph $T_{k}$ with $|V(T_{k})|=\lfloor 5^{r/2}\rfloor$ with no triangles, and the same idea suggest to define $T_{k+1} $ to be five copies of $T_{k}$ just as last time, but the problem is that with this construction $|V(T_{k+1})|=5\lfloor 5^{r/2} \rfloor$, which may be strictly less than $\lfloor 5^{r/2 +1} \rfloor$
The Ramsey number $R_k(3)$ is the least integer $n$ such that any $k$-coloring of the edges of the complete graph $K_n$ contains a monochromatic triangle. To save typing I define $R_k=R_k(3)$.
Let $Q_k=R_k-1$, the greatest $n$ such that the edges of $K_n$ can be $k$-colored without making a monochromatic triangle; so $Q_2=5$.
Lemma 1. $Q_{k+2}\ge5Q_k$.
Proof. Let $T$ be a complete graph on $Q_k$ vertices whose edges are $k$-colored with no monochromatic triangles. Color the edges of $K_5$ with $2$ new colors so that there is no monochromatic triangle, and replace each vertex of $K_5$ with a copy of $T$.
Lemma 2. $Q_k\ge\left\lceil5^{k/2}\right\rceil$ for all $k\ge2$.
Proof. Since $Q_k$ is an integer, it will suffice to show that $Q_k\ge5^{k/2}$. We use induction on $k$. For the basis, $Q_2=5$ and $Q_3=16\gt5^{3/2}$. If $k\ge1$ and $Q_k\ge5^{k/2}$, then by Lemma 1 $$Q_{k+2}\ge5Q_k\ge5\cdot5^{k/2}=5^{(k+2)/2}.$$
Theorem 1. $R_k\ge\left\lceil5^{k/2}\right\rceil+1$ for all $k\ge2$.
Theorem 2. $R_k=Q_k+1\ge16\cdot5^{(k-3)/2}+1$ for odd $k\ge3$.
Remark. In the same way as Lemma 1, it is easy to see that $Q_{h+k}\ge Q_hQ_k$, so that $Q_{k+1}\ge2Q_k$, $Q_{k+2}\ge5Q_k$, $Q_{k+3}\ge16Q_{k}$, etc. From $Q_{k+3}\ge16Q_{k}$ it follows that: $$Q_{3k}\ge2^{4k},$$ $$Q_{3k+1}\ge2^{4k+1},$$ $$Q_{3k+2}\ge5\cdot2^{4k}.$$