Minimum number of ways to color each integer I have seen this problem floating around for a while but with no answer. 
Since the USAMTS deadline has passed, I would really like to see an answer for this. The farthest I got with this was that $n \geq 4$ (proved this by showing that there must be more than $3$). I'm inclined to say that $4$ is the answer.
Find the smallest positive integer $n$ that satisfies the following: We can color each positive integer with one of $n$ colors such that the equation $w+6x=2y+3z$ has no solutions in positive integers with all of $w,x,y,z$ the same color. (Note that $w,x,y,z$ need not be distinct: for example, 5 and 7 must be different colors because $(w,x,y,z)=(5,5,7,7)$ is a solution to the above equation.
 A: My full solution that will hopefully get a 5. 

$$ \begin {eqnarray} w + 6x &=& 2y + 3z \qquad (1) \end {eqnarray} $$
First, we prove that $ n \ge 4 $. 
Lemma: 
For any positive integer $m$, the integers $m$, $2m$, and $3m$ must have pairwise distinct colors. 
Proof: 
Note that, for any positive integer $m$, the following statements hold. 


*

*The ordered quadruple $(w,x,y,z)=(2m,m,m,2m)$ satisfies (1), so $m$ and $2m$ cannot have the same color.

*The ordered quadruple $(w,x,y,z)=(3m,m,3m,m)$ satisfies (1), so $m$ and $3m$ cannot have the same color.

*The ordered quadruple $(w,x,y,z)=(3m,2m,3m,3m)$ satisfies (1), so $2m$ and $3m$ cannot have the same color. 


Hence, $m$, $2m$, and $3m$ must all have different colors. $\Box$
By lemma 1, $ n \ge 3 $. Now, say, for sake of contradiction, that $n=3$ is enough. Let the color of the positive integer $u$ be $c(u)$. Then, say that $c(m)=r$, $c(2m)=s$, and $c(3m)=t$, where $r$, $s$, and $t$ are pairwise distinct. Then, by lemma 1, we have that $ c(6m) \ne c(2m) $ and $ c(6m) \ne c(3m) $, so $ c(6m) = c(m) = r $. This also gives $ c(12m) = c(2m) $, by substituting $ m \mapsto 2m $ into the equation $ c(6m) = c(m) $. And $ c(2m) = s $, so $ c(12m) = s $. 
Lemma: 
For any positive integer $m$, the integers $2m$, $9m$, and $12m$ cannot all have the same color. 
Proof: 
Note that, for any positive integer $m$, the ordered quadruple $(w,x,y,z)=(12m,2m,9m,2m)$ satisfies (1), so $2m$, $9m$, and $12m$ cannot all have the same color. 
By lemma 2, and the fact that $c(2m)=c(12m)=s$, we have $c(9m)\ne s$. By lemma 1, $c(9m)\ne c(3m)$, so $c(9m) \ne t$. Therefore, $c(9m)=r$. But, by lemma $1$, substituting $6m$ and $9m$ cannot have the same color, so we have a contradiction to the fact that $n=3$ colors suffices. Therefore, $ n \ge 4 $. 
Now, we claim that the following construction for $n=4$ colors works: 


*

*Color the positive integer $a$ $\textbf{red}$ if and only if $ \nu_3 (a) $ is even and $ \frac {a}{3^{\nu_3(a)}} \equiv 1 \pmod {3} $. 

*Color the positive integer $a$ $\textbf{blue}$ if and only if $ \nu_3 (a) $ is even and $ \frac {a}{3^{\nu_3(a)}} \equiv 2 \pmod {3} $. 

*Color the positive integer $a$ $\textbf{green}$ if and only if $ \nu_3 (a) $ is odd and $ \frac {a}{3^{\nu_3(a)}} \equiv 1 \pmod {3} $. 

*Color the positive integer $a$ $\textbf{yellow}$ if and only if $ \nu_3 (a) $ is odd and $ \frac {a}{3^{\nu_3(a)}} \equiv 2 \pmod {3} $. 


Here, $v_3(r)$ represents the largest positive integer $k$ such that $3^k \mid r$. 
To prove that this construction is a valid coloring, first assume, for sake of contradiction, that we have a quadruple $(w,x,y,z)$ such that $w$, $x$, $y$, and $z$ all possess the same color. 
Consider the quadruple $ \left( \frac {w}{k}, \frac {x}{k}, \frac {y}{k}, \frac {z}{k} \right) $, where $ k = 3^{\text{min} \left( \nu_3 (w), \nu_3(6x), \nu_3(y), \nu_3(3z) \right)} $, which also equals $$ 3^{\text{min} \left( \nu_3 (w), 1 + \nu_3 (x), \nu_3 (y), 1 + \nu_3 (z) \right)}. $$ We allow $x$ and $z$ to turn into rationals, since the denominator must be either $1$ or $3$, making $6x$ and $3z$ still integral. What we have done is divided $(w,x,y,z)$ a number of times sufficient to make at least one of $\frac{w}{k}$, $6\frac{x}{k}$, $2\frac{y}{k}$, and $\frac{z}{k}$ not divisible by $3$. So we split it up into two cases. 


*

*If one of $w$ and $2y$ is not divisible by $3$, then the other one is not as well. Since $ \nu_3 (6x) $ and $ \nu_3 (3z) $ are both of the parity opposite to that of $ \nu_3 (w) $ and $ \nu_3 (y) $, both $ \nu_3 (6x) $ and $ \nu_3 (3z) $ are odd and so $6x$ and $3z$ are both divisible by $3$. Therefore, taking both sides mod $3$ gives us $ w \equiv 2y \pmod 3 $. Therefore, since one of $w$ or $2y$ isn't divisible by $3$, the other isnt either, so $ w \equiv y \pmod 3 $. This cannot be satisfied with $ w \equiv 2y \pmod {3} $ and $ w \not\equiv 0 \pmod 3 $, so we have a contradiction. 

*If one of $6x$ and $3z$ is not divisible by $3$, then the other one is not as well. Since $\nu_3(x)$ and $\nu_3(2y)$ have opposing parity, they must be divisible by $3$. Let $ x_1 = 3x $ and $ z_1 = 3z $. Then, $ w + 2x_1 = 2y + z_ 1 $. Taking mod 3, we have $ 2x_1 \equiv z_1 \pmod 3 $. Hence, one of them is not divisible by 3, so the other one isn't as well. But $ x_1 $ and $ z_1 $ are both factors of $x$ and $z$, and we find that $$ x_1 \equiv z_1 \pmod 3, $$which is, again, a contradiction. 
$$\blacksquare$$
