<s>[INST] Integer solutions of $2x+3y=n$ What is the smallest $m$ such that for all $n\geq m$, the equation $2x+3y=n$ has solutions with $x,y \in \mathbb{Z}$ and $x,y\geq2$?
My approach. We can write the solutions in terms of the parameter $t$ as:
$$x(t)=-n-3t$$ and 
$$y(t) = n+2t$$
Setting both of those greater than or equal to $2$, we have $-n-3t\geq2$ and $n+2t\geq2$. Solving for $t$, we find $\dfrac{-n-2}{3} \geq t \geq\dfrac{2-n}{2}$. Solving the resulting inequality, we get
$$-2n-4\geq6-3n$$
So $n\geq10$. This seems to be right. For $n=10$, take $(2,2)$. But, suppose $n=11$. Then, $2x+3y=11$ clearly has no solution where both $x,y\geq2$. Is my solution incorrect? Can someone point me in the right direction? 
EDIT: Thanks for the great answers!
 [/INST] We consider two cases: $n$ is even and $n$ is odd. 
If $n$ is even then $2x+3y = n \implies$ $y$ is an even number. The minimum possible value of even $n$ would be then $2(2)+3(2)=10$. For all even $n \ge 10$, we can set $x = n/2-3$ and $y=2$ to satisfy $2x+3y=n$.
If $n$ is odd then $2x+3y=n$ implies that $y$ is odd. Consequently, $y \ge 3$ and the minimum possible value of odd $n$ is $2(2)+3(3) = 13$. For odd $n \ge 13$, we set $y=3$ and $x= (n-9)/2$ to satisfy $2x+3y=n$.
For above, we conclude that for $n \ge 12$, we can find an integer solution to $2x+3y=n$ for $x\ge2$ and $y \ge 2$.
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