<s>[INST] If the equation $2x+2y+z=n$ has $28$ solutions, find the possible values of $n$ 
Let $n$ be a positive integer. If the equation $2x+2y+z=n$ has $28$ solutions, find the possible values of $n$.

I tried taking $2x$ as $a$, and then $2y$ as $b$, and then finding the possibilities. However, I am not sure about the approach I've used.
A little hint would be appreciated.
$x$, $ y$ and $z$ belong to whole numbers.
 [/INST] I assume that the equation $2x+2y+z=n$ has $28$ solutions when $x,y,z$ are positive integers.
Note that this is same as finding the powers of $t$ with coefficient $28$ in the following expansion.
$$(t^2+t^4+t^6+t^8+.......)^2(t+t^2+t^3+t^4+......)$$
$$\left(\dfrac{t^2}{1-t^2}\right)^2\times\dfrac{t}{1-t}=t^5\times\left(\dfrac{1}{1-t}\right)^3=t^5\times(1+t)^3\times\left(\dfrac{1}{1-t^2}\right)^3$$
$$=t^5(1+t)^3\left(\dbinom{2}{2}+\dbinom{3}{2}t^2+\dbinom{4}{2}t^4+\dbinom{5}{2}t^6+..........\right)$$
Note that the last part has the general term of $\dbinom{k}{2}t^{2k-4}$. Since $\dbinom{8}{2}=28$, the last part contains the term $28t^{12}$. So we have $28t^{17}+28t^{18}$
So, the value of $n$ can be either $17$ or $18$
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