If the equation $2x+2y+z=n$ has $28$ solutions, find the possible values of $n$

Question

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.

Answer 1

Assume $x,y,z$ are required to be positive integers.

Consider two cases . . .

Case $(1)$:$\;n$ is even.

From the equation $2x+2y+z=n$, it follows that $z$ is also even.

Writing $z=2w$, and $n=2m$, for some positive integers $w,m$, we get \begin{align*} &2x+2y+z=n\\[4pt] \iff\;&2x+2y+2w=2m\\[4pt] \iff\;&x+y+w=m\\[4pt] \end{align*} By the Stars-and-Bars formula, the equation $$x+y+w=m$$ has exactly ${\large{\binom{m-1}{2}}}$ positive integer solution triples $(x,y,w)$, for any given positive integer value of $m$, hence \begin{align*} &{\small{\binom{m-1}{2}}}=28\\[4pt] \implies\;\;&{\small{\frac{(m-1)(m-2)}{2}}}=28\\[4pt] \implies\;\;&m^2-3m+2=56\\[4pt] \implies\;\;&m^2-3m-54=0\\[4pt] \implies\;\;&(m-9)(m+6)=0\\[4pt] \implies\;\;&m=9\\[4pt] \implies\;\;&n=18\\[4pt] \end{align*}

Case $(2)$:$\;n$ is odd.

From the equation $2x+2y+z=n$, it follows that $z$ is also odd.

Writing $z=2w-1$, and $n=2m-1$, for some positive integers $w,m$, we get \begin{align*} &2x+2y+z=n\\[4pt] \iff\;&2x+2y+(2w-1)=2m-1\\[4pt] \iff\;&x+y+w=m\\[4pt] \end{align*} As for case $(1)$, since $m$ is a positive integer such that the equation $$ x+y+w=m \qquad\qquad $$ has exactly $28$ positive integer solution triples $(x,y,w)$, it follows that $m=9$, hence $n=2m-1=17$.

Note:$\;$If $x,y,z$ are only required to be nonnegative integers, then using arguments analogous to the ones given above, we get $n=12$ or $n=11$.

Answer 2

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 $18$