Find m, n, p of given expression with some conditions Find $m, n, p$ such that in the expansion of the expression $ \left( x^m + \frac{1}{x^p} \right)^n$ the 12th and the 24th terms contain $x$, respectively $x^5$, and furthermore, this expansion also contains one free term (without $x$).
By using what we are given, I got the following:
$$mn - k(m + p) = 0$$
$$mn - 11(m + p) = 1$$
$$mn - 23(m + p) = 5$$
From these expressions you can easily conclude that $mn = -\frac{8}{3}$ and $m + p = -\frac{1}{3}$. By using these in the first equation, we get $k = 8$.
From this point we have three equations with 3 unknown variables. I have no idea what to do next.
Thank you in advance!
 A: 
Expanding the binomial we obtain
  \begin{align*}
\left( x^m + \frac{1}{x^p} \right)^n&=\sum_{k=0}^n\binom{n}{k}x^{mk}x^{-p(n-k)}\\
&=\sum_{k=0}^n\binom{n}{k}x^{(m+p)k-pn}
\end{align*}

According to the requirements we derive three equations:
\begin{align*}
11(m+p)-pn&=1\tag{1}\\
23(m+p)-pn&=5\tag{2}\\
k(m+p)-pn&=0\tag{3}\\
\end{align*}
Subtracting  (2) from (1) and (3) from (2) gives
\begin{align*}
12(m+p)&=4\\
(k-23)(m+p)&=-5
\end{align*}
From the first we  derive $m+p=\frac{1}{3}$ and obtain by putting this value in the second equation according to OPs result
\begin{align*}
k&=8
\end{align*}
With  $k=8$ we obtain  from (3)
\begin{align*}
p\cdot  n=k(m+p)=\frac{8}{3}\tag{4}
\end{align*}

Since we have only to find a proper tripel $(n,m,p)$ we are free to specify appropriate values of $p$ and $n$ in (4). According to the requirement that the exponent of $x$ of the $24$-th term is $5$, we have to set $n$ at least to $24$ and we just take this value for $n$.
We obtain with $n=24$ from (4)  and since $m+p=\frac{1}{3}$
  \begin{align*}
p=\frac{1}{9},
m=\frac{2}{9}
\end{align*}
  We finally get a solution
  \begin{align*}
\left(x^{\frac{2}{9}}+x^{-\frac{1}{9}}\right)^{24}=x^{-\frac{8}{3}}(1+x^{\frac{1}{3}})^{24}
\end{align*}

