Solve (k+n)^(k-n)x=(x)

$$( k + n ) ^ { ( k - n ) } x = ( x )$$

Solve for k (complex solution)

$$\left\{\begin{matrix}\\k=1-n\text{, }&\text{unconditionally}\\k=n\text{, }&n\neq 0\\k=-\left(n+1\right)\text{, }&Numerator(2n+1)\text{bmod}2=0\\k\in \mathrm{C}\text{, }&x=0\end{matrix}\right.$$

Solve for n (complex solution)

$$\left\{\begin{matrix}\\n=1-k\text{, }&\text{unconditionally}\\n=k\text{, }&k\neq 0\\n=-\left(k+1\right)\text{, }&Numerator(2k+1)\text{bmod}2=0\\n\in \mathrm{C}\text{, }&x=0\end{matrix}\right.$$

Solve for k

$$\left\{\begin{matrix}\\k=1-n\text{, }&\text{unconditionally}\\k=n\text{, }&n\neq 0\\k=-\left(n+1\right)\text{, }&Numerator(2n+1)\text{bmod}2=0\text{ and }Denominator(2n)\text{bmod}2=1\\k\in \mathrm{R}\text{, }&x=0\text{ and }n<-k\text{ and }Denominator(k-n)\text{bmod}2=1\\k=-n\text{, }&n<0\text{ and }x=0\\k>-n\text{, }&x=0\end{matrix}\right.$$

Solve for n

$$\left\{\begin{matrix}\\n=1-k\text{, }&\text{unconditionally}\\n=k\text{, }&k\neq 0\\n=-\left(k+1\right)\text{, }&Numerator(2k+1)\text{bmod}2=0\text{ and }Denominator(2k)\text{bmod}2=1\\n\in \mathrm{R}\text{, }&x=0\text{ and }k<-n\text{ and }Denominator(k-n)\text{bmod}2=1\\n=-k\text{, }&k>0\text{ and }x=0\\n>-k\text{, }&x=0\end{matrix}\right.$$