Selected Problems from the Final Round of National Olympiad

Consider the following cases. If $w=1$, then no solution can exist, since the l.h.s. of the equality equals $2$ while the r.h.s. equals $1$. If $w\ge 2$, then $x<z$ and $y<z$, i.e., $x\le z-1$ and $y\le z-1$. Thus $w^x+w^y\le w^{z-1}+w^{z-1}=2\cdot w^{z-1}\le w\cdot w^{z-1}=w^z$. To satisfy the equation, equalities must hold in both inequalities and thus $x=y=z-1$ and $w=2$. This gives the solutions $(w,x,y,z)=(2,n,n,n+1)$, where $n$ is an arbitrary positive integer.

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Alternative solution.

In the case $w=1$ there are no solutions because $1+1=2$ is not a power of $1$. Assume in the rest that $w>1$. W.l.o.g., assume $x\le y<z$. Then the equation takes the form $w^x(1+w^{y-x})=w^z$, whence $1+w^{y-x}=w^{z-x}$. Consequently, $1+w^{y-x}$ is a positive power of $w$ and is divisible by $w$. If $y-x$ were positive, then $w^{y-x}$ would also be divisible by $w$, whence $1$ should be divisible by $w$, which is impossible. The remaining case $y-x=0$ leads to $2=w^{z-x}$ that gives $w=2$ and $z-x=1$ as the only possibility. Hence the solutions of the equation are of the form $(w,x,y,z)=(2,n,n,n+1)$, where $n$ is any positive integer. Checking shows that all these quadruples satisfy the equation.