Since $x,y < z$ we can safely WLOG $x \le y < z$. Now clearly $w = 1$ cannot work. Thus we assume $w > 1$. Now suppose $x=y$. Then $2w^x = w^z , 2 = w^{z-x}, w = 2$. We conclude $(w,x,y,z) = (2, n, n, n+1)$ works. Now we take the case $x<y $. Let $v_p(w) = k$ for some arbitrary prime $p|w$. Then $v_p (w^x+w^y) = \min(v_p (w^x), v_p(w^y)) = v_p (w^x) = kx$. (Note: We are using the property that for all $a,b$ such that $v_p (a) \neq v_p (b)$, we have $v_p (a+b) = \min(v_p (a), v_p(b))$. ) Thus we must also have $v_p (w^z) = kx$ but we see that $v_p (w^z) = kz$. Then $kx = kz \implies x = z$, but clearly $x<z$. To conclude, all such quadruples $(w,x,y,z)$ are of the form $(2 , n, n, n+1)$

Clearly this is impossible for $w>2$, which is obvious after it is written out in base $w$. Now, since $w\neq1$, we set $w=2$, and writing in base 2, we find that for all $x\neq y$, this is impossible. Finally, we set $x=y$, and find that for $z=x+1$, this always holds, so the sets are of the form $\{2, x, x, x+1\}$ Vinny stop sniping me

Here is my solution WLOG assume $x>y$ $$w^y(w^{x-y}+1)=w^z$$$$\implies w^{x-y}+1=w^{z-y}$$$$\implies w|w^{x-y}+1$$$$w^{x-y}>1$$as $$w\neq1$$$$\implies w|w^{x-y}$$which gives us no solutions Now let's take $x=y$ It gives $2=w^{z-y}$ So$z=y+1$ and $w=2$ So $(w, x, y,z)=(2,x,x,x+1)

We divide both sides of the equation by w^x. Since y and z are arbitrary, y - x and z - x are likewise arbitrary. Thus we have 1 + w^p = w^q (at the end we can just assume all positive integers as solutions for x). Since w^p ≥ 1, 0 < w^-p ≤ 1. Thus p must be a non-negative integer as if w^q < 1 the inequality would have no real solutions. Clearly, the only solution is when w = p + 2 = q + 1 = 2. However, all values of x satisfy this. Thus we have w = 2, -∞ < x < ∞, y = x, z = y + 1

tastymath75025 wrote: Then $v_p (w^x+w^y) = \min(v_p (w^x), v_p(w^y))$ It's not true in general that $v_p(a+b)=\min(v_p(a),v_p(b))$ if $p\mid a,b$ (take $p=2,a=6,b=10$), so why is this true?