Question:
Find all integer quadruples $(a,b,x,y)$ satisfying $a^x-b^y=(ab)^{x-y}$. An example quadruple that works is $(1,0,x,x)$ for every integer $x$.
I gave the problem to my Math teachers and friends, and they said to try it with special cases and search for solutions answer. I tried with $x=y$, but I didn’t make progress in the general case. I’ve noticed that in positive cases the factorization of $a^x-b^y$ is a very big help. Also in some cases the Mihăilescu's theorem is helpful. I know that this is a hard problem, so I will be thankful for every help.
This is an unfinished draft, two difficult cases remain, but my machine is freezing up so I'm posting it here for now.
Let $a$, $b$, $x$ and $y$ be integers such that $$a^x-b^y=(ab)^{x-y}.\tag{1}$$ The case $ab=0$ is mostly a matter of conventions and some cumbersome casework. So I'll assume that $a$ and $b$ are nonzero. First I'll cover a few special cases:
Special case 1: If $x=y$ then $(a,b,x,y)$ is either $$(2,-2,-1,-1)\qquad\text{ or }\qquad (a,a-1,1,1),$$ for some integer $a$.
Proof. If $x=y$ then $(1)$ simplifies to $$a^x-b^x=(ab)^{x-x}=(ab)^0=1.$$ Clearly $x=0$ does not yield any solutions. Two nonzero perfect powers cannot differ by $1$, so we see that $x\leq1$. If $x=1$ then $b=a-1$ yields a solution $(a,b,x,y)=(a,a-1,1,1)$. If $x<0$ then we can clear denominators and rearrange to get $$(a^{-x}-1)(b^{-x}+1)=-1.$$ Because $ab\neq0$ it follows that $a^{-x}=2$ and $b^{-x}=-2$, yielding the solution $(a,b,x,y)=(2,-2,-1,-1)$.
Special case 2: If $x=0$ then $(a,b,x,y)=(-2,-1,0,-1)$.
Proof. If $x=0$ then $(1)$ simplifies to $$1-b^y=(ab)^{-y}.$$ Clearly $y=0$ then does not yield a solution. If $y<0$ then the right hand side is an integer, hence so is the left hand side, and therefore $|b|=1$. Because $ab\neq0$ we have $b=-1$ and we quickly find the solution $$(a,b,x,y)=(-2,-1,0,-1).$$ If $y>0$ then the left hand side is an integer, hence so is the right hand side, and therefore $|ab|=1$. Because $b\neq0$ it follows that $ab=-1$ and so $b^y=2$, a contradiction. So there is no solution with $x=0$ and $y>0$.
Special case 3: If $y=0$ then $(a,b,x,y)$ is either $$(-1,2,1,0)\qquad\text{ or }\qquad(2,-1,-1,0).$$
Proof. If $y=0$ then $(1)$ simplifies to $$a^x-1=(ab)^x.$$ Then $x\leq1$ because otherwise we have two nonzero perfect powers that differ by $1$. If $x=1$ then $a-1=ab$ and so $|a|=1$, and because $ab\neq0$ we find that $a=-1$ and $b=2$, yielding the solution $$(a,b,x,y)=(-1,2,1,0).$$ Clearly $x=0$ does not yield a solution. If $x<0$ then clearing denominators yields $$b^{-x}-(ab)^{-x}=1,$$ and again we have two nonzero perfect powers that differ by $1$ if $x<-1$, which is impossible. Finally, for $x=-1$ we find that $|b|=1$ and because $ab\neq0$ we must have $b=-1$, and we quickly find the solution $$(a,b,x,y)=(2,-1,-1,0).$$
From now on we may assume that $x-y\neq0$ and $xy\neq0$. We distinguish six cases:
We recall from the top that $a$, $b$, $x$ and $y$ are integers such that $$a^x-b^y=(ab)^{x-y}.\tag{1}$$
Case 1: If $y>x>0$ then the left hand side of $(1)$ is an integer. Then the right hand side $(ab)^{x-y}$ is also an integer, where $x-y<0$, and so $|ab|=1$. In particular $a$ and $b$ are both odd, and so $a^x-b^y$ is even, a contradiction. So this yields no solutions.
Case 2: If $y>0>x$ then clearing denominators yields $$a^yb^{y-x}-a^{y-x}b^{2y-x}=1.$$ Note that all terms are integers, because $y-x>0$ and $2y-x>0$. We see that both $a$ and $b$ divide $1$ and so the left hand side is even, a contradiction.
Case 3: If $0>y>x$ then we distinguish further between $2y<x$ and $2y\geq x$. If $2y\geq x$ then clearing denominators yields $$b^{y-x}-a^{-x}b^{2y-x}=a^{-y}.\tag{2}$$ Because $-y<-x$ we see that $a^{-y}$ divides $b^{y-x}$, and becayse $2y-x<y-x$ we see that $b^{2y-x}$ divides $a^{-y}$. In particular $a$ and $b$ have the same prime factors, i.e. any prime $p$ divides $a$ if and only if it divides $b$. Let $c:=\gcd(a,b)$ so that $a=cA$ and $b=cB$ for coprime integers $A$ and $B$. Note that every prime that divides either $A$ or $B$ also divides $c$. Now $$c^{2y-x}B^{y-x}-c^{3y-2x}A^{-x}B^{2y-x}=A^{-y}.$$ Because $-y<-x$ we see that $A^{-y}$ divides $c^{2y-x}$, and because $2y-x<3y-2x$ we see that $c^{2y-x}$ divides $A^{-y}$, and so $c^{2y-x}=\pm A^{-y}$. It follows that $$B^{y-x}-c^{y-x}A^{-x}B^{2y-x}=\pm1.$$ In particular this shows that $B$ and $c$ are coprime. We noted that every prime that divides $B$ must also divide $c$, so we see that $|B|=1$. It follows that $c^{y-x}A^{-x}=\pm2$. Because every prime that divides $A$ also divides $c$, we see that $|c|=2$. It quickly follows that $y-x=1$ and $|A|=1$, and so $|a|=|b|=2$. Then $$(ab)^{x-y}=(\pm4)^{-1}\qquad\text{ and }\qquad a^x-b^y=(\pm2)^x-(\pm2)^{x+1}=(\pm2)^x(1\mp2),$$ and so by unique factorization we must have the $-$-sign in $(1\mp2)$, and then we must have $x=-2$ and the $-$-sign in $(ab)^{x-y}=(\pm4)^{-1}$. This means $a=-b$ and we get the solution $$(a,b,x,y)=(-2,2,-2,-1).$$
If $2y<x$ then clearing denominators yields $$b^{-y}-a^{-x}=a^{-y}b^{x-2y}.$$ Because $-y<-x$ we see that $a^{-y}$ divides $b^{-y}$, and so $a$ divides $b$, say $b=ac$. Because $-y>x-2y$ we see that $b^{x-2y}$ divides $a^{-x}$, so every prime that divides $c$ also divides $a$. Then $$c^{-y}-a^{y-x}=a^{x-2y}c^{x-2y},$$ and because $-y>x-2y$ we see that $c^{x-2y}$ divides $a^{y-x}$.
[To be continued.]
Case 4: If $x>y>0$ then we see that $a^{x-y}$ divides $b^y$ and $b^z$ divides $a^x$, where $z:=\min(y,x-y)$. In particular we see that $a$ and $b$ have the same prime factors, i.e. that $\operatorname{rad}(a)=\operatorname{rad}(b)$. [To be continued.]
Case 5: If $x>0>y$ then $(ab)^{x-y}$ and $a^x$ are integers, hence so is $b^y$. Because $y<0$ this means $|b|=1$ and so $$a^x\pm1=\pm a^{x-y},$$ which means that also $a$ divides $1$. This yields no solutions.
Case 6: If $0>x>y$ then clearing denominators yields $$b^{-y}-a^{-x}=a^{-y}b^{x-2y},$$ where the left hand side is an integer, and both $a^{-y}$ and $b^{x-2y}$ are integers because $y<0$ and $x>2y$. Because $-x<-y$ we see that $a^{-x}$ divides $b^{-y}$, and because $-y<x-2y$ we see that $b^{-y}$ divides $a^{-x}$, and hence $b^{-y}=\pm a^{-x}$. Because $ab\neq0$ we must have $b^{-y}=-a^{-x}$ and so $$-2a^{-x}=b^{-y}-a^{-x}=a^{-y}b^{x-2y},$$ from which it follows that $-2=a^{x-y}b^{x-2y}$, where both factors on the right hand side are integers. But this contradicts the fact that $b^{-y}=-a^{-x}$, so this yields no solutions.