How many solutions to each of the equations $2^x+3^y=5^z$, $2^x+5^y=3^z$, $3^x+5^y=2^z?$

Question

Let $x,y,z\in\Bbb{N}$ How many total solutions are there to each of the three distinct equations below? $$2^x+3^y=5^z \tag 1$$ $$2^x+5^y=3^z \tag 2$$ $$3^x+5^y=2^z \tag 3$$ I found 3 solutions to equation (1), 4 solutions to equation (2), 5 solutions to equation (3),(12 altogether) which might be all of them. $$3^0+5^0=2^1$$ $$5^0+2^1=3^1$$ $$5^0+3^1=2^2$$ $$3^0+2^2=5^1$$ $$2^1+3^1=5^1$$ $$3^1+5^1=2^3$$ $$5^0+2^3=3^2$$ $$2^2+5^1=3^2$$ $$3^2+2^4=5^2$$ $$2^1+5^2=3^3$$ $$5^1+3^3=2^5$$ $$3^1+5^3=2^7$$

Answer 1

Here is a partial answer for equation $(1)$. If $y=0$ then we get $5^z-2^x=1$ which by Mihăilescu's theorem only has solution $(x,z)=(2,1) \implies (x,y,z)=(2,0,1)$ otherwise reducing $\mod{}3$ we see that $x,z$ are of the same parity. Now if $x=0$ we get $5^z-3^y=1$ which again by Mihăilescu's theorem has no solutions and for $x=1$ we get $5^z-3^y=2$. For $x>1$ we can reduce $\mod{}4$ to show that $y,z$ are both even and thus $x,y,z$ are all even. We then have $2^{x}=(5^{z'}-3^{y'})(5^{z'}+3^{y'}) \implies 2^{x_1}=5^{z'}+3^{y'}, 2^{x_2}=5^{z'}-3^{y'}$ with $x_1>x_2$ obviously. We then get $5^{z'}=2^{x_1-1}+2^{x_2-1} \implies x_2=1$ and thus $5^{z'}-2^{x_1-1}=1$ which as means as before that $(x_1-1,z')=(2,1) \implies (x_1,z')=(3,1)$. We thus get $x=x_1+x_2=4, z=2z'=2$ which gives the solution $(x,y,z)=(4,2,2)$.

Now we need to check $5^z-3^y=2$. We have $5(5^{z-1}-1)=3(3^{y-1}-1)$. Suppose $z>7,y>6$. Notice that $3*5=1\mod{}7$ and thus $5^{z-1}=3^{y-1}\mod{}7 \implies 5^{z-y}=1\mod{}7 \implies z=y\mod{}6 \implies z=y\mod{}3$.

If $z=y=0\mod{}3$ then we have an equation of the form $u^3+v^3=2$ which is known to give $(u,v)=(1,1) \implies (z,y)=(0,0)$.

If $z=y=1\mod{}3$ then notice that $6|z-1 \implies 3^6-1|3^{y-1}-1$ and $3^6-1=728=7*104 \implies 7|3^{y-1}-1 \implies 7|5^{z-1}-1 \iff 5^{z-1}=1\mod{}7$. The order of $5$ in ${\mathbb{Z}_7}^*$ is 6 however and thus $6|z-1 \implies 5^6-1|5^{z-1}-1 \implies 9*1736|5^{z-1}-1 \implies 9|5^{z-1}-1 \implies 3|3^{y-1}-1$ contradiction. This only leaves us to check the cases $y-1<6$ and $z-1<7$ thus $y<6$ or $z<8$ which is easy with a computer search and we only get $(x,y,z)=(1,1,1)$.

If $z=y=2\mod{}3$ then we get $5^{3u}25-3^{3v}9=2$. Setting $t=2*5^{3u}25-2=2*3^{3v}9+2$ we get $t^2=900(5^u3^v)^3+4 \iff (t/2)^2=(15)^2(5^u3^v)^3+1 \iff ((15)^2t/2)^2=((15)^2(5^u3^v))^3+(15)^4$. Thus a solution to our equation would imply an integral point on the elliptic curve $y^2=x^3+(15)^4$ which you can check in LMFDB has integral points: $(-36,63),(0,(15^2)),(100,41*25)$ and thus $((15)^2(5^u3^v),(15)^2t/2)\in\{(-36,63),(0,(15^2)),(100,41*25)\}$ which gives a contradiction.

So for equation $(1)$ you found every solution. Perhaps you could apply a similar argument to your other equations too.

Answer 2

Mahler proved this equation has only finite solutions.$a^x+b^y=c^z$