well, if one looks modulo something differente from 2^n or 3^n you wont get any help, always get solution. so locking mod 3^5=243 you find out that 2^mis different from 5 always. so y=1,y=2,y=3,y=4 and is easy to evaluate this cases. Maybe i misscalculated something, just let me now.

For $x=1$ we have $2=3^y+5$ what is a contradiction. For $x=2$ we have $4=3^y+5$ what is a contradiction. For $x=3$ we have $8=3^y+5\Leftrightarrow 3^y=3$ what means that one solution is \[(x,y)=(3,1)\] For $x=4$ we have $16=3^y+5\Leftrightarrow 3^y=11$ what is a contradiction. For $x=5$ we have $32=3^y+5\Leftrightarrow 3^y=27$ what means that another one solution is \[(x,y)=(5,3)\] Now, suppose that $x\geqslant 6$. We have $64|2^x$ so $3^y\equiv 59\ (mod\ 64)$. Now consider the following table: $3^1\equiv 3\ (mod\ 64)$ $3^2\equiv 9\ (mod\ 64)$ $3^3\equiv 27\ (mod\ 64)$ $3^4\equiv 17\ (mod\ 64)$ $3^5\equiv 51\ (mod\ 64)$ $3^6\equiv 25\ (mod\ 64)$ $3^7\equiv 11\ (mod\ 64)$ $3^8\equiv 33\ (mod\ 64)$ $3^9\equiv 35\ (mod\ 64)$ $3^{10}\equiv 41\ (mod\ 64)$ $3^{11}\equiv 59\ (mod\ 64)$ $3^{12}\equiv 49\ (mod\ 64)$ $3^{13}\equiv 19\ (mod\ 64)$ $3^{14}\equiv 57\ (mod\ 64)$ $3^{15}\equiv 43\ (mod\ 64)$ $3^{16}\equiv 1\ (mod\ 64)$ and everything repeats. That means \[y=16k+11\] Note that $3^2\equiv -8\ (mod\ 17)$ $3^3\equiv -24\equiv -7\ (mod\ 17)$ $3^4\equiv -21\equiv -4\ (mod\ 17)$ $3^8\equiv -18\equiv -1\ (mod\ 17)$ $3^{16}\equiv 1\ (mod\ 17)$ $3^{16k}\equiv 1\ (mod\ 17)$ $3^{11}\equiv 3^8\cdot3^3\equiv (-1)\cdot(-7)\equiv 7\ (mod\ 17)$ \[3^y\equiv 3^{16k+11}\equiv 3^{16k}\cdot 3^{11}\equiv1\cdot 7\equiv 7\ (mod\ 17)\] On the other hand, we have: $2^1\equiv 2\ (mod\ 17)$ $2^2\equiv 4\ (mod\ 17)$ $2^3\equiv 8\ (mod\ 17)$ $2^4\equiv 16\equiv -1\ (mod\ 17)$ $2^5\equiv -2\ (mod\ 17)$ $2^6\equiv -4\ (mod\ 17)$ $2^7\equiv -8\ (mod\ 17)$ $2^8\equiv -16\equiv 1\ (mod\ 17)$ and everything repeats. We have that RHS is always congruent to $7\ (mod\ 17)$, and LHS is never congruent to $7\ (mod\ 17)$, what is a contradiction. The only solutions are \[(x,y)\in\left\{(3,1),(5,3)\right\}\]

if $y>1$ $2^{x}= 5 (mod9)$ so $x=5 (mod6)$ hence $2^{x}=5(mod7)$ and $3^{y}=0 (mod 7)$contradiction.

anonymous1173 wrote: if $y>1$ $2^{x}= 5 (mod9)$ so $x=5 (mod6)$ hence $2^{x}=5(mod7)$ and $3^{y}=0 (mod 7)$contradiction. What about $y=3$?

2^x=3^y+5 2^x=3^y+2+3 2(2^(x-1)-1)=3(3^(y-1)+1) (x,y)=(3,1),(5,3)[/quote]

lukeKP wrote: 2^x=3^y+5 2^x=3^y+2+3 2(2^(x-1)-1)=3(3^(y-1)+1) (x,y)=(3,1),(5,3) How do you prove that $ (x,y) = (3,1),(5,3)$ is unique?

2(2^(x-1)-1)=3(3^(y-1)+1) 2(K-1)=3(L+1) (K=(2^(x-1)) , L=(3^(y-1)) gcd(2.3)=1 3|(K-1) , 2|(L+1) K-1=3d , L+1+2d 2^x=6d+2 , 3^y=6d-3 6d=2^x-2 and 6d=3^y+3

I still don't get it... lukeKP wrote: ...6d=2^x-2 and 6d=3^y+3 And what is the problem with that???