<s>[INST] Cool little system of equations. Solving the system of equations for integers: 
$2^a+3^b=5^b$
$3^a+6^b=9^b$
How is it done? I tried substituting the $2^a$ from the first equation into the second, and dividing the two equations by $2^a$. 
 [/INST] The second equation gives you $b\le a\lt2b$, because $3^b+6^b\le9^b\lt3^{2b}+6^b$ when $b\gt0$ (and the equations obviously have no solutions when $b\le0$).
By diving it by $3^b$, you get $3^{a-b}+2^b=3^b$.
It means $2^b=3^b-3^{a-b}=3^{a-b}(3^{2b-a}-1)$.
By unicity of decomposition in prime factors, we must have $a-b=0$, so $a=b$, and $2^b=3^b-1$.
This is only possible if $a=b=1$. Then the two equations are verified. Notice that the first one is not needed to find the necessary condition $a=b=1$.
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