$5^m = 2 + 3^n$ help what to do how to solve this for natural numbers $5^m = 2 + 3^n$ i did this 
$5^m = 2 + 3^n \Rightarrow 5^m \equiv 2 \pmod 3 \Rightarrow m \equiv 1 \pmod 2$
now  if   i put it like  this
$   5^{2k+1} = 2 + 3^n $
what to do ?? 
now is this right another try : 
$ m = n \Rightarrow 5^n - 3^n = 2 = 5 - 3 \Rightarrow (m_1,n_1)= (1,1) $
we can prove by induction :
$5^n - 3^n > 2 \quad \forall n > 1  \Rightarrow m = n = 1 $
another case :
$m > n \Rightarrow 5^m > 5^n \geq 3^n+2 \quad \forall n \geq 1 \Rightarrow 5^m - 3^n > 2 \Rightarrow \emptyset$
$m < n \Rightarrow no \ sol. $ by putting some values

Assume there is a solution with $n>1$. Then $ 5^m \equiv 2 \pmod 9 \Rightarrow m \equiv 5 \pmod 6$.
Then $ 5^m \equiv 3 \pmod 7 \Rightarrow 3^n \equiv 1 \pmod 7 \Rightarrow n \equiv 0 \pmod 3.$
Also, $ 5^m \equiv 5 \text { or } 8 \pmod {13} \Rightarrow 3^n \equiv 3\text { or } 6 \pmod {13} \Rightarrow n \equiv 1 \pmod 3.$
This contradiction means that $n$ cannot be greater than $1$ and so $m=n=1$.
