Find the all the pairs $(n,m) \in \mathbb{N} \times \mathbb{N}$ with the property that $ 2^n+3^m $ is divisible by $23$. Find the all the pairs $(n,m) \in \mathbb{N} \times \mathbb{N}$ with the property that $ 2^n+3^m $ is divisible by $23$.
I'm not really sure how to start this one, but since I found it in a book on group theory, such an approach might also be possible.
 A: Powers of two modulo $23$: $1,2,4,8,16,9,18,13,3,6,12,1 \ldots$
Powers of three modulo $23$: $1,3,9,4,12,13,16,2,6,18,8,1\ldots$
Find all the pairs $(n,m)$ such that $n$ is in the first set and $m$ is in the second, with $n+m=23$.
A: As $\displaystyle 2^3\cdot3=24\equiv1\pmod{23},$
$\displaystyle 2^n+3^m\equiv0\pmod{23}\iff 2^{n+3m}\equiv-3^m\cdot2^{3m}\equiv-(3\cdot2^3)^m$
$\displaystyle\implies2^{n+3m}\equiv-1\pmod{23}$
Now as $23$ is prime, $\displaystyle a^2\equiv1,a\equiv\pm1\pmod{23}$ for $(a,23)=1$
and $\displaystyle2^5=32\equiv9,2^{10}\equiv9^2=81\equiv12,2^{11}\equiv2\cdot12\equiv1$
So, there is no integer exponent$(e)$ of $2$ such that $2^e\equiv-1\pmod{23}$
Hence, the system admits no solution in integers

Alternatively,
HINT:
As $3^3\equiv2^2\pmod{23},$
Case $1:$ If $\displaystyle m=3r+1,2^n+3^m=2^n+3^{3r+1}\equiv 2^n+3\cdot2^{2r}\pmod{23}$
So, we need $\displaystyle 2^{n-2r}\equiv-3\pmod{23}$
Observe that there is no integer exponent$(e)$ of $2$ such that $2^e\equiv-3\pmod{23}$
Case $2:$ If $\displaystyle m=3r+2$
Case $3:$ If $\displaystyle m=3r$
A: Hint: 
Look up Fermat's little theorem.
Solution:
We are looking for $n,m$ such that $2^n+3^m\equiv 0$ modulo $23$. By Fermat's little theorem $2^{22}\equiv 3^{22}\equiv 1$. Thus it only remains to check the pairs $(n,m)$ with $0\le n,m<22$. This is just $484$ numbers therefore it can be done with a computer, e.g. using Wolfram alpha with Table[mod(2^n+3^m,23)=0,{n,0,21},{m,0,21}]

It returns a huge table with the entries only being False, therefore there are no natural numbers $n,m$ such that $2^n+3^m$ is divisible by $23$.
A: Suppose a solution to $2^{n}\equiv-3^m$ (everything in this answer is $\pmod {23}$.) I'll show that this implies $-2$ is a quadratic residue $\pmod {23}$, which is false.
Lemma. $n$ is odd.
Firstly, we know $3\equiv 7^2$, so $2^{n}\equiv-7^{2m}$. Now, the inverse of two $2^{-1}$ is $\equiv 12$, so $-1\equiv7^{2m}12^{n}$. If $n$ were even, $-1$ would be a quadratic residue $\pmod {23}$, which is false.

Set $n=2k+1$. This implies $-2\equiv 7^{2m}12^{2k}=\left(7^m12^k\right)^2$, so $-2$ is a quadratic residue $\pmod {23}$, but this is false. Therefore, there are no solutions $(m, n)$.
A: Notice that $\ {\rm mod}\ 23\!:\  3\equiv 1/8\equiv \color{#0a0}{2^{-3}},\ \ 2\equiv 5^2\Rightarrow\ 2^{11}\equiv 5^{22}\equiv \color{#c00}1\,$ by $\,\mu$Fermat.  Therefore
$$ 2^{\large n} \equiv -3^{\large m} \equiv -\color{#0a0}{2^{\large -3\color{black}{\large m}}}\ \Rightarrow\ 2^{\large n+3m}\equiv -1\smash{\overset{\large\ (\ \ )^{11}}\Rightarrow} \color{#c00}1 \equiv -1 \,\Rightarrow\,\Leftarrow\qquad\  $$
Remark $\ $ Alternatively, note $\, 3\equiv 7^2$ so $\ 3^{11}\equiv 7^{22}\equiv 1$ hence $\ 2^n \equiv -3^m\smash{\overset{\large\ (\ \ )^{11}}\Rightarrow}1\equiv -1.\ $ Therefore, the proof essentially boils down to the fact that the product of the two squares $2^n$ and $3^{-m}$ cannot equal $-1$ since it is not a square (by Euler's Criterion)
