prove that $\,5\,$ is factor of $\,\,3^{2n+1}+ 2^{2n+1}$ Actually I have done this problem by induction.(trivial)
Please tell me another method to do this problem instead of induction.
 A: Note: This solution assumes that the edit made by user Souvik Dey is in line with the OP's intent.
Note that $$
3^{2n+1}=3\cdot 9^n\equiv 3\cdot(-1)^n\pmod{5}
$$
and
$$
2^{2n+1}=2\cdot 4^n\equiv 2\cdot(-1)^n\pmod{5}.
$$
So, we have
$$
3^{2n+1}+2^{2n+1}\equiv 5\cdot(-1)^n\equiv0\pmod{5}.
$$
A: HINT:
As for integer $a,b$ and integer $n\ge0,$
  $a^n-b^n$ is divisible by $(a-b)$( Proof 1,2)
$a^{2n+1}+b^{2n+1}=a^{2n+1}-(-b)^{2n+1}$ is divisible by $a-(-b)=a+b$  
A: Let $A=3^{2n+1}+2^{2n+1}$. 


*

*$3=-2\pmod{5}$ hence $A=(-2)^{2n+1}+2^{2n+1}\pmod{5}$ 

*$(-2)^{2n+1}=-2^{2n+1}$ hence $A=0\pmod{5}$


Likewise, $3^{2n}+2^{2n}=(-2)^{2n}+2^{2n}=2\cdot2^{2n}$ hence $5$ does not divide $3^k+2^k$ when $k$ is even, only when $k$ is odd.
A: By induction: For $n=0$ the result is clear.
Suppose the result true for $n$.
$$3^{2(n+1)+1}+2^{2(n+1)+1}=9.3^{2n+1}+4.2^{2n+1}=4\left(3^{2n+1}+2^{2n+1}\right)+5.3^{2n+1}$$
A: Since $$3\equiv-2\mod5,$$ we have $$3^{2n+1}\equiv(-2)^{2n+1}\equiv-2^{2n+1}\mod5,$$
and therefore
$$2^{2n+1}+3^{2n+1}\equiv2^{2n+1}(1-1)\equiv0\mod5$$
