<s>[INST] Prove that $13^n -4^n$ is divisible by $9$ for every integer $\geq $ 
Prove that $13^n - 4^n$ is divisible by $9$ for every integer $n$ is greater than or equal to $1$. 

I have written down my answer but I don't know if it is correct.
$    13(13^k) - 4^{(k+1)}$
$=    9(13^k) + 4(13^k) - 4{(4^k)}$
$=    9(13^k) + 4(13^k - 4^k)$
$=    9(13^k) + 4(9a)$ where $9a = 13^k -4^k$
$=    9(13^k + 4a)$
$=    9b$ where $b=13^k + 4a$
 [/INST] You can prove this by induction.

First, show that this is true for $n=1$:
$13^{1}-4^{1}=9$
Second, assume that this is true for $n$:
$13^{n}-4^{n}=9k$
Third, prove that this is true for $n+1$:
$13^{n+1}-4^{n+1}=$
$13\cdot13^{n}-4\cdot4^{n}=$
$13\cdot13^{n}-(13-9)\cdot4^{n}=$
$13\cdot13^{n}-13\cdot4^{n}+9\cdot4^{n}=$
$13\cdot(\color\red{13^{n}-4^{n}})+9\cdot4^{n}=$
$13\cdot\color\red{9k}+9\cdot4^{n}=$
$9\cdot13k+9\cdot4^{n}=$
$9\cdot(13k+4^{n})$

Please note that the assumption is used only in the part marked red.
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