<s>[INST] Proving $7|13\cdot 6^{n}+8\cdot 13^{n}$ for all natural numbers. I would like to verify whether my proof is correct. The answer sheet used a much more intuitive and logical approach but I think mine is correct also.

To prove:  $7|13\cdot 6^{n}+8\cdot 13^{n}$ for all natural numbers
Proof: We proceed by induction and show the base case holds.
$13+8=21 $. Since 7 divides 21 the base case holds.
We assume $7|13\cdot 6^{m}+8\cdot 13^{m}$ and we need to show $7|13\cdot6^{m+1}+8\cdot 13^{m+1}$. 
Since, $7|13\cdot 6^{m}+8\cdot 13^{m}$, we have,  $13\cdot 6^{m}+8\cdot 13^{m} =7x, \space x \in \mathbb{N}$. 
We rewrite $13\cdot 6^{m+1}+8\cdot 13^{m+1}$ as $6(13\cdot 6^m)+13(8\cdot 13^m)$ and notice, $13\cdot 6^m=7x-8\cdot13^m$.
We substitute and find,
$6(7x-8\cdot13^m)+13(8\cdot13^m)=42x-48\cdot13^m+104\cdot13^m = 42x+56\cdot13^m=7(6x+8\cdot13^m)$ 
Since $6x+8\cdot13^m \space  \in \mathbb{N}$ 
$\\ \therefore $ By the principle of mathematical induction,   $7|13\cdot 6^{n}+8\cdot 13^{n}$ for all natural numbers $n \space \blacksquare$.
 [/INST] $$13\cdot 6^{n}+8\cdot 13^{n}\equiv  (-1)(-1)^n+(1)(-1)^n =0 \text { mod (7) }$$
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