Prove that for all $n \in \mathbb{N}_0: \ 4^{2n + 1} + 3^{n+2} $ is divisible by 13.

Question

Prove that for all $$n \in \mathbb{N}_0: \ 4^{2n + 1} + 3^{n+2} $$ is divisible by 13.

Base case: $n = 0$ $$4 + 3^2 = 13$$ which is divisible by 13.

We assume it is true for a given $n$. For $n+1$ we have $$4^{2(n+1) + 1} + 3^{(n+1)+2} = 4^{2n + 2 + 1} + 3^{n+1+2} = 4^{2n + 1 +2} + 3^{n+2+1} = 4^{2n + 1}*4^2 + 3^{n+2}*3^1 = $$

What should I do now ?

$\mod 13$, $4^{2n + 1} + 3^{n+2} = 4 (16^n)+ 9(3^n) = 4(3^n) + 9(3^n) = 3^n (13) = 0$ — sku, Sep 23 '23 at 20:31
Does this answer your question? $13\mid4^{2n+1}+3^{n+2}$ - found using an Approach0 search. Note there are other duplicates, e.g., How do I show that for any natural number n, there exists a natural number m such that $4^{2n+1} + 3^{n+2} = 13m$?, Is this proof by induction that $13 \mid 4^{2n+1}+3^{n+2}$ correct/sufficient?, etc. — John Omielan, Sep 23 '23 at 20:32
@JohnOmielan I swear I did not find it when searching, I pasted the exact same title in the search bar — wengen, Sep 23 '23 at 20:34
@sku I did not think about doing modulo 13, that makes things easier in fact, thank you ! — wengen, Sep 23 '23 at 20:35
Continuing using induction:
We write $4^{2n + 3} + 3^{n+2} = 16 (4^{2n+1} )+ 3(3^{n+2}) = 3(4^{2n+1} + 3^{n+2}) + 13(4^{2n+1}) = 0 \bmod 13$ — sku, Sep 23 '23 at 20:35
@wengen Thanks for letting me know. That negative result doesn't surprise me very much. I've found, although admittedly from only several attempts, that searching for formulas using the site's search function often does not work particularly well. This is why I usually use Approach0 instead, as I indicated in my duplicate question comment. — John Omielan, Sep 23 '23 at 20:37
See the Remark here for a very simple way to derive such inductive proofs. — Bill Dubuque, Sep 23 '23 at 21:07
From my perspective, the key to the problem is that $~4^2 \equiv 3 \pmod{13}.$ — user2661923, Sep 23 '23 at 22:30

Prove that for all $n \in \mathbb{N}_0: \ 4^{2n + 1} + 3^{n+2} $ is divisible by 13.

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