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How can I prove that $4^{2n+1}+3^{n+2}$ is always divisible by 13?

Parcly Taxel
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Luis Armando
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5 Answers5

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$$4^{2n+1}+3^{n+2}=16^n\cdot 4+3^n\cdot 9\\16^n\cdot4+3^n\cdot 9\equiv3^n\cdot4+3^n\cdot 9\pmod {13}\\3^n(4+9)\equiv3^n\cdot13\equiv0\pmod{13}$$ This can also be solved with induction,for $n=0$ $$4+3^2=13$$ Assume it holds for $n=k$ $$4^{2k+1}+3^{k+2}$$ Prove it holds for $n=k+1$ $$4^{2k+3}+3^{k+3}=16\cdot4^{2k+1}+3\cdot3^{k+2}=16\cdot4^{2k+1}+16\cdot3^{k+2}-13\cdot3^{k+2}=16(4^{2k+1}+3^{k+2})-13\cdot3^{k+2}$$ by the inductive hypothesis $4^{2k+1}+3^{k+2}$ is divisible by $13$ because of that the whole expression is

kingW3
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    Remark that the inductive proof is just the congruence proof unwound, i.e. it arises by inlining the proof of the Congruence Product Rule. This is explained at length in this answer. – Bill Dubuque Jul 07 '14 at 23:59
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You can use congruence: $$ 4^{2n+1}+3^{n+2}\equiv 4(16^n)+9(3^n) \equiv 4(16^n)+(13-4)(3^n) \equiv 4(16^n-3^n) \equiv 4(3^n-3^n) \equiv 0\ (\text{mod }13). $$

Also you can prove it directly: $$ 4^{2n+1}+3^{n+2} = 4(16^n-3^n)+(13)(3^n) = 13[4(16^{n-1}+16^{n-2}3+16^{n-3}3^2+\ldots+3^{n-1})+3^n] $$ which proves that $4^{2n+1}+3^{n+2}$ is dividable by $13$.

Mohammad Khosravi
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Hint $\ $ Specialize $\, a = 3\,$ below. $ $ See this post for an inductive proof and further discussion.

Theorem $\quad a^2\!+a+1\mid a^{n+2}+(a\!+\!1)^{2n+1}\! =: b$

Proof $\, \ {\rm mod}\,\ a^2\!+a+1\!:\ \color{#0a0}{a(a\!+\!1)\equiv -1}$ and $\,\color{#c00}{a^3\equiv 1}\ $ by $\,0\equiv (a\!-\!1)(a^2\!+a+1) = a^3\!-1,\,$ so

$\qquad\quad\! a^{2n+1}b = a^{3n+3} + (\color{#0c0}{a(a\!+\!1)})^{2n+1} \equiv\, (\color{#c00}{a^3})^{n+1}\!-1\equiv 0$

Thus $\,\ a^{2n+1}b \equiv 0\ $ times $\,a^{n-1}\,$ yields $\ b \equiv (\color{#c00}{a^3})^n b\equiv 0\ \ $ QED

Community
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Bill Dubuque
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By induction: the base case $n=0$ is obvious. Assume true for $n=k$, so all that remains to show is for $n=k+1$.

Since we know the claim is true for $n=k$:

$$A_k : = 4^{2k+1}+3^{k+2}$$

is divisible by 13. We proceed by showing that $A_{k+1} - 3 A_k$ is divisible by 13, from which the claim follows: $$= 4^2 \cdot4^{2k+1} + 3\cdot3^{k+2} - 3(4^{2k+1} + 3^{k+2}) = 4^{2k+1}\cdot 13$$

So we are done.

dmh
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  • This is equivalent (a rearrangement) of the common inductive proof in KingW3's $2014$ answer. They prove $!\bmod 13!:\ A_{k+1} \equiv 16 A_k\ (\equiv 3 A_k$ as in your proof). All these common inductive proofs are special cases of a standard proof of the Congruence Product Rule - as I explain at length here. It's best for site health not to duplicate prior answers (esp. for FAQs like this). – Bill Dubuque Sep 26 '23 at 03:11
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let's using mod function $$ \left(4^{2n+1}+3^{n+2} \right)\text{ mod }13=\left((4-13)^{2n+1}+3^{n+2} \right)\text{ mod }13=\left(-3^{4n+2}+3^{n+2} \right)\text{ mod }13 $$ $$ \because \text{ }3^3\text{ mod }13=1 \text{ }\therefore\text{ }3^n\text{ mod }13=3^{n\text{ mod }3} $$ $$ \to \left(-3^{4n+2}+3^{n+2} \right)\text{ mod }13 =\left(-3^{n+2}+3^{n+2} \right)\text{ mod }13=0$$ $$ \therefore 4^{2n+1}+3^{n+2} \text{ is divisible by }13$$

Faoler
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