Prove $133\mid \left(11^{n+2}+12^{2n+1}\right)$

Question

Prove $133\mid \left(11^{n+2}+12^{2n+1}\right)$, where $n$ is a non-negative integer.

So, I went about proving this using Fermat's theorem. So I had $11^{n}\cdot 11^2+(12^2)^n= 0\ (\mod 133)$ then $11^n\cdot 11^2+1728^n=0\ (\mod133)$ and finally $1^n+132^n=0\ (\mod 133)$. Then I said $133^n=0\ (\mod133)$ because $133$ raised to any power will have a remainder of $0$. Did I apply Fermat's theorem correctly?

Parentheses missing? Is this $133, |, \left(11^{n+2} + 12^{2n+1}\right)$? If so, please edit your question to reflect this. — Nicholas Stull, May 12 '14 at 21:09
I am unable to do exponents for some reason, but yes that is my question — Lil, May 12 '14 at 21:11
possible duplicate of Show that $11^{n+1}+12^{2n-1}$ is divisible by $133$. — Ayman Hourieh, May 12 '14 at 21:15
@Ayman Not a dupe of that since that one restricts to inductive proofs. Please be more careful. — Bill Dubuque, May 12 '14 at 21:22
@BillDubuque Still useful to reference the other question here. — Ayman Hourieh, May 12 '14 at 21:23
@Ayman But now most likely it will be wrongly closed as a dupe of that. — Bill Dubuque, May 12 '14 at 21:23
@BillDubuque Retracted the vote for now. Still, I think the questions should be linked eventually. — Ayman Hourieh, May 12 '14 at 21:25
You have to specify the exact conditions in which you want the statement to be true. The statement as stated would normally imply that $n\in\mathbb R$ or even $n\in\mathbb C$ (though usually the former one). — user26486, May 12 '14 at 21:30

score 2 · Answer 1 · answered Oct 25 '18 at 22:38

2

See that $12^2 = 144 \equiv 11 \pmod{133}.$

So we have

$$11^{n+2} + 12^{2n+1} \equiv 121(11)^n +12(144)^n $$

$$\equiv 121(11)^n +12(11)^n \equiv 133(11)^n \equiv 0 \pmod{133}.$$

answered Oct 25 '18 at 22:38

B. Goddard

1

Bill Dubuque · Answer 2 · 2020-12-13T05:24:56.467

2

Notice $\ 133 = 11^2\!+\!11\!+1\ $ so we may apply

Lemma $ \bmod\ a^2\!+a+1\!:\,\ b := a^{n+2}\!+(a\!+\!1)^{2n+1}\!\equiv 0\,$ in any ring, $ $ for all $\,n\in\Bbb Z$

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

$\ a^{2n+1}b\, =\, a^{3n+3}\! + (\color{#0c0}{a(a\!+\!1)})^{2n+1} \equiv\, (\color{#c00}{a^3})^{n+1}\!-1\equiv 0\ $ so $\ b\equiv 0,\,$ by $\,a$ $\rm\color{#c00}{unit}$ so cancellable.

Or use $\ \color{#90f}{a\!+\!1}\equiv -a^2\,\Rightarrow\, (\color{#90f}{a\!+\!1})^{2n+1}\equiv -a^{4n+2}\equiv -(\color{#c00}{a^3})^n a^{n+2}\equiv -a^{n+2}$

Remark $ $ Both are essentially special cases of the method used here, which is a special case of the method of simpler multiples.

edited Dec 13 '20 at 05:24

answered May 12 '14 at 21:47

Bill Dubuque

272,048

This idea is in one of my old answers, but it wil be hard to find by search. – Bill Dubuque May 12 '14 at 21:48
You should specify the conditions under which your shown equivalences hold. E.g., $n$ is a non-negative integer. – user26486 May 12 '14 at 21:51
@mathh The tag is elem-number-theory so that should be clear. – Bill Dubuque May 12 '14 at 21:55
The tag shows that the variables being used are most likely of a similar to the 'integer' type. It does not, however, determine the sign of the variables - they are often able to be negative in many problems. – user26486 May 12 '14 at 22:03
@mathh Actually the proof works fine for negative $,n,$ too if you interpret the assertion of the Lemma as a modular congruence. – Bill Dubuque May 12 '14 at 22:10
I couldn't find the old answer I had in mind, but I did find this one which is similar (same problem, with similar solution). – Bill Dubuque May 13 '14 at 03:46
@user26486 I edited the answer so your above comments are now obsolete. – Bill Dubuque Oct 25 '18 at 22:29

Fermat · Answer 3 · 2014-05-12T22:08:03.840

since $133=7*19$, first show that $$7\, |\, \left(11^{n+2} + 12^{2n+1}\right)$$ then $$19\, |\, \left(11^{n+2} + 12^{2n+1}\right)$$

Since $gcd(7,19)=1$, the result follows. $$ 11^{n+2}\equiv 11^{n}\cdot 11^2\equiv 2\cdot11^{n}\bmod 7$$ $$12^{2n+1}\equiv 5\cdot 144^{n}\equiv 5\cdot 11^{n}\bmod 7$$ So $$11^{n+1} + 12^{2n+1} \equiv 7\cdot11^n\equiv 0 \bmod 7$$ $$..........................................................................$$ $$11^{n+2}\equiv 121\cdot 11^n\equiv 7\cdot11^{n}\bmod 19$$ $$12^{2n+1}\equiv 12\cdot 144^{n}\equiv 12\cdot 11^{n}\bmod 19$$ So $$11^{n+1} + 12^{2n+1} \equiv 19\cdot11^n\equiv 0 \bmod 19$$

Prove $133\mid \left(11^{n+2}+12^{2n+1}\right)$

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