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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?

Bill Dubuque
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Lil
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3 Answers3

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

So we have

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

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

B. Goddard
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Generally we have $\bmod\, a^2\!+\!a\!+\!1\!:\,\ \color{#c00}{a^3-1_{\phantom{|}}\!} =(a\!-\!1)(a^2\!+\!a\!+\!1)\equiv \color{#c00}0,\,$ so $\ \left[\,\color{#0af}{a\!+\!1}\equiv -\color{darkorange}a^2\:\!\right]^{2n+1}\!\Rightarrow\, {(\color{#0af}{a\!+\!1})^{2n+1}} \equiv { {-}{\overbrace{\color{#c00}a^{\color{#c00}3\:\!n}\!}^{\bf\color{#c00} 1}\,} \color{darkorange}a^{n+2}}\!$ $\underset{\large \color{darkorange}{a\,=\,11}^{\phantom{|}}\!\!}\Longrightarrow \color{#0af}{12}^{2n+1}\equiv -\color{darkorange}{11}^{n+2}\pmod{\!133^{\phantom{|^{|^|}}}\!\!\!\!}$

Remark $ $ This is a special case of the method of simpler multiples. Similarly here.

Bill Dubuque
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  • 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
  • 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
  • @user26486 The proof is valid for negative exponents too since $,\color{#0af}{a+1},$ and $,\color{darkorange}a,$ are invertible, by $,-a(a+1)\equiv 1.,$ The proof is valid in any ring for any integer $,n.\ \ $ – Bill Dubuque Jul 11 '24 at 18:40
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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 $$11^{n+1} + 12^{2n+1} \equiv 19\cdot11^n\equiv 0 \bmod 19$$

Fermat
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