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I need to prove that $5^{12n+3} \equiv 21 \pmod{26}.$ I probably need to use Fermat's little theorem but don't know where to start.

J. W. Tanner
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Ori Yehuda
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    It helps to notice $5^2 = 25 \equiv -1\pmod{26}$ – JMoravitz Nov 25 '22 at 20:43
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    Fermat’s little theorem is for prime numbers, and $26$ is not prime – J. W. Tanner Nov 25 '22 at 20:44
  • @J.W.Tanner But OP's statement is clear mod 2, so only need mod 13. – coffeemath Nov 25 '22 at 21:03
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    @coffeemath: I agree. That’s a start – J. W. Tanner Nov 25 '22 at 21:04
  • Clearly $5^{12n+3} \equiv 21 \pmod{26}\iff 5^{12n+3}-21=26M$ and it is evident that $5^{12n+3}-21$ is even so divisible by $2$. It follows that it is enough to prove that $5^{12n+3}-21$ is divisible by $13$ or $$5^{12n+3} \equiv 21 \pmod{13}\iff 5^{12n+3} \equiv 8 \pmod{13}\5^3\cdot5^{12n}\equiv8\pmod{13}$$ Since $5^{12n}\equiv1\pmod{13}$ (by $FLT$) and $5^3=125\equiv 8\pmod{13}$, we are done. – Piquito Nov 25 '22 at 21:45
  • $!\bmod 26!:\ \color{#c00}{5^2\equiv -1}\Rightarrow a := 5^{3+12n} = 5( \color{#c00}{5^2})^{1+6n}\equiv 5( \color{#c00}{-1})^{1+6n}\equiv -5,,$ so $,a-21\equiv -26\equiv 0,$ by the congruence power & sum rules in the linked dupe. Or, equivalently, use $,5^{ \color{#0a0}4}\equiv ( \color{#c00}{5^2})^2\equiv (\color{#c00}{-1})^2\equiv 1,$ so exponents on $5$ can be reduced mod $ \color{#0a0}4$ by mod order reduction in the 2nd dupe. – Bill Dubuque Nov 26 '22 at 03:14

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Fermat's little theorem requires that your modulus is a prime number. Since $26$ is not prime, it won't quite work here (at least, not straight up as is, that is).

Consider, however, that $5^{12n + 3} = 5^{12n} \cdot 5^3$. Taking JMoravitz's hint into consideration that $5^2 \equiv -1 \pmod{26}$, we see that \begin{align*} 5^{12n + 3} \equiv 5^{12n} \cdot (-5) \equiv 5^{12n} \cdot 21 \pmod{26}. \end{align*} If this is, in fact, congruent to $21 \pmod{26}$, then it is necessary to show that $5^{12n} \equiv 1 \pmod{26}$. Can you take it from here?

Chris Uchizono
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Hint

By Euler's theorem, $5^{\varphi (26)}\equiv1 \pmod{26}$. And $\varphi (26)=12$.

calc ll
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