I would like to calculate $ 2^{5^{77}}\bmod 113 $. I solved similar problems but with smaller exponent e.g. $ 2^{185}\bmod 113 $ with successive squaring method. Is there any way to use this method to solve it?
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2$113$ is prime, so I would reduce $5^{77}$ modulo $112$ first – J. W. Tanner May 12 '20 at 19:52
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@J.W.Tanner why I can do this? – kekereke May 12 '20 at 20:13
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@JohnOmielan no – kekereke May 12 '20 at 20:13
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1@kekereke Actually, the other question will help give you the answer once you use J.W.Tanner's suggestion, with it being useful since by Fermat's little theorem you have $2^{112} \equiv 1 \pmod{113}$. – John Omielan May 12 '20 at 20:18
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@JohnOmielan Thanks. But what if it wasn't prime number? How can i deal with it? – kekereke May 12 '20 at 21:09
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1@kekereke You're welcome. Note Fermat's little theorem is actually a special case (i.e., for primes only) of Euler's totient function, which deals with relatively prime numbers modulo $n$, even when $n$ itself is not prime. – John Omielan May 12 '20 at 21:12
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to reduce $5^{77}$ modulo $112$, note that $5^{12}\equiv1\bmod112$ – J. W. Tanner May 13 '20 at 00:16
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Rather than successive squaring, try successive raising to the $5^{th}$ power:
$2^{5^0}\equiv2\bmod 113$
$2^{5^1}\equiv32\bmod113$
$2^{5^2}\equiv32^5\equiv99\bmod113$
$2^{5^3}\equiv99^5\equiv56\bmod113$
$2^{5^4}\equiv56^5\equiv60\bmod113$
$2^{5^5}\equiv60^5\equiv105\bmod113$
$2^{5^6}\equiv105^5\equiv\color{red}2\bmod113$.
Can you take it from here? $\;$ Note that $77\equiv5\bmod6$.
J. W. Tanner
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