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Find with $0≤<624$ such that $2^{82}≅$ mod 625.

I have figured out that maybe we can use Fermat's little theorem to solve this question in which $r^{624}≅1$ mod 625. But I am kinda stuck in here as I'm not sure how to keep solving this question when the mod is larger than p in $a^p$. Can anyone leave me a tip? Any helps would be appreciated!

markvs
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sasaki085
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    Easiest, I think, to use iterated squaring. We quickly get $2^{16}\equiv 536$ and $2^{64}\equiv 366$ (all congruences $\pmod {625}$). which is practically enough to get the job done. – lulu Oct 21 '20 at 15:48
  • \equiv makes the $\equiv$ – Randall Oct 21 '20 at 17:17
  • "I have figured out that maybe we can use Fermat's little theorem" Can't. $625$ is not prime. But can use Eulers thereom that $\phi(5^4)= 4*5^3=500$ so $2^{500}\equiv 1 \pmod {625}$ which is not particularly useful. – fleablood Oct 21 '20 at 22:49

3 Answers3

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You cannot use Fermat since 625 is not prime. You cannot use Euler either because $\phi(625)=500$. But $2^{82}\equiv 329 \mod 625$ which can be easily computed by repeated squaring.

markvs
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By here $\,a\equiv b\pmod{\!5^k}\Rightarrow a^5\equiv b^5\pmod{\!5^{k+1}}\ $ if $\,k\ge 1\,$ so

$\qquad\qquad \qquad\begin{align} \bmod 25\!:\,\ &2^3\equiv 8\\ \Rightarrow\ \ \bmod 125\!:\,\ &2^{15}\equiv 8^5\equiv 18\\ \Rightarrow\ \ \bmod 125\!:\,\ &2^{16}\equiv 36\\ \Rightarrow\ \ \bmod 625\!:\,\ &2^{80}\equiv 36^5\equiv -74\\ \Rightarrow\ \ \bmod 625\!:\,\ &2^{82}\equiv 4(-74)\equiv 329 \end{align}$

Bill Dubuque
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$\require{cancel}$Just to be different.

$2^{82}=4^{41}=(5-1)^{41} =\sum_{k=0}^{41} 5^{k}(-1)^{41-k}{41\choose k}$

Now $625= 5^4$ so for all $k\ge 4$ we have $5^{k}(-1)^{41-k}{41\choose k} \equiv 0\pmod{625}$.

So $2^{82} \equiv \sum_{k=0}^3 5^k(-1)^{41-k}{41\choose k}\equiv $

$-1 + 5\cdot 41 - 25\cdot\frac {41\cdot \color{red}{\cancel{40}}^{8\cdot 5}}2 +125\frac {41\cdot \color{red}{\cancel{40}}^{8\cdot 5}\cdot 39}{6}\equiv$

$-1+(\color{blue}{205}) -125\cdot\frac {41\cdot 8}2 + 625\frac {41\cdot 8\cdot 39}6\equiv$

$-1 + (\color{blue}{125 + 80)} - 125\cdot 164 \equiv $

$-1+80 -125\cdot 163 \equiv$

$79 - 125(165 - 2)\equiv $

$79 -625\frac {165}5 + 250\equiv $

$79+250 \equiv 329\pmod {650}$

.....

That.... probably wasn't the easiest way.

fleablood
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