How to compute $2^{2475} \bmod 9901$?

Question

My work: $$2^{2475} = 2^{5^2\cdot 9\cdot 11} = 1048576^{5\cdot 9\cdot 11} = (-930)^{5\cdot 9\cdot 11} \bmod 9901$$

but I got stuck after this. Any further computation continuing from where I got stuck results in numbers that are too large for me to work with.

As $9901$ is prime and $\equiv1\pmod4,$ $$\left(\dfrac2p\right)=(-1)^{(p-1)/4}$$ — lab bhattacharjee, Aug 22 '16 at 04:23
By Euler's theorem: $\gcd(2,9901)=1\implies2^{\phi(9901)}\equiv1\pmod{9901}$.
Since $9901$ is prime, $\phi(9901)=9900$, hence $2^{9900}\equiv1\pmod{9901}$.

Hence $2^{9900}\equiv(2^{2475})^{4}\equiv1\pmod{9901}$.

Now we need to solve the equation $x^{4}\equiv1\pmod{9901}$. — barak manos, Aug 22 '16 at 04:30
@barakmanos, By my hint above, we have found $$x^{(9901-1)/4}\equiv1\pmod{9901}$$ — lab bhattacharjee, Aug 22 '16 at 05:01

score 1 · Answer 1 · edited Apr 13 '17 at 12:20

Use the fast exponentiation algorithm (see description in my answer to this question. It requires $11$ squarings and $6$ multiplications. $$\begin{array}{cccc} \hline n&S\bmod9901&&P\bmod9901\\ \hline 2475&2&\to&2\\ 1237&4&\to&8\\ 618&16&&8\\ 309&256&\to&2048\\ 154& 6130&&2048\\ 77&2605&\to& 8302\\ 38& 3840&& 8302\\ 19&3011&\to&7198\\ 9&6706&\to&2413\\ 4& 94&&2413\\ 2&8836&&2413\\ 1&5511&\to&\color{red}{1000}\\ \hline \end{array}$$

score 0 · Answer 2 · answered Aug 22 '16 at 06:22

First of all, you made a mistake early. Remember your rules of exponents.

$$2^{5^2\times9\times11}=(2^5)^{5\times9\times11}=32^{5\times9\times11}$$

Lab bhattacharjee's method is probably the best for this specific case. Failing that, it shouldn't be too hard to calculate by a method like exponentiation by squaring with a computer program or even just a calculator. If we're squaring numbers at most, the products won't exceed $9900^2<10^8$. I believe this can be done in as few as $15$ multiplications.

score 0 · Accepted Answer · answered Aug 22 '16 at 12:30

0

$9901$ is prime and $9900=4\cdot2475$ then $$(2^{2475})^4=2^{9900}\equiv 1\pmod{9901}$$ Hence we can solve in the field $\Bbb F_{9901}$ the equation $$X^4=1$$ Wolfram gives $X=1000,X=8901,X=9900$ and we have to pick $\color{red}{X=1000}$ because it corresponds to $2^{2475}$.

answered Aug 22 '16 at 12:30

Ataulfo

32,657

$X=1$ is also a solution to $X^4\equiv1\pmod{9901}$. 2. Why does $1000$ "corresponds to $2^{2475}$"??? 3. I'm pretty sure that there's a mathematical way (other than WolframAlpha) for solving $X^4\equiv1\pmod{9901}$. 4. I think that it's worth explaining the statement "$(2^{2475})^4=2^{9900}\equiv 1\pmod{9901}$" (see my comment to the question).

barak manos

Aug 23 '16 at 05:39

$1$ and $-1=9900$ are trivial solutions and note that, similarly, $1000=-8901$, however here an explanation: $2475=3^3\cdot 5^2\cdot 11$. You have in the field $\Bbb F_{9901}$ the following partial calculations $2^{11}= 2048$, $2048^3=5210$; $5210^3=5718$; $5718^5=7047$ and finally $7047^5-1000=0$. Thanks for your comment. Best regards. – Ataulfo Aug 24 '16 at 12:03

Note ,please, that I have taken power $3$ just twice (with $2048$ and $5210$). This shows that in $2475=3^3\cdot 5^2\cdot 11$ I made a lapse with exponent $3$. The true exponent of $3$ is $2$ of course. – Ataulfo Aug 24 '16 at 14:04

How to compute $2^{2475} \bmod 9901$?

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