How do I calculate
- $2^{65536} \pmod{2^{31} -1}$
- $3^{256} \pmod{2^8 +1}$
Please help?
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It's quite easy to do with a computer; these numbers aren't really that large. – May 31 '14 at 02:43
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@user61527 this from a past paper I am solving for my telecommunications exam about RSA encryptions.. I am very confused, the answer must be an integer. Help will be appreciated. I need to do this by hand ( or a handheld calculator). – user8572 May 31 '14 at 02:48
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@amzoti i still dont get it. Could i get a more detailed explanation , please? – user8572 May 31 '14 at 03:00
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I've removed the reference to RSA in the title, since it has nothing to do with the question asked. – May 31 '14 at 03:16
1 Answers
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$(1)$
As $65536=2114\cdot31+2\pmod{31}$
$2^{65536}=2^{31\cdot2114+2}=(1+2^{31}-1)^{2114}\cdot2^2$
Now $(1+2^{31}-1)^{2114}=1+\sum_{r=1}^{2114}\binom{2114}r(2^{31}-1)^r\equiv1\pmod{2^{31}-1}$
$(2)$
Fortunately $2^8+1=257$ is prime, so we can safely use Fermat's Little Theorem
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