I was requested to find the congruence of $15^{81}\mod{13}$ without using Fermat's theorem (since that is covered in the chapter that follows this exercise). Of course I know that by property $15^{81} \equiv 2^{81} \pmod{13}$, but how could I find what is the congruence of $2^{81}$ without using Fermat? Needless it is to say that an exhaustive iterative method would be extremely long.
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2Hint: What is $2^6\bmod13$? – J. W. Tanner Jul 23 '20 at 15:06
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Hint :Make progressive solving $2^{4×20}×2$.. – EDX Jul 23 '20 at 15:07
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4Iterated squaring works well, even without using Fermat. – lulu Jul 23 '20 at 15:07
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Are you at least allowed to reduce the base ? In this case, you can calculate the exponent for which $2^n\equiv 1\mod 13$ holds by hand and apply the trick with fermat's little theorem without actually having to use the theorem. In some cases (not here), you will find an even smaller exponent which can further simplify the task. – Peter Jul 23 '20 at 15:30
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@lafinur I believe you'll find at least a few of the answers in How do I compute $a^b,\bmod c$ by hand?, as well as in some of its many linked posts, to be helpful. – John Omielan Jul 23 '20 at 18:56
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$15\equiv2\bmod13$, so we have $$15^{81}\equiv2^{81}\bmod{13}$$$$\equiv512^9\equiv5^9\bmod{13}$$$$\equiv125^3\equiv8^3\bmod{13}$$$$\equiv512\equiv5\bmod13$$
Rushabh Mehta
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We can for example use $15=3 \cdot 5$ and notice $3^3=27 \equiv 1 \bmod 13$, and $5^2=25 \equiv -1 \bmod 13$. So $$ 15^{81}=3^{81}5^{81}=(3^{3})^{27}(5^2)^{40}5\equiv 1^{27}(-1)^{40}5=5 \pmod{13}. $$
Sil
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