The title is a question itself.
Does exist any fast algorithm to get $$k^n \pmod{ka}$$ ?
($k, a, n > 1 $, natural number)
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what has been tried ? – Aug 11 '17 at 11:59
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@RoddyMacPhee any suggestions would be welcomed. I checked wikipedia, but I didn't get any clues. – plhn Aug 11 '17 at 12:02
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hint Euler's totient. – Aug 11 '17 at 12:05
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can it be applied when numbers are not coprime? – plhn Aug 11 '17 at 12:07
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How fast do you want it? Exponentiation by squaring is requires $O(\log n)$ multiplications modulo $ka$. – sbares Aug 11 '17 at 12:16
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@SBareS oh I've lost a very basic thing... – plhn Aug 11 '17 at 12:24
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you can reduce the problem to $k(k^{n-1} \pmod a)\pmod{ka}$ – Aug 11 '17 at 12:31
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1We have $\ k^n\bmod ka, =, k(k^{n-1}\bmod a)\ $ by the mod Distributive Law, and the latter power can be computed quickly by repeated squaring. – Bill Dubuque Aug 11 '17 at 12:35
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If we have $k,a>n$ we would probably just have to do the exponentiation, rather than reduce the exponent using Euler/Carmichael etc. - but it might be worth checking that (although probably not for an coded system). In particular if $a$ is composite with no small factors it may not be worth the time to determine its factors, since the exponentiation calculation is not very time-consuming anyway.
Assuming that any reduced exponent has been determined, calculate $k^{n-1} \bmod a$ using exponentiation by squaring, then multiply by $k$.
Note that you may need to handle intermediate values as large as $a^2$, if using simple multiplication.
Joffan
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