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I can seem to find any formal proof that would help me derived a knowing $2^a \pmod{13} = 5$ ?

The following is given

$2^a \pmod{13} = 5$

Soham
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John
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  • If there is such an $a$, there must be such an $a$ less than $12$, so only a small amount of computation is needed. In general, for large primes, such problems seem to be computationally difficult. – André Nicolas May 11 '16 at 23:58
  • See this-http://math.stackexchange.com/questions/1775788/solve-for-b-in-2b-bmod11-7/1776801#1776801 – Soham May 14 '16 at 05:42

3 Answers3

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I notice that $2^3 = 8 \equiv -5 \pmod{13}$. Furthermore, $2^6 = 64 \equiv -1$. Thus, $2^{3+6}$ should work.

I'm not sure if that's what you meant by a formal proof. It's not really something you can figure out without calculating some powers of $2$ and reducing modulo $13$.

G Tony Jacobs
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These numbers are small, so the simplest approach is entirely feasible: start with $2^1=2$ and keep doubling modulo $13$ until you get $5$. The first few doublings (with reduction modulo $13$) yield $2,4,8,3,6,12,11$. Continue, and you’ll get to $5$ in short order. Note that there are only $13$ equivalence classes modulo $13$, and $2^n$ is never going to be $0$ modulo $13$, so you have at most $11$ doublings to try: if none of them gives you $5$, there is no solution. (But in this case there is.)

Brian M. Scott
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You should find $a$ such that $2^a(mod 13) = 5$, ie, $a$ such that there is $k \in \mathbb{Z}$ such that $5 -2^a = 13k$

Luísa Borsato
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