How to find fast modular exponentiation?

Question

I need to find $$5448^{5^3} \pmod{11251}$$ and $$6909^{5^3} \pmod{11251}$$ fast? I couldn't calculate it with normal tricks.

It is not too difficult to find $5448^5 \pmod{11251}$, then do that again, and again — Henry, Dec 19 '14 at 09:55
Just about the fastest way I know to calculate $5448^{5^3} \pmod{11251}$ is to type pow(5448, 5**3, 11251) into the Python shell. But I assume you're after something else? — Ilmari Karonen, Dec 19 '14 at 10:04
@IlmariKaronen I'm assuming he wants a fast way to do it by hand. I'm working on it right now, but I've hit a block. If we take both expressions to the $90^\text{th}$ power then we can use Fermat's little theorem, to find that both of the expressions are solutions $x$ to $$x^{90}\equiv 1 \pmod{11251}$$
I'm not sure how to find the relevant roots of unity. — Zubin Mukerjee, Dec 19 '14 at 10:13
@ugradmath It seems that you got this question from this book. — Zubin Mukerjee, Dec 19 '14 at 10:41

score 1 · Answer 1 · edited Apr 13 '17 at 12:21

Notice that $11251$ is prime and equal to $5^3 \cdot 90 + 1$.

$$\left(5448^{5^3}\right)^{90} \equiv 5448^{11250} \pmod{11251}$$

$$\left(6909^{5^3}\right)^{90} \equiv 6909^{11250} \pmod{11251}$$

$$\left(6909^{5^3}\right)^{90} \equiv 1 \pmod{11251}$$

$$\left(5448^{5^3}\right)^{90} \equiv 1 \pmod{11251}$$

Now we have that your two expressions are solutions $x$ to the equation

$$x^{90} \equiv 1 \pmod{11251}$$

See this answer by André Nicolas on another Math.SE question for how to finish the beast. Reducing the problem to an equation to the above form probably isn't the best way to do it, but it should work.

1 Answers1