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Calculate $17^{14} \pmod{71}$

By Fermat's little theorem:
$17^{70} \equiv 1 \pmod{71}$
$17^{14} \equiv 17^{(70\cdot\frac{14}{70})}\pmod{71}$

And then I don't really know what to do from this point on. In another example, the terms were small enough that I could just simplify down to an answer, but in this example, I have no idea what to do with that $17^{(70\cdot\frac{14}{70})}$

What do I do from here?

  • 0
    Fermat is true, but the point of using it is to subtract an integer multiple of 70 from the exponent. I fail to see how that would help in this case. The others have already shown how to do this.2011-11-07

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$17$ isn’t particularly close to a multiple of $71$, but as Ragib Zaman pointed out, $17^2=289$ is: $289=4\cdot71+5$. Thus, $17^{14}=(17^2)^7=289^7\equiv 5^7\pmod {71}$. At that point you can use brute force, or you might notice that $5^4=625$ is pretty close to $9\cdot71=639$. In fact $625=639-14$, so $5^4\equiv -14\pmod{71}$, $5^5\equiv -70\equiv 1\pmod{71}$, and finally $17^{14}\equiv 5^7\equiv 5^2\equiv 25 \pmod{71}\;.$

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    @Arvin: $5^7=5^5\cdot 5^2\equiv 1\cdot25\pmod {71}$.2011-11-10
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We have that

$ 17^{14}\equiv 17^{70} \cdot 17^{14} \equiv 17^{84} $

or

$ 17^{14}\equiv 17^{-70} \cdot 17^{14} \equiv 17^{-56} $

I don't see any of these as especially easy to calculate, as $17^{14} = 289^{7} \equiv 5^{7}$. If you don't have to use Fermat's, I'd suggest going with that.

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Here is the computation using an addition chain.

$\qquad 17^2\: \equiv\ 5$

$\qquad 17^3\: \equiv\ 17\cdot 17^2\:\equiv\ 5\cdot 17\ \equiv\ 14$

$\qquad 17^5\: \equiv\ 17^2\cdot 17^3\: \equiv\ 5\cdot 14\ \equiv\: -1$

$\qquad 17^7\: \equiv\ 17^2\cdot 17^5\: \equiv\ 5\:(-1)\ \equiv\: -5$

$\qquad 17^{14} \equiv\ 17^7\cdot 17^7\: \equiv\ (-5)^2\: \equiv\ 25$

See the leftmost path in the tree below.

enter image description here

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In this curious problem, the subgroup of order 5 of $(Z_{71}^*, \cdot)$ is formed by these 5 numbers (5 equivalence classes modulo 71):

$ \{1, 5, 25, -17, -14\} $

(When multiplying two, another one comes out of the same subgroup).

This group is isomorphic to the group $ (Z_5, +) $ and contains 4 generators.

The 4 different elements of 1 are generators of $\{1, 5, 25, -17, -14\}$

Using (-17) as a generator of the subgroup of order 5, gives us this order:

$[1, -17, 5, -14, 25]$

So that:

$ 5 ^ 3 = -17 $ $(-17) ^ 2 = 5$ $(17) ^ {14} = (-17) ^ {14} = (-17)^{14 (mod \space 5)} = (-17) ^ 4 = 25$