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$$2851^{604^{20}}\equiv 9^{604^{20}}\equiv(-5)^{604^{20}}\equiv 5^{604^{20}}\pmod{14}$$ $$5^{\phi(14)}\equiv 1\pmod{14}$$ (Thanks for the remark in comments) $$\gcd(k,14)=1,k\in \mathbb N, k<14 ,k\in\{1,3,5,9,11,13\}\implies\phi(14)=6$$ $$604^{20}\equiv 4^{20}\equiv 4\pmod{6}$$ $$5^x\equiv 1\pmod{14}\implies 5^{x+4}\equiv5^4\pmod {14}\implies5^{604^{20}}\equiv5^4\equiv 9\equiv\pmod{14}$$ Therefore: $$2851^{604^{20}}\equiv 5\pmod{14}$$ Is this correct?

Michael Hardy
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PinkyWay
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  • $\phi(14) = \phi (2)\phi(7) = 6$ and you forget $1$ as one of the integers relative to $14$. Then numbers are $1,3,5,9,11,13$ and there are six of them, not $5$. – fleablood Jan 04 '20 at 17:27
  • So in terms of $6$ and not five we get $604^{20}\equiv 4^{20}\equiv (-2)^{20}\equiv 2^{20}\equiv 4^{10} \equiv 2^{10} \equiv 4^{5} =(-2)^5 = -32 \equiv -2\equiv 4\pmod 6$. – fleablood Jan 04 '20 at 17:31
  • So $2851^{604^{20}}\equiv 5^{604^{20}}\equiv 5^4\equiv 25^2 \equiv 11^2 \equiv (-3)^2 \equiv 9\pmod {14}$. But, yes, that was the correct idea. – fleablood Jan 04 '20 at 17:33
  • @fleablood, Thank you for correction, I wouldn't notice at the moment. – PinkyWay Jan 04 '20 at 17:37
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    That is the correct way of doing it. – fleablood Jan 04 '20 at 17:43
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    BTW.... The only time $\phi(M)$ is odd is if $M = 2$ and $\phi(M) = 1$. – fleablood Jan 04 '20 at 17:51
  • @fleablood I appreciate your attention to the question. Thank you very much! – PinkyWay Jan 04 '20 at 17:53

1 Answers1

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Close....

but in calculating $\phi(14)$ you forgot to include $1$ as one of the values relative prime to $14$.

It's a good idea to get used to the idea of calculating $\phi$ using it's multiple distributive property over primes.

$\phi(14) =\phi(2*7) =\phi(2)\phi(7) = (2-1)(7-1) = 6$.

Then we have $604^{20}\equiv 4^{20}\pmod 6$.

Calculating $4^{20}\pmod 6$ can be done are a variety of ways. But if you had a really difficult case you could use the Chinese Remainder theorem $4^{20}\equiv 0 \pmod 2$ and $4^{20}\equiv 1^{20\%\phi(3)=0}\equiv 1\pmod 3$ so $4^20\equiv 0\pmod 2$ and $1\pmod 3$ and the only option for that is $4^{20}\equiv 4 \pmod 6$.

.... or we could have simply done a few successiv squarings. $4^2\equiv 16\equiv 4\pmod 6$ so $4^k\equiv 4 \pmod 6$.

So we have $ 2851^{604^{20}}\equiv 5^4\pmod {14}$

And $5^4\equiv 25^2 \equiv 11^2 \equiv (-3)^2 \equiv 9\pmod {14}$.

fleablood
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