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$$x^{59} \equiv 604 \pmod{2013}$$

Could somebody give me any clue? I have no idea how to start.

I see that $59$ is prime.

Rebecca J. Stones
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xawey
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    Hint: $2013 = 3\cdot 11 \cdot 61$. Now if $x^{59} \equiv 604$ mod $2013$ then $x^{59} \equiv 604$ mod $61$. Now use Fermats little theorem which says that $x^{60} \equiv 1$ mod $61$ (if $x$ is not divisible by $61$) since $61$ is prime. – Winther Aug 28 '14 at 22:44

2 Answers2

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Let $t$ be some integer such that $t^{59} \equiv 604 \pmod{2013}$; then, $t^{59} = 604+2013k$ for some integer $k$. Since $\gcd(604,2013) = 1$, Euclidean Algorithm tells us $\gcd(t^{59},2013)=1$ and thus $\gcd(t,2013)=1$. Thus, we can apply Euler's Theorem.

Note that $\varphi(2013) = 1200$ and $\gcd(1200,59) = 1$. By Bézout's Lemma, there exist integers $n,m$ such that $1200n+59m = 1$. Euler's Theorem says $t^{1200} \equiv 1 \pmod{2013}$, so $t^{1200n} \equiv 1 \pmod{2013}$.

$t^{59} \equiv 604 \pmod{2013}$

$t^{59m} \equiv 604^m \pmod{2013}$

$t^{59m} \equiv t^{59m}t^{1200n} \equiv t^{59m+1200n} \equiv t\equiv 604^m \pmod{2013}$

All you have to do is explicitly find $m$ using the Extended Euclidean Algorithm.

Ben Chen
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As $(604,2013)=1,x$ must be co-prime to $2013$

As $2013=3 \cdot 11 \cdot 61,$ using Carmichael Function,

$$x^{60}\equiv1\pmod{2013}$$

$$x^{59}\equiv604\pmod{2013}\implies x^{60}\equiv604x$$

So, we have $$604x\equiv1\pmod{2013}$$

Can you find $x$ using Euclid's Algorithm?

lab bhattacharjee
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  • See also : http://math.stackexchange.com/questions/407478/solving-a-linear-congruence/407486#407486 or http://math.stackexchange.com/questions/361336/how-to-solve-the-diophantine-equation-8x-13y-1571/361343#361343 – lab bhattacharjee Aug 29 '14 at 04:31