Not sure how to go about this. Law of quadratic reciprocity and Euler's Criterion is recently learned material but I'm not sure how this applies.
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Note that by a Legendre symbol calculation, we have $(3/83)=1$. Since $3$ is a quadratic residue of $83$, it follows that $3^{(83-1)/2}\equiv 1\pmod{83}$. So $83$ is a prime divisor of your number.
Alternately, you could use Euler's Criterion to show that $3$ is a quadratic residue of $83$. That involves somewhat more calculation than the calculation of the Legendre symbol (but less theory).
André Nicolas
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Why do you choose 83? – Anonymoose Apr 19 '16 at 03:15
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1Why 83? Maybe because $83 = 2\cdot 41+1$ is prime. Mathematica claims that the next prime factor is 2526913, and the third and last one is 86950696619. – Catalin Zara Apr 19 '16 at 03:16
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Because $2(41)+1$ is the prime $83$. – André Nicolas Apr 19 '16 at 03:16
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One answer is 83, as found by Andre (sorry for the accent .. )
Here is a way to educatedly "guess" it: If $p = 41a+1$ is prime, then both $p$ and $3^{41}-1$ divide $3^{41a} -1= 3^{p-1}-1$. There is a chance that $p$ divides $3^{41}-1$. Since $a=1$ doesn't work, the next to try is $a=2$. Then $(3^{41})^2 -1 = (3^{41}-1)(3^{41}+1)$ is a multiple of 83. As $3^4 \equiv -2 \pmod{83}$, it follows that $3^{40} \equiv 1024 \equiv 28 \pmod{83}$, hence $3^{41} \equiv 1 \pmod{83}$.
Remark: Mathematica claims that the prime factorization of $3^{41}-1$ is $2\cdot 83\cdot 2526913\cdot 86950696619$. All three odd primes are $\equiv 1\pmod{41}$
Catalin Zara
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