Proving x^6 ≡ 1 (mod 7) Using Fermat's Little Theorem

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SUMMARY

The discussion focuses on proving that if \( (x, 7) = 1 \), then \( x^6 \equiv 1 \mod 7 \) using Fermat's Little Theorem. The theorem states that for a prime \( p \) and an integer \( a \) such that \( (a, p) = 1 \), it holds that \( a^{p-1} \equiv 1 \mod p \). In this case, applying the theorem with \( p = 7 \) confirms that \( x^6 \equiv 1 \mod 7 \). The user also explores the implications of this result for \( (x^3)^2 \equiv \pm 1 \mod 7.

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Homework Statement


I'm suppose to prove that if (x,7)=1, then x to the 6th is congruent to 1 mod 7.


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The Attempt at a Solution


Now, i have the proof by induction when (a,p)=1 but how do i apply this to prove it when a=x and p=7?
 
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What does Fermat's Little Theorem state?
 
it states that if (a,p)=1 then a^(p-1) is congruent to 1 (mod p)
 
7 is a prime, and you have that (x,7), so use Fermat's Little Theorem on x7-1 = x6
 
so how is this for an answer?:

since 7 is a prime and the gcd(x,7) =1, then by Fermat's Little Theorem,

x^(7-1)=x^6 is congruent to 1(mod7)
 
Yes.
 
so now, if I want to show that (x^3)^2 is congruent to +/-1 (mod 7) would my work be correct? (please see the attachment).
 

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