Answer:
(3/13) = -1 ≡ 3^6 (mod 13)
Explanation:
We can use Euler's criterion theorem to prove or disprove the statement.
Euler's criterion states that for an odd prime p and integer a not divisible by p,
a^((p-1)/2) ≡ (a/p) (mod p)
where (a/p) is the Legendre symbol. If (a/p) = 1, then a is a quadratic residue modulo p, and if (a/p) = -1, then a is a quadratic non-residue modulo p.
In this case, we have p = 13 and a = 3. Therefore, we need to evaluate:
3^6 ≡ (3/13) (mod 13)
To evaluate (3/13), we need to use the properties of the Legendre symbol. First, we note that:
(3/13) ≡ (-1/13) (3/13)
since (-1/13) = (-1)^((13-1)/2) = 1 by Euler's criterion. Next, we use the quadratic reciprocity law to simplify (-1/13):
(-1/13) = (-1)^((13-1)/2) * (13-1)/2 = 1
Therefore, we have:
(3/13) ≡ (-1/13) (3/13) ≡ 1 * (3/13) ≡ (3/13)
So we need to evaluate (3/13) to determine if it is equal to 1 or -1.
To do this, we can use the quadratic reciprocity law again:
(3/13) = (13/3) * (-1/3)
Since 13 ≡ 1 (mod 3), we have:
(13/3) = (1/3) = 1
Since 3 ≡ 3 (mod 4), we have:
(-1/3) = -1
Therefore, we have:
(3/13) = (13/3) * (-1/3) = 1 * (-1) = -1
So (3/13) = -1.
Therefore, the statement is true, and we have:
(3/13) = -1 ≡ 3^6 (mod 13)