I am relatively new to Legendre symbols.
For the following problem, I feel that what I have is correct thus far:
(103/1009) -> (1009/103) -> (82/103) -> (2/103)(41/103)
103 is congruent to 3 (mod 4) and 1009 is congruent to 1 (mod 4), I flipped (103/1009) without negating it.
1009 is congruent to 82 (mod 103).
Since 82 is composite, I split it into two prime numbers.
(2/103)(41/103) -> (103/2)(103/41) -> (1/2)(21/41) -> (1/2)(3/41)(7/41)
103 is congruent to 1 (mod 2) 103 is congruent to 21 (mod 41)
(21/41) was split into (3/41)(7/41) since 21 is composite.
(1/2)(3/41)(7/41) -> (2/1)(41/3)(41/7) -> (1/1)(2/3)(6/7)
41 is congruent to 2 (mod 3) 41 is congruent to 6 (mod 7)
(1/1)(2/3)(6/7) -> (1/1)(2/3)(2/7)(3/7)
(6/7) became ((2/7)(3/7) since 6 is composite.
(1/1)(2/3)(2/7)(3/7) -> (1/1)(2/3)(7/2)(7/3) -> (1/1)(2/3)(1/2)(1/3)
7 is congruent to 1 (mod 2) 7 is congruent to 1 (mod 3)
2 is a quadratic non-residue of 3, so (2/3) evaluates to -1. 1 is a quadratic residue of 3, so (1/2) evaluates to 1. 1 is a quadratic residue of 3, so (1/3) evaluates to 1.
I am not sure what (1/1) evaluates to. Please help me understand this.
(1/1)(-1)(1)(1)
Please let me know if I am on the right track. Did I go too far in trying to solve this? If what I am doing is correct, where do I go from here? Thanks! (1/1)(-1)(1)(1)
I tried to reduce each symbol when possible by flipping when both the top and bottom numbers are prime.
The Legendre symbol $(\frac{a}{ p})$ is only defined when $p$ is prime, so $(\frac{1}{ 1})$ shouldn't arise. $(\frac{1}p) = 1$ for any prime $p$, since $1 = 1^2$ is a square.