# Math Help - explanation of quadratic residue

1. ## explanation of quadratic residue

I'm having trouble understanding the concept of quadratic residue. I know you are supposed to take the integers lower than your mod number and square them and reduce via the mod number. My question is how do you know if you have a solution? For example 2^2 \equiv 4(mod 13) is not a solution but 3^2 \equiv 9(mod 13) is a solution. I don't understand why one is and not the other. Both 4 and 9 are squares of an integer. I know I'm misunderstanding something fundamnetal but I can't find any simple answers in my book or on the net. Thanks for any help.
David

2. Originally Posted by drounds67
I'm having trouble understanding the concept of quadratic residue. I know you are supposed to take the integers lower than your mod number and square them and reduce via the mod number. My question is how do you know if you have a solution? For example 2^2 \equiv 4(mod 13) is not a solution but 3^2 \equiv 9(mod 13) is a solution. I don't understand why one is and not the other. Both 4 and 9 are squares of an integer. I know I'm misunderstanding something fundamnetal but I can't find any simple answers in my book or on the net. Thanks for any help.
David
Both 4 and 9 are quadratic residues modulo 13. For a complete list, see Quadratic residue.

3. ## explanataion

I know I can look them up on a list but that didn't answer my question. Let me try to phrase it better. I'm trying to determine the procedure to know what values are the quadratic residues of (mod 13). so I can repeat the method on a test or quiz with a different modulus.
thanks
david

4. Originally Posted by drounds67
I know I can look them up on a list but that didn't answer my question. Let me try to phrase it better. I'm trying to determine the procedure to know what values are the quadratic residues of (mod 13). so I can repeat the method on a test or quiz with a different modulus.
thanks
david
Considering mod 13,

$1^2\equiv1$

$2^2\equiv4$

$3^2\equiv9$

$4^2\equiv3$

$5^2\equiv12$

$6^2\equiv10$

$7^2\equiv(-6)^2\equiv10$
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$12^2\equiv(-1)^2\equiv1$

Therefore, the quadratic residues modulo 13 are 1, 4, 9, 3, 12 and 10.

5. thanks. I got it now, it makes sense. I think the way the notes were written was throwing me off.