Is there any specific reason in not including non-units of $\mathbb{Z}_n$ as quadratic residues? As an examples, we say that in $\mathbb{Z}_8$, the set of quadratic residues is just {1} and not {1,4}.
Asked
Active
Viewed 137 times
0
-
1If we were more inclusive, too many theorems, proofs would have to say "quadratic residue relatively prime to $n$" instead of "quadratic residue." Also, for many years the focus was almost entirely on prime $n$. – André Nicolas May 23 '13 at 07:50
1 Answers
0
It is a nice idea to associate with unit quadratic residues the number $1$ and with unit quadratic non-residues the number $-1$ (this is the logic behind the Legendre and Jacobi symbols).
The reason this is a nice idea is that you have the following properties of unit residues and non-residues:
residue $\cdot$ residue $=$ residue
residue $\cdot$ non-residue $=$ non-residue
non-residue $\cdot$ non-residue $=$ residue
If you replace "residue" by $1$ and "non-residue" by $-1$ above, you get truthful statements about integers.
When you include non-unit residues, the above formulas are no longer correct. For example (taken from Wikipedia), if you're working modulo $15$, then $3\cdot 5 = 9 \cdot 5 = 9 \cdot 10 = 0$. But $3, 5$ are non-residues and $9, 10$ are residues.
Yoni Rozenshein
- 6,817
-
But this sort of thing happens only with $0$ and not other numbers, isn't it? – Shiva May 27 '13 at 12:06
-
Try looking at modulo $30$ instead of modulo $15$... Then $9 \cdot 5 = 15$ is an example of a residue times a non-residue giving a residue. – Yoni Rozenshein May 27 '13 at 12:10
-