quadratic character of 2
For any odd prime , Gauss’s lemma quickly yields
But there is another way, which goes back to Euler, and is worth seeing, inasmuch as it is the prototype of certain more general arguments about character sums.
By the binomial formula, we have
If , this implies . If , we get instead . In both cases, we get , proving (1) and (2).
A variation of the argument, closer to Euler’s, goes as follows. Write
Both are algebraic integers. Arguing much as above, we end up with
which is enough.
|Title||quadratic character of 2|
|Date of creation||2013-03-22 13:58:03|
|Last modified on||2013-03-22 13:58:03|
|Last modified by||mathcam (2727)|