Re: Math -- Re: quadratic quadratic non-residue

In article
<5798da9e-93b7-4e01-97f1-78c85503a792@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Tonico <Tonicopm@xxxxxxxxx> wrote:

Very nice, neat proof. I wonder whether this could be of any help in
the next problem a student asked:
Let q be a prime > 2 power, and let f(x) = x^2 + ax + b be an
irreducible pol. in F q[x] (and thus a^2 - 4b is not a square in
F q) . Let X be the quadratic (residue) character on F q: X(w) = 1
if w=/=0 is a quadratic q-residue, X(w) = -1 if w is not a quadratic
q-residue, X(0) = 0 , then:
SUM[X(f(x))] = -1, when the sum is over all the elements x in F q (of
course, the same would be true for the sum over (F q)*)

Since clearly there is, just as in F p, the same number of non-zero
quadratic residues as of non-zero quad. non-residues in F q, lemma 1
above applies, and since the discriminant of f is clearly non-zero
then (B) =/=> (A), from which we already know that range(f) is not a
subset of T.
But range(f) is not a subset of S, either:
Prop.: With the same notation as above, for r,s,t in F q:
1) f(s) = f(r) <==> r = s or r + s = -a ;
2) f(r + m) = f(r) + f(m) + 2rm - b ;
3) f(r + m) = f(r + t) <==> m = t or m + t = - a - 2r
Proof: Simple evaluation. Even (3) is only (1) with a little twist.
From the above prop., since we already know that f(r) is in S for
some r and the right side in (3) above has at most one solution for
fixed m, we get that for some m in F q f(r + m) will be in T.
Now the big question: is there any way to count the points
x in F q (or (F q)*) s.t. f(x) is in S, and/or in T?
I've tried the above with several primes and even with some rather
simple non prime fields, like F 9 and F 25, but that's all so far...
I in fact suspect that it is enough to require just a^2 - 4b =/= 0,
without requiring f(x) to be irreducible. Again, I've checked some
simple cases.
Any input will be appreciated.

You may want to look at exercises 6, 7, and 8
of Chapter 5 of Ireland & Rosen, A Classical Introduction
to Modern Number Theory

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.

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