Exercise in algebra with modulo

Question

I'm studying Cassels' book Elliptic Curves for a week now, and I'm at the local global principle. I'm trying to prove the first exercise of this chapter, which says

Let $p > 2$ be prime and let $b, c \in \Bbb Z$ and $ p $ does not divide $b$. Show that $bx^2 + c$ takes precisely $\frac {p + 1}{2}$ distinct values $p$ for $x \in \Bbb Z$.

The problem here is that i don't understand what "values $p$" means. At first i thought that it means $\mod(p)$ values, but if that was the case, then for $x=0\in \Bbb Z$,we have that the $b\cdot 0^2+c=c$ can take $p-1$ values $\mod(p)$, $\{0,1,...,p-1\}$, and then we should have that $p-1=\frac {p + 1}{2}=>p=3$ which restricts the exercise.

So what does "values $p$" mean?

Thank you!

No, $b$ and $C$ are given. You may only let run $x$ through the integers. — Dietrich Burde, May 23 '14 at 11:48
@DietrichBurde ,yeap,i got confused,but does values p mean modp values? — user113576, May 23 '14 at 12:01
It's likely the problem should read "$bx^2+c$ takes precisely $(p+1)/2$ distinct values mod $p$ for $x \in \mathbb{Z}.$" That is, the word "mod" should be inserted into the phrase immediately after the word "values" in the reading you have typed. — coffeemath, May 23 '14 at 12:19
$x\rightarrow bx^2+c$ is a composition of three functions. Two are 1-1. The third isn't quite. Need more clues? — Dan Piponi, May 23 '14 at 13:52
@user113576 I put up an "answer" which gives some outline, but would need a bit more work to write it up in a convincing form. — coffeemath, May 23 '14 at 13:59
See http://math.stackexchange.com/questions/799023/help-needed-in-understanding-proof-every-odd-prime-p-has-exactly-p-1-2-qu for the hard part. — lhf, May 23 '14 at 15:08

coffeemath · Accepted Answer · 2014-05-23T20:47:36.227

Since $b$ is not divisible by $p$ it has an inverse under multiplication, call it $b'$ for now, such that $b'b=1$ mod $p$. Then the equation $bx^2=c$ may be multiplied on both sides by $b'$ to get $b'bx^2=b'c,$ which since $b'b=1$ is the same as the equation $$x^2=b'c \tag{1}$$ Thus since $b,c$ are fixed we may as well define $d=b'c$ and we're looking at the equation $x^2=d$. Now note that $x^2=(-x)^2$ for any $x$, and note that if $d \neq 0$ the values $x,-x$ are distinct, otherwise $2x=0$ and then $x=0$ since $p$ is odd. The solution to $x^2=0$ is only $x=0$, so that counts as one, and the other solutions can be paired up as $x,-x$ as indicated above. The "values taken on" are then the various possible $b$ which can appear on the right side of $x^2=b$, so by the 2-to-1 ness of the solutions for nonzero $b$ there are $(p-1)/2$ with nonzero $b$, and one more solution when $b=0$, for a total of $(p-1)/2+1=(p+1)/2.$

Once you've arrived at $x^2=b$ you are looking for values of $b$ which are either $0$ or else are what is called "quadratic residues". If you google the latter you can likely find sites where this count of the quadratic residues is done more carefully than outlined above.

Exercise in algebra with modulo

1 Answers1