13
$\begingroup$

I was working out some problems. This is giving me trouble.

  • If $p$ is a prime number of the form $4n+1$ then how do i show that:

$$ \sum\limits_{i=1}^{p-1} \Biggl( \biggl\lfloor{\frac{2i^{2}}{p}\biggr\rfloor}-2\biggl\lfloor{\frac{i^{2}}{p}\biggr\rfloor}\Biggr)= \frac{p-1}{2}$$

Two things which i know are:

  • If $p$ is a prime of the form $4n+1$, then $x^{2} \equiv -1 \ (\text{mod} \ p)$ can be solved.

  • $\lfloor{2x\rfloor}-2\lfloor{x\rfloor}$ is either $0$ or $1$.

I think the second one will be of use, but i really can't see how i can apply it here.

  • 0
    can you edit this to have a proper title using more searchable words, and move the equation into the question?2011-04-23
  • 0
    @Jeff: I don't know whether that can be done. I think this is perfect.2011-04-23
  • 0
    It is fine the way it is I can assure you Jeff was probably having a drug induced psychosis again2018-09-28

2 Answers 2

4

Here are some more detailed hints.

Consider the value of $\lfloor 2x \rfloor - 2 \lfloor x \rfloor$ where $x=n+ \delta$ for $ n \in \mathbb{Z}$ and $0 \le \delta < 1/2.$

Suppose $p$ is a prime number of the form $4n+1$ and $a$ is a quadratic residue modulo $p$ then why is $(p-a)$ also a quadratic residue?

What does this say about the number of quadratic residues $< p/2$ ?

All the quadratic residues are congruent to the numbers $$1^2,2^2,\ldots, \left( \frac{p-1}{2} \right)^2,$$ which are themselves all incongruent to each other, so how many times does the set $\lbrace 1^2,2^2,\ldots,(p-1)^2 \rbrace$ run through a complete set of $\it{quadratic}$ residues?

Suppose $i^2 \equiv a \textrm{ mod } p$ where $i \in \lbrace 1,2,\ldots,p-1 \rbrace$ and $a$ is a quadratic residue $< p/2$ then what is the value of

$$ \left \lfloor \frac{2i^2}{p} \right \rfloor - 2 \left \lfloor \frac{i^2}{p} \right \rfloor \quad \text{?}$$

  • 0
    @Chandru1 Please don't change maths in my answer, although I'm very happy for you to correct typos. Please put your ideas in the comments instead. Many thanks.2011-01-28
  • 0
    sorry Derek.2011-01-28
  • 1
    @Chandru1: We know $x^2 \equiv a$ and since $p=4n+1$ we have $y^2 \equiv -1$ and so $(xy)^2 \equiv -a \equiv p-a.$ I hope this helps.2011-01-28
  • 0
    wow incredible response infinitely impressed Sir2018-09-28
2

Without giving everything away: when is $\lfloor2x\rfloor - 2\lfloor x\rfloor$ equal to $0$, and when is it equal to $1$? Can you find some bijection between values of $i$ in your sum that fall into the first camp, and those that fall into the second? (You may find the other fact you gave to be useful for finding this bijection!)

  • 0
    $[2x]-2[x]$ is $0$ when $x \in \mathbb{Z}$, so using this i think it will be $0$ when $p \mid i^{2}$ or something like that.2011-01-28
  • 0
    You can have a more precise description of when [2x]-2[x] = 0 or 1..2011-01-28
  • 0
    What Soarer said - what can you say about [2x]-2[x] w.r.t. x mod 1? If this isn't enough hint, incidentally (and assuming this isn't homework, of course!), I can give you the full answer tomorrow, but I think it's worth trying to puzzle through on your own...2011-01-28