Are there infinitely many finite fields of non-two prime order with consecutive quadratic residues? I'm curious whether there are non-two prime numbers $p$ where such that fields of order $p$ have $\left\{1, 2, \dots \frac{p-1}{2}\right\}$ as their quadratic residues.
This paper Real Analysis is a Degenerate Case of Discrete Analysis by Doron Zeilberger was linked in an answer on Stack Exchange and someone commented that taking a large finite field as your universe leads to positive and negative numbers being indistinguishable, since all elements of the field are expressible as a finite sum of $1$s. My question is only loosely inspired by the paper and the comment since I'm not qualified to comment on either of them. 
I'm wondering if there are infinitely many fields where the Legendre symbol and a signum function "line up" ... by having $\{1,2,\dots,\frac{p-1}{2}\}$ be the quadratic residues.
The finite field of order 3 has this property. Here's its multiplication table for the nonzero elements. It's clear if $1$ were considered positive and $2$ were considered negative our property would hold.
   1  2
 +-----
1| 1  2
2| 2  1

The finite field of order $5$ does not have this property, since its perfect squares are $1$ and $4$.
   1  2  3  4
 +-----------
1| 1  2  3  4
2| 2  4  1  3
3| 3  1  4  2
4| 4  3  2  1

So, are there finitely or infinitely many finite fields with this property?

no, not possible. The residues make a multiplicative subgroup; there must be exactly $\frac{p-1}{2}$ of them, no more. You can check for, say, primes between $3$ and $19.$
For prime $p \geq 23,$ there will be a number of the form $n^2 - n$ (where $n-1$ and $n$  are consecutive quadratic residues) between $\frac{p-1}{2}$ and $p-1.$ For $p=23,$ $n(n-1) = n^2 - n  = 12$ works. But then $12$ is a residue, and there are too many
For $p=29, $ we take $n=5,$ so $n(n-1) = 20$ is forced to be a residue, too many.
For a pattern, we may take $$n = \left\lfloor \sqrt p \right\rfloor  $$
so that $n$ and $n-1$ are both smaller than $\sqrt p,$ there product is an integer that is now forced to be a residue
Actually, now that I think of it, we may just take $n^2,$ where $n = \left\lfloor \sqrt p \right\rfloor  $ For large enough $p,$ this square (already below $p$) will also be larger than $\frac{p-1}{2}$
