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Let $P$ be the set of all primes of the form $8k+1$. Can we say anything about the density of primes $p$ in this set such that if $x^2=2$ in $\mathbb F_p$, then $x+1$ is a square in $\mathbb F_p$? (Note that there is no ambiguity of $x$ and $-x$ in this particular case, because $x+1$ is a square iff $-x+1$ is a square, as $(x+1)(-x+1)=1-x^2=-1$, which is a square in $\mathbb F_p$.)

In general, Suppose we have an equation $x^2=a$ in $\mathbb F_p$. For such $x$ values, if we look at the values $bx+c$ for fixed $b\ne 0$ and $c$, then is their being quadratic residue or non-quadratic residue equally distributed among the primes such that $x^2=a$ has a root in $\mathbb F_p$?

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For your first question, the easiest way to do it is to note that primes $p=8k+1$ correspond to split primes of $\mathbb Q(\sqrt{-1},\sqrt{2})$, and since a density $1$ set of primes is split, it's equivalent to ask for the fraction of primes $\mathfrak p$ of $\mathbb Q(\sqrt{-1},\sqrt{2})$ such that $1-\sqrt{2}$ is a quadratic residue mod $\mathfrak p$. This density is $1/2$ unless $1-\sqrt{2}$ is a square in $\mathbb Q(\sqrt{-1},\sqrt{2})$.

It is not a square, as if it were equal to $(a+ib)^2$ for $a,b\in \mathbb Q(\sqrt{2})$, we would have $2ab=0$ so $a=0$ or $b=0$ and it would equal either $a^2$ (always positive) or $-b^2$ (always negative), but $1+\sqrt{2}$ is sometimes positive and sometimes negative.

So the answer is always yes.


For your second question, the answer you seek is the distribution of the number of roots of the equation $ ((y^2-c)/b )^2 = a$. When this equation has four roots, then $x^2=a$ has two roots, both with $bx+c$ quadratic residues. When it has no roots, then $x^2=a$ has two roots, one with $bx+c$ quadratic residue and one not. The density of the remaining set, where $x^2=a$ has two roots, neither a quadratic residue, can be found by subtracting $1/2$ from the proceeding equation.

So you want to apply Chebotarev to the Galois group of this equation. So for instance if the Galois group is full $D_4$, as should happen for generic $a,b,c$, the probability of two quadratic residues is $1/8$, one quadratic residue is $1/4$, two roots of $x^2=a$ but no quadratic residues is $1/8$.

However the Galois group is not always very large. If $x^2=a$, then $(dx+e)^2 = 2de x+ (e^2+ d^2a)$, so if we happen to have $b=2de$, $c=e^2+d^2a$ then all the roots are quadratic residues.

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    $\begingroup$ "$1+\sqrt2$ is sometimes positive and sometimes negative." This line made my day. BTW, isn't it $-b^2$ that is always negative? $\endgroup$
    – Fan Zheng
    Dec 12 '16 at 18:20
  • $\begingroup$ @FanZheng Fixed. $\endgroup$
    – Will Sawin
    Dec 12 '16 at 18:59

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