Let $$p_1$$, $$p_2$$, $$p_3$$ be distinct primes satisfying $$p_1 \equiv p_2 \equiv p_3 \equiv 5 \pmod 8$$, such that $$p_i$$ is a quadratic residue modulo $$p_j$$ for any $$i\neq j$$. Prove that, for any prime $$p$$, there exist integers $$x,y$$ such that $$y^2 \equiv p_1(x^2 - p_2p_3)(x^2 + p_2p_3) \pmod p$$.

There are surely not many $$x$$ for which $$x^2 \pm p_2p_3$$ are both squares, but still I get confused in cases of "this is a square mod this prime" whatever I try. Any help appreciated!

• Notice that, in general, you don't need both $x^2-p_2p_3$ and $x^2+p^2p_3$ to be squares, since one of them could be $0$. Apr 15, 2019 at 11:41
• Yes, I count $0$ as a square, too. Anyway, in case $p_1$ is a square $mod p$, they must either both be squares, or both not be squares. To attack the latter seems much harder, though. Apr 15, 2019 at 11:47
• Curious problem. Where does it come from, please? Apr 15, 2019 at 13:12
• A friend saw it in an Oxford past paper in Elliptic Curves. The original one wants to find a solution in $\mathbb{Q}_p$ and he told me that he somehow figured out it is enough to show it in the rational integers (I am still not much into p-adics). Apr 16, 2019 at 9:56

The cases where $$p\in\{p_1,p_2,p_3\}$$ are easy, and I assume that $$p\ne p_i$$, $$i\in\{1,2,3\}$$.

For brevity I write $$P:=p_2p_3$$; also, let $$\left(\frac{\cdot}p\right)$$ denote the Legendgre symbol.

If either $$P$$ or $$-P$$ is a quadratic residue mod $$p$$, then we can take $$y=0$$ and choose $$x$$ appropriately; suppose thus that both $$P$$ and $$-P$$ are quadratic non-residues. Notice that this implies $$\left(\frac{-1}p\right)=1$$.

If $$p_1$$ is a quadratic residue mod $$p$$, then we can take $$x=0$$; suppose therefore that $$p_1$$ is a quadratic non-residue mod $$p$$.

Clearly, with all the assumption made, it suffices to show that there exists $$x\in\mathbb Z/p\mathbb Z$$ such that one of $$x^2-P$$ and $$x^2+P$$ is a quadratic residue, while another is a quadratic non-residue mod $$p$$.

Suppose for a contradiction that for each $$x\in\mathbb Z/p\mathbb Z$$ we have $$\left(\frac{x^2-P}p\right)=\left(\frac{x^2+P}p\right)$$. Replacing $$x$$ with $$Px$$ and using multiplicativity of the Legendre symbol, we conclude that $$\left(\frac{Px^2-1}p\right)=\left(\frac{Px^2+1}p\right)$$, for all $$x\in\mathbb Z/p\mathbb Z$$. Recalling that $$P$$ is a quadratic non-residue, we further conclude that $$\left(\frac{z+1}p\right)=\left(\frac{z-1}p\right)$$ for any quadratic non-residue $$z$$, and also for $$z=0$$.

Let $$N_0\subset\mathbb Z/p\mathbb Z$$ be the set containing all quadratic non-residues and $$0$$. Since $$|N_0|=(p+1)/2>p/2$$, this set contains two consecutive elements of $$\mathbb Z/p\mathbb Z$$; say, $$z-1$$ and $$z$$. But then also $$z+1\in N_0$$ and, consecutively, $$z+2\in N_0$$ etc. As a result, all elements of $$\mathbb Z/p\mathbb Z$$ are in $$N_0$$, meaning that there are no quadratic residues mod $$p$$. This is a clear nonsense, showing that $$\left(\frac{x^2-P}p\right)=\left(\frac{x^2+P}p\right)$$ cannot hold for all $$x\in\mathbb Z/p\mathbb Z$$.

Notice that we have not used the assumption $$p_1\equiv p_2\equiv p_3\equiv 5\pmod 8$$ and $$\left(\frac{p_i}{p_j}\right)=1$$, $$i\ne j$$.

• I think there is a problem. If $p_1$ is a QR mod $p$ then $x = 0$ to work is equivalent to $p \equiv 1 \pmod 4$ which is not always the case. Apr 16, 2019 at 10:04
• @DesmondMiles: There is no problem with it, this is actually addressed in my solution: notice the sentence in the beginning of the proof starting with "Notice that..." Apr 16, 2019 at 11:35
• Yep, my bad, thanks! Apr 16, 2019 at 12:53