Interesting number theory question, which I feel should be reasonably straight forward, but I can't seem to crack it.

Show that 7 is a quadratic residue for any prime p of the form 28k + 1 and 28k + 3.

Now, I know that a quadratic residue $a$ (mod m) means that there exists $x$ such that $x^2\equiv a$(modm). So if 7 is a quadratic residue mod 28k+1 or 28k+3 then $x^2\equiv 7$ (mod 28k+1 or 28k +3). But I am unsure of how to tackle this problem. Perhaps, I could go on to say that for the first case, $x^2\equiv 6$ (mod 28k) $\equiv$ 3 (mod 14k)??

Or perhaps I am going about this wrong.

I also know the laws of quadratic reciprocity, so I need to show $(\frac{7}{28k+1})$ =1 and likewise for 28k +3

Help would be greatly appreciated. Thank you


Hint: If $p=28k+1$ then $\frac{7-1}{2}\cdot \frac{p-1}{2}$ is even. If $p=28k+3$ then $\frac{7-1}{2}\cdot\frac{p-1}{2}$ is odd.

  • $\begingroup$ that's a very well-judged hint, Thomas! $\endgroup$ – David Holden Dec 20 '13 at 12:37

Using Quadratic Reciprocity Theorem,

$$\left(\frac p7\right)\left(\frac 7p\right)=(-1)^{\frac{3(p-1)}2}$$

Now, if $p\equiv1\pmod4,\left(\frac p7\right)\left(\frac 7p\right)=1\iff \left(\frac 7p\right)=\left(\frac p7\right)$

Now, $(\pm1)^2\equiv1\pmod 7,(\pm2)^2\equiv4,(\pm3)^2\equiv2\implies p$ needs to be $1,2,4\pmod 7$

$p\equiv1\pmod4$ and $p\equiv1\pmod7\implies p\equiv1\pmod {4\cdot7}$

Using CRT, we can show

$p\equiv1\pmod4$ and $p\equiv2\pmod7\implies p\equiv9\pmod {4\cdot7}$

$p\equiv1\pmod4$ and $p\equiv4\pmod7\implies p\equiv25\pmod {4\cdot7}\equiv-3$

Now, if $p\equiv-1\pmod4,\left(\frac p7\right)\left(\frac 7p\right)=-1\iff \left(\frac 7p\right)=-\left(\frac p7\right)$

and $p$ needs to be $3,5,6\pmod 7$

$p\equiv-1\pmod4\equiv3$ and $p\equiv3\pmod7\implies p\equiv3\pmod {4\cdot7}$

$p\equiv-1\pmod4$ and $p\equiv6\pmod7\equiv-1\implies p\equiv-1\pmod {4\cdot7}$

Using CRT, we can show

$p\equiv-1\pmod4$ and $p\equiv5\pmod7\implies p\equiv19\pmod {4\cdot7}\equiv-9$

So, $7$ is a quadratic residue for any prime $p\equiv\pm1,\pm3,\pm9\pmod{28}$

  • 1
    $\begingroup$ It's obvious that $\left(\frac{28k+1}{7}\right) $ = $\left(\frac{1}{7}\right) $ $\endgroup$ – Zack Ni Apr 11 '16 at 13:36

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