# Thread: Find all primes p such that $$\frac{10}{p}$$=1.

1. ## Find all primes p such that (10/p)=1.

Find all primes p such that $(\frac{10}{p})=1$

2. First: $
\left( {\tfrac{{10}}
{p}} \right)_L = \left( {\tfrac{2}
{p}} \right)_L \cdot \left( {\tfrac{5}
{p}} \right)_L

$

You should know that: $
\left( {\tfrac{2}
{p}} \right)_L = \left( { - 1} \right)^{\tfrac{{p^2 - 1}}
{8}}
$
( If you don't, this can be deduced easily from Gauss' Lemma )

By the Quadratic Reciprocity Theorem: $
\left( {\tfrac{5}
{p}} \right)_L \cdot \left( {\tfrac{p}
{5}} \right)_L = \left( { - 1} \right)^{\tfrac{{5 - 1}}
{2} \cdot \tfrac{{p - 1}}
{2}} = \left( { - 1} \right)^{p - 1} = 1

$
hence $
\left( {\tfrac{5}
{p}} \right)_L = \left( {\tfrac{p}
{5}} \right)_L
$

And by Euler's Criterion: $
\left( {\tfrac{p}
{5}} \right)_L \equiv p^2 \left( {\bmod .5} \right)
$

So what we actually want is: $
p^2 \cdot \left( { - 1} \right)^{\tfrac{{p^2 - 1}}
{8}} \equiv 1\left( {\bmod .5} \right)
$

Let's go over all the possible remainders module 5:

• Case: $
p = 5k + 1
$
here we must have: $
\left( { - 1} \right)^{\tfrac{{5^2 k^2 + 10k}}
{8}} = 1
$
$
\Rightarrow 16\left| {\left( {5k^2 + 2k} \right)} \right.
$
so k is even: $
k = 2k_1 \Rightarrow 5^2 k^2 + 10k = 4 \cdot \left( {5 \cdot k_1 ^2 + k_1 } \right)
$
$
\Rightarrow 4\left| {\left[ {k_1 \cdot \left( {5 \cdot k_1 + 1} \right)} \right]} \right.
$
and this can happen if and only if $
{4\left| {k_1 } \right.}
$
or $
{k_1 = 4k_2 +3}
$
and so we have either: $
\left. 8 \right|k
$
or $
k = 8k_2 + 6
$
this means that we must have: $
p \equiv 1\left( {\bmod .40} \right) \vee p \equiv 31\left( {\bmod .40} \right)
$
• Case: $
p = 5k - 1
$
( the 'similar' case, just change the sign - check it-), we get: $
p \equiv -1\left( {\bmod .40} \right) \vee p \equiv -31\left( {\bmod .40} \right)
$
• Case: $p=5k-3$: here we require $
\left( { - 1} \right)^{\tfrac{{p^2 - 1}}
{8}} = \left( { - 1} \right)^{\tfrac{{5^2 k^2 - 30k + 8}}
{8}} = - 1
$
so we must have: $
\left( { - 1} \right)^{\tfrac{{5^2 k^2 - 30k}}
{8}} = 1
$
and for this to happen: $
\left. {16} \right|\left[ {k\left( {5k - 6} \right)} \right]
$
so k must be even, and then: $
k = 2k_1
$
and we must have: $
\left. 4 \right|\left[ {k_1 \left( {5k_1 - 3} \right)} \right]
$
and for this either: $
\left. 4 \right|k_1
$
or $
5k_1 \equiv k_1 \equiv 3\left( {\bmod .4} \right) \Rightarrow k_1 = 4k_2 + 3
$
then we must have ( substitute back to p): $
p \equiv 37\left( {\bmod .40} \right) \vee p \equiv 27\left( {\bmod .40} \right)
$
• Case $p=5k+3$ here we get: $
p \equiv -37\left( {\bmod .40} \right) \vee p \equiv -27\left( {\bmod .40} \right)
$
-check it-

So the solution is the set of all primes $p$ satisfying one of the following congruences: $
p \equiv \pm 1, \pm 27, \pm 31, \pm 37\left( {\bmod .40} \right)
$

3. Originally Posted by mancillaj3
Find all primes p such that $(\frac{10}{p})=1$
Here is an alternate approach but not much of a difference.

First, $(10/p) = (2/p)(5/p)$ therefore, $(2/p)=(5/p)=1\text{ or }(2/p)=(5/p)=-1$.

You should know that $(2/p) = 1$ iff $p\equiv 1,7(\bmod 8)$.

Now $(5/p) = (p/5)$ (since $5\equiv 1(\bmod 4)$).
If $p\equiv 1,4(\bmod 5) \implies (p/5) = (\pm 1/5) = 1$.
If $p\equiv 2,3(\bmod 5)\implies (p/5) = (\pm 2/5) = -1$.
Thus, $(5/p) = 1$ iff $p\equiv 1,4(\bmod 5)$.

Hence, $(2/p)=(5/p)=1$ iff:
$p\equiv 1(\bmod 5) \text{ and }p\equiv 1(\bmod 8)$
$p\equiv 1(\bmod 5) \text{ and }p\equiv -1(\bmod 8)$
$p\equiv -1(\bmod 5)\text{ and }p\equiv 1(\bmod 8)$
$p\equiv -1(\bmod 5)\text{ and }p\equiv -1(\bmod 8)$
Solving this:
$p\equiv \pm 1, \pm 31(\bmod 40)$

Do same computation for the second part, I am too lazy to write all the details.