$a^2 \equiv 1 \pmod{311}$

Question

Is easy to see that we can solve $a^2 \equiv 1 \pmod{311}$ with $a \equiv 1 \pmod{311}$ and $a \equiv -1 \pmod{311}$ but how do we prove that there is no other solution without doing a table of 311 values of $a$.

$311$ is a prime, hence $\mathbb{F}_{311}$ is a field and the only roots of $x^2-1$ are $1$ and $-1$. — Jack D'Aurizio, Feb 18 '19 at 15:52

score 3 · Accepted Answer · answered Feb 18 '19 at 09:17

If $a\neq 1$ then $a-1\neq 0$ so it is invertibile with respect the product on $\mathbb{Z}_{311}$ because $311$ is a prime number.

If

$a^2=1 $ (mod 311)

then

$(a-1)(a+1)= 0 $ (mod 311)

but

$a-1$ is invertibile so

$a+1=0$

then

$a=-1$

So the solutions of your modular equation are only $a=1,-1$

cansomeonehelpmeout · Answer 2 · 2019-02-18T09:42:02.687

$311$ is a prime, so $\Bbb Z_{311}$ is a field (and hence an integral domain). This means in particular that if $(a+1)(a-1)\equiv_{311}0$ then either $a+1\equiv_{311}0$ or $a-1\equiv_{311}0$

More generally,

for a field $k$ we have that $k[x]$ is a Euclidean domain (for a polynomial $f(x)\in k[x]$, define $N(f)=\deg f$).
The Polynomial Remainder Theorem holds in any Euclidean domain, and therefore also The Factor Theorem
with this we deduce that the number of roots of $f(x)\in k[x]$ is $\leq \deg f$.

$a^2 \equiv 1 \pmod{311}$

2 Answers2