# Solving $x^2\equiv1\pmod{7}$

I'm in a number theory course and my teacher explained Fermat's little theorem, but I'm not sure how to find solutions to the congruence.

Don't necessarily need an answer, a method would be much more helpful! Thank you for all help!

-
My first inclination is to observe the squares of the members of $[7]$ and find the reduced residues to see if they equal 1, but I figure there must be a generalized method for large values. –  Neurax Apr 11 '13 at 3:48
a method for this sort of situations: Try factoring the polynomial first. If the polynomial is reducible, then find its factorisation as products of irreducible oplynomials; to find the irreducible polynomials, one usually employs some specialized skills like FLT or trial and error, etc... . –  awllower Apr 12 '13 at 4:45

Hint $\rm\,\ 7\mid x^2\!-1 = (x\!-\!1)(x\!+\!1)\:\Rightarrow\: 7\mid \ldots\ \ \ or\ \ 7\mid \,\ldots,\:$ since $\,7\,$ is prime.
@Neurax, in general you can show that $x^2\equiv1(\mod p)\Leftrightarrow x=\pm1(\mod p)$. –  Easy Apr 11 '13 at 4:00