I split this into $x^2\equiv 1\pmod {7}$ and $x^2\equiv 1\pmod {13}$.
For $x^2\equiv 1\pmod {7}$, i did:
$$ (\pm1 )^2\equiv 1\pmod{7}$$ $$(\pm2 )^2\equiv 4\pmod{7}$$ $$(\pm3 )^2\equiv 2\pmod{7}$$ Which shows that the solutions to $x^2\equiv 1\pmod {7}$ are $\pm1$.
For $x^2\equiv 1\pmod {13}$, i did:
$$ (\pm1 )^2\equiv 1\pmod{13}$$ $$(\pm2 )^2\equiv 4\pmod{13}$$ $$(\pm3 )^2\equiv 9\pmod{13}$$ $$ (\pm4 )^2\equiv 3\pmod{13}$$ $$(\pm5 )^2\equiv {-1}\pmod{13}$$ $$(\pm6 )^2\equiv 10\pmod{13}$$Which shows that the solutions to $x^2\equiv 1\pmod {13}$ are $\pm1$.
Thus, I concluded that the solutions to $x^2\equiv 1\pmod {91}$ must be $\pm1$. I thought that $\pm1$ were the only solutions, but apparently I am incorrect! How do I go about finding the other solutions to this congruence?
Best Answer
You have $x\equiv\pm1\mod7$ and $x\equiv\pm1\mod13$. For all the solutions, you have to consider the systems: $$x\equiv1\mod7$$ $$x\equiv1\mod13$$ and $$x\equiv-1\mod7$$ $$x\equiv1\mod13$$ and $$x\equiv1\mod7$$ $$x\equiv-1\mod13$$ and $$x\equiv-1\mod7$$ $$x\equiv-1\mod13$$ as each system will get you a valid answer. I think you only had the first and the last systems and that you only considered the cases where the signs were similar.