# Thread: let a be a solution of x^2\equiv 1 (mod m). Show that m-a is also a solution.

1. ## let a be a solution of x^2\equiv 1 (mod m). Show that m-a is also a solution.

let $\displaystyle a$ be a solution of $\displaystyle x^2\equiv 1 \ \mbox{(mod m)}$. Show that $\displaystyle m-a$ is also a solution.

$\displaystyle a^2\equiv 1 \ \mbox{(mod m)}$

$\displaystyle (m-a)^2\equiv 1 \ \mbox{(mod m)}\rightarrow m^2-2am+a^2\equiv 1 \ \mbox{(mod m)}\rightarrow m^2-2am+1\equiv 1 \ \mbox{(mod m)}$

$\displaystyle \rightarrow m-2a\equiv 0 \ \mbox{(mod 1)}$

Not sure if this is going anywhere.

2. Originally Posted by dwsmith
let $\displaystyle a$ be a solution of $\displaystyle x^2\equiv 1 \ \mbox{(mod m)}$. Show that $\displaystyle m-a$ is also a solution.

$\displaystyle a^2\equiv 1 \ \mbox{(mod m)}$

$\displaystyle (m-a)^2\equiv 1 \ \mbox{(mod m)}\rightarrow m^2-2am+a^2\equiv 1 \ \mbox{(mod m)}\rightarrow m^2-2am+1\equiv 1 \ \mbox{(mod m)}$

$\displaystyle \rightarrow m-2a\equiv 0 \ \mbox{(mod 1)}$

Not sure if this is going anywhere.
m-a is congruent to -a (mod m). Square this and you get a^2.

3. Originally Posted by dwsmith
let $\displaystyle a$ be a solution of $\displaystyle x^2\equiv 1 \ \mbox{(mod m)}$. Show that $\displaystyle m-a$ is also a solution.

$\displaystyle a^2\equiv 1 \ \mbox{(mod m)}$

$\displaystyle (m-a)^2\equiv 1 \ \mbox{(mod m)}\rightarrow m^2-2am+a^2\equiv 1 \ \mbox{(mod m)}\rightarrow m^2-2am+1\equiv 1 \ \mbox{(mod m)}$

$\displaystyle \rightarrow m-2a\equiv 0 \ \mbox{(mod 1)}$

Not sure if this is going anywhere.
I prefer undefined's method more, but if you want to solve this the way you were, then $\displaystyle (m-a)^2=m^2-2am+a^2\equiv a^2\bmod{m}$

Now, we know what $\displaystyle a^2$ is equivalent to modulo $\displaystyle m$ .