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

#### dwsmith

MHF Hall of Honor
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.

#### undefined

MHF Hall of Honor
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.

#### [email protected]

MHF Hall of Honor
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 $$\displaystyle m$$ (Wink).