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

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July 10th 2010, 04:55 PM
dwsmith

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

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

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

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

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

Not sure if this is going anywhere.
July 10th 2010, 04:57 PM
undefined

Quote:

Originally Posted by dwsmith

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

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

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

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

Not sure if this is going anywhere.

m-a is congruent to -a (mod m). Square this and you get a^2.
July 10th 2010, 05:38 PM
chiph588@

Quote:

Originally Posted by dwsmith

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

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

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

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

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 $(m-a)^2=m^2-2am+a^2\equiv a^2\bmod{m}$

Now, we know what $a^2$ is equivalent to modulo $m$ (Wink).