if $\displaystyle a\equiv b \ \mbox{(mod m)}$, then $\displaystyle a^2\equiv b^2 \ \mbox{(mod m)}$

Here is what I tried but it didn't go anywhere.

$\displaystyle m|(a-b)\rightarrow (mx)^2=(a-b)^2\rightarrow m^2x^2=a^2-2ab+b^2$

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- Jun 26th 2010, 12:12 PMdwsmithMore Congruence
if $\displaystyle a\equiv b \ \mbox{(mod m)}$, then $\displaystyle a^2\equiv b^2 \ \mbox{(mod m)}$

Here is what I tried but it didn't go anywhere.

$\displaystyle m|(a-b)\rightarrow (mx)^2=(a-b)^2\rightarrow m^2x^2=a^2-2ab+b^2$ - Jun 26th 2010, 12:32 PMundefined
- Jun 26th 2010, 12:33 PMdwsmith
- Jun 26th 2010, 12:36 PMundefined
- Jun 26th 2010, 12:38 PMdwsmith
- Jun 26th 2010, 12:49 PMundefined
I never wrote a=b. "a turning into b" is not the same as a=b, if you meant to convey this then you need to be more clear.

All we need is $\displaystyle ba \equiv bb$.

It's perhaps a little confusing since the letters overlap.

Let's rewrite the mathworld property as follows

Let $\displaystyle x\equiv x'$ (mod m) and $\displaystyle y \equiv y'$ (mod m), then

$\displaystyle xy \equiv x'y'$ (mod m)

So here

x=b

x'=b

y=a

y'=b - Jun 26th 2010, 01:09 PMTwistedOne151An alternative
dwsmith,

An alternative, more along the lines of your original argument is take where you noted that if $\displaystyle a\equiv{b}\,\pmod{m}$, then m divides a-b; and then note that if m divides a-b, it divides $\displaystyle (a-b)(a+b)=a^2-b^2$, so $\displaystyle a^2-b^2$ is then a multiple of m as well.

--Kevin C. - Jun 26th 2010, 01:10 PMdwsmith
- Jun 26th 2010, 01:36 PMdwsmith
Never mind, I have it.

- Jun 26th 2010, 01:48 PMundefined
I like TwistedOne151's approach, but it's good to know what kinds of rules you can apply with congruences, which can make things easier in general, for example

$\displaystyle ab+cd\equiv a'b'+c'd'\ (\text{mod}\ n)$

as long as

$\displaystyle a\equiv a'\ (\text{mod}\ n)$

$\displaystyle b\equiv b'\ (\text{mod}\ n)$

etc.

This will help later on; for example, using Euler's theorem, you will be able to do this manipulation:

$\displaystyle a^{2\varphi(n)}\equiv \left(a^{\varphi(n)}\right)^2 \equiv 1^2 \equiv 1\ (\text{mod}\ n)$

given that gcd(a,n)=1.