Math Proof help

[coldize]

Hey I'm trying to prove a biconditional statement: Let a be an integer. a is congruent to 2 modulo 5 if and only if a squared is congruent to 4 modulo 5.

 

I proved it to the right and and I'm working on proving it to the left. I thought a proof by contradiction would be best but what I run into is the equation 5(b/a+2) = a-2 for some integer b. Now, if I could somehow prove that (b/a+2) is an integer, I would be set because that would contradict our assumption that 5 does not divide a-2 (because we're doing a proof by contradiction).

 

So if anyone could help or point me in the right direction, that would be wonderful. I was hoping these forums would have LaTeX embedding capabilities but it appears that isn't possible so I've just attached the LaTeX code I do have as a code snippet.

 

Thanks again!

 

\begin{proof}[Part 2] 
We will prove that if $a^2 \equiv 4$ (mod 5), then $a \equiv 2$ (mod 5) by using a proof by contradiction. That is, we will prove that  if $a^2 \equiv 4$ (mod 5)and $a \not \equiv 2$ (mod 5), 
then there will exist a logical fallacy.\\ \\Assume that $a^2 \equiv 4$ (mod 5) and $a \not \equiv 2$ (mod 5). By the definition of congruency, we know that $5|a^2-4$ and $5 \nmid a-2$.
By the definition of divides, we know that  $\exists b \in \mathbb{Z}$ such that $5b=a^2-4$. Observe that $a^2-4 = (a+2)(a-2)$. By dividing both sides of the equation $5b=(a+2)(a-2)$ by $(a+2)$, 
we obtain the equation $5(b/a+2)=(a-2)$. Since $b/a+2$ can be rewritten ..... ? 
\end{proof}

Edited February 27, 2011 by coldize

[Schrödinger's hat]

Poor coldize, noone wants to answer you because it's ages since we've done any discrete and remembering how to do it takes effort.

Perhaps you can appease our laziness a bit and use the left square bracket math right square bracket tag to rewrite your working with the forum's LaTeX embedding (there's a tutorial around somewhere too, use the search function)?

On that note, sleepy time. May be able to summon the effort when I get up.

 

(Alright, I gave in and played with it for a tiny while) Simplest (not shortest) way I can think of it is to just show directly that if a is not congruent to 2 mod 5 then a^2 is not congruent to 4 mod 5

There are only four cases to go through and you're done. (either that or you've disproven it)

too brain-fried to go for anything shorter atm

 

Edit: There's a trivial counter-example. try some single digit integers.

Edited March 3, 2011 by Schrödinger's hat