Prove or disprove: For every integer a, if a is not congruent to 0 (mod 3), the a^2 is congruent to 1 (mod 3)
SO this is for abstract algebra and I am really struggling with this. Here are some of the definitions and theorems I think would apply.
-an integer a divides and integer b if there is an integer q such that b=aq -Let a and b be integers, with a>0. Then there exist unique integers q and r such that b= aq+r and 0<=r
The question says hint: use 2 cases but I'm not really sure what that means. I'm guessing it wants me to look at it when a is congruent to 0 and when it's not? so i tried to do that but just ended up stuck. below is as far as i got with that attempt
Proof: Case 1 There should be some information about letting variables exist in certain number sets here
assume a is not congruent to 0 mod 3, then 3 does not divide a-0, therefore a-0 does not equal 3q for some q in the integers
I was trying to get this to wind down to thus 3 divides a^2-1 but even tho it seems true (I couldn't find a counter example) I couldn't figure out how to prove it with what I have.
Next i tried reversing it thinking I could contradict it or something. I think this would be case 2? assume a is congruent to 0 mod 3, then 3 divides a-0 and a-0=3q for some q in the integers, then 3q=a since 0 is the additive identity, therefore 3 divides a,
and again stuck and my brain hurts a little.. Can anyone help me figure out how to work this.