proof

**a.katerina** · Mar26-09, 03:33 PM

Hi,

i have just registered to the forum, because this time i study number theory and in some problems i can't figure out how to solve them.

This time i have to prove: If two integers x,y doesn't divided with 3 then the (x^2 - y^2) always is divided with 3.

Does any one has a clue how to start?

Thank you!

**Office_Shredder** · Mar26-09, 03:37 PM

Break it up into what x and y modulo 3 can be (so four cases)

**a.katerina** · Mar26-09, 03:51 PM

what?? Can you explain it a little bit, please?

**HallsofIvy** · Mar26-09, 06:07 PM

If x is not divisible by 3 then it can be written as x= 3k+1 or 3k+2 for some integer k.
If y is not divisible by 3 then it can be written as y= 3j+1 or 3j+2 for some integer k.

That gives 4 cases to consider:
1) x= 3k+1 and y= 3j+1.
2) x= 3k+1 and y= 3j+2.
3) x= 3k+2 and y= 3j+1.
4) x= 3k+2 and y= 3j+2.

Calculate

for each of those cases.

**ramsey2879** · Mar26-09, 08:36 PM

Originally Posted by HallsofIvy

If x is not divisible by 3 then it can be written as x= 3k+1 or 3k+2 for some integer k.
If y is not divisible by 3 then it can be written as y= 3j+1 or 3j+2 for some integer k.

That gives 4 cases to consider:
1) x= 3k+1 and y= 3j+1.
2) x= 3k+1 and y= 3j+2.
3) x= 3k+2 and y= 3j+1.
4) x= 3k+2 and y= 3j+2.

Calculate for each of those cases.

Since +/- 0 = 0 are not cases 2 and 3 equivalent?

Since 2 = -1 and -1*-1 = 1 , (-1)^2 = (+1)^2 are not cases 1 and 4 equivalent?

Even better you can rewrite each term of an expression mod n before evaluating the expression mod n. Thus each of the terms can be rewriten mod 3 by substituting 0 for 3n and 1 for -1^2 (or 2^2) to get the equivalent expressions 1-1 = 0 mod 3 which is clearly correct.

**a.katerina** · Apr1-09, 07:16 PM

Thank you very much for the help!! I'm grateful to you!