Expert: Paul Klarreich - 2/14/2011

Question
Hello, I'm really stuck on these 2 proof problems. Can you show me how to do them? ANY help on ANY part of these is greatly appreciated!!

1. Prove: Let a and b be integers, a ≠ 0 and b ≠ 0. If a |b and b| a then a = ± b

2. Prove Let a and b be integers. Prove that a²+2b²=0 (mod 3), then either a and b are both congruent to 0 modulo 3 or neither is congruent to 0 modulo 3.

Thank you,
Blake

Answer
Questioner: Blake
Category: Advanced Math
Private: No
Subject: Math 3034: Mathematical Proofs
Question: Hello, I'm really stuck on these 2 proof problems. Can you show me how to do them? ANY help on ANY part of these is greatly appreciated!!

1. Prove: Let a and b be integers, a ≠ 0 and b ≠ 0. If a |b and b| a then a = ± b

2. Prove Let a and b be integers. Prove that a²+2b²=0 (mod 3), then either a and b are both congruent to 0 modulo 3 or neither is congruent to 0 modulo 3.

Thank you,
Blake
................................................
I like your plus-or-minus signs.  Some day I will learn to make them.

1. Prove: Let a and b be integers, a ≠ 0 and b ≠ 0. If a |b and b| a then a = ± b

At least 90% of the time, a proof can be handled by just looking at the definition of a term, such as:

Definition:  a | b  means there is an integer  n such that

b = na

And b | a means there is  m such that  a = mb

So  a = mb = m(na) = mn a

But if a = mn a, then mn = 1, BECAUSE  a /= 0, so m,n are divisors of 1.

The only divisors of 1 are +- 1. (Oh, well, some day, and the 'not-equals' thing, too.)

So m = +- 1 and  a = +- b.

......................

2. Let a and b be integers. Prove that (IF?) a²+2b²=0 (mod 3), then either a and b are both congruent to 0 modulo 3 or neither is congruent to 0 modulo 3.

I am going to write  == to mean '= .. (mod 3) to save typing.

Case:  Suppose a == 0.

Then a² == 0.  [Need proof?]

Could we have  b == +-1 ?  Then b² == 1.

And then 2b² == 2  and  a² + 2b² == 0 + 2 = 2

So we cannot have b == +- 1, therefore  b == 0.

Case:  Suppose  b == 0.

Then  b² == 0  and  2b² == 0

So  a² == 0.

Could we have  a == +- 1 ?  Then a² == 1

and a² + 2b² == 1 + 0 = 0

Now what about your 'neither' part?  We just proved that if one of a,b is == 0, so is the other.  You can supply the rest, I think.