Discrete Math: For a^2+b^2=c^2, prove a, b, or c is even

[Migz]

I have a couple of questions that are on a test I have tomorrow.

1. Prove that for all integers a, b, and c, a^2 + b^2 = c^2 implies that at least one of a, b, or c is even.

2. Prove that for any integers a, b and c if a divides b, but not c^2, then a does not divide (b + c)(b ? c).

If someone could help with solutions to these proofs it would be much appreciated.
Thanks!

 

[mmm4444bot]

... If someone could help with solutions to these proofs it would be much appreciated.

Hello Mig Z:

How nice for you to get the questions before the test tomorrow. How long have you had them? (This is a rhetorical question.)

What kind of help are you expecting to receive at this site? You have not made a single statement, yet, to let us know what it is that you need.

This place is neither a classroom nor an answer service. In general, we need you to ask questions about what you do not understand OR to show some work and reasoning so that we can guess.

(Please read the post titled, "Read Before Posting", if you have not already done so.)

I'm not going to write a proof for you. If this is what you are expecting, then you will need to wait for someone else. Writing proofs is somewhat like telling a story; there are different ways to get to the end of the same story, and some ways are longer than others.

Here are some things for you to think about.

To start on the first one exercise, can you define even and odd numbers?

Make a list of some even integers squared.

Make a list of some odd integers squared.

What do you notice about the squares in one list compared to the other?

Are you familiar with the expressions 2n and 2n-1 (where n is any whole number) to represent even and odd integers, respectively?

In other words, two times any whole number is even.

Using this notation, the expression (2n - 1)^2 is the square of an odd integer.

Can you write it as 2 times some integer + 1? If you do, then you show that the square of an odd integer is odd.

If the symbol A is some arbitrary whole number, and the symbol B is another, then together the expressions 2A-1 and 2B-1 represent two arbitrary odd integers.

Can you write the expression (2A-1)^2 + (2B-1)^2 factored as 2 times an integer? If you do, then you show that the sum of the squares of two arbitrary odd integers is an even number.

Can you understand how this information shows that there is no way that c can be odd when both a and b are odd? (Note, here I mean a, b, and c, as given in your exercise. I do not mean A and B, as given in my example.)

This is the sort of reasoning one must do when writing a proof. You need to come up with some way of accounting for all possibilites. What if b and c are odd -- can a also be odd?

If none of this makes any sense to you, then you may want to spend your remaining time concentrating on the other questions on your test.

There is no simple explanation for how to write proofs. Please feel free to ask any specific questions.

Cheers,

~ Howard I. Noe

 

[PAULK]

I have a couple of questions that are on a test I have tomorrow.

1. Prove that for all integers a, b, and c, a^2 + b^2 = c^2 implies that at least one of a, b, or c is even.

2. Prove that for any integers a, b and c if a divides b, but not c^2, then a does not divide (b + c)(b ? c).

If someone could help with solutions to these proofs it would be much appreciated.
Thanks!

Here is an idea for you:

If x divides two integers, then it divides their difference.

 

[Denis]

1. Prove that for all integers a, b, and c, a^2 + b^2 = c^2 implies that at least one of a, b, or c is even.

a = 2x +1 , b = 2y + 1, c = 2z +1
(2x + 1)^2 + (2y + 1)^2 = (2z+ 1)^2
Expand:
4(z^2 + z - x^2 - x - y^2 - y) = 1

Not enough? See your teacher...

 

[mmm4444bot]

a = 2x +1 , b = 2y + 1, c = 2z +1

(2x + 1)^2 + (2y + 1)^2 = (2z+ 1)^2

Expand:

4(z^2 + z - x^2 - x - y^2 - y) = 1

Hey Migz:

Denis shows, with experience, how to write a short story.

Good luck on your test.