MHB Can there be two positive integers x and y that satisfy x^2 - 4y^2 = 14?

[nano1]

Hey, I've been stuck on these questions for awhile. They're bonus/ extra practice questions and I have a midterm coming up and I'm not quite comfortable with the process. If anyone can help me that'd be great!

Prove the following theorem: for all integers a, b and c, if a does not divide b - c
then a does not divide b or a does not divide c. Hint: an indirect proof would work
well.

Prove that there do not exist two positive integers x and y such that x^2 - 4y^2 = 14.
Additional information: you have all known for a long time that there are many
solutions to the equation x^2 + y^2 = z^2, where x, y and z are all positive integers.
This problem considers a slightly dierent, but similar looking, type of equations.
Hint: use an indirect proof, and start by factoring x^2 - 4y^2.I don't fully understand how to write a proper proof. Thanks for anyone's help!

 

[chisigma]

Re: Extra practice help!

Hey, I've been stuck on these questions for awhile. They're bonus/ extra practice questions and I have a midterm coming up and I'm not quite comfortable with the process. If anyone can help me that'd be great!

Prove the following theorem: for all integers a, b and c, if a does not divide b - c
then a does not divide b or a does not divide c. Hint: an indirect proof would work
well.

Wellcome on MHB nano!...

The first theorem can be proved as follows... let's suppose that a|b and a|c with $b \ne c$, so that $b = \beta\ a$ and $c = \gamma\ a$. In such case is $b - c = (\beta - \gamma)\ a\ \implies a|(b-c)$ which constrasts the hypothesis, so that it must be false or a|b or a|c...

Kind regards

$\chi$ $\sigma$

 

[MarkFL]

Re: Extra practice help!

For the first problem, I would first assume that $a$ does not divide $b-c$, hence:

$$a\ne k_1(b-c)$$

Then I would assume the complement of "$a$ does not divide $b$ or $a$ does not divide $c$" is true, i.e $a$ does divide $b$ AND $a$ does divide $c$ is true:

$$a=k_2b$$

$$a=k_3c$$

where $k_i\in\mathbb{N}$.

Can you show this leads to a contradiction?

For the second problem, I would factor as suggested, and consider the different integral factorizations of 14. What do you find?

 

[nano1]

Re: Extra practice help!

I don't quite understand what the k is supposed to mean in your proof. Am I supposed to pick some number b and c and multiply it by a different number k? Sorry, math easily confuses me

 

[MarkFL]

Re: Extra practice help!

The parameters $k_i$ represent arbitrary natural numbers. If $a|b$ then we may write:

$$b=ka$$

I apologize, as I wrote things backwards in my post above, I should have written (as chisigma did):

For the first problem, I would first assume that $a$ does not divide $b-c$, hence:

$$b-c\ne k_1a$$

Then I would assume the complement of "$a$ does not divide $b$ or $a$ does not divide $c$" is true, i.e $a$ does divide $b$ AND $a$ does divide $c$ is true:

$$b=k_2a$$

$$c=k_3a$$

where $k_i\in\mathbb{N}$.

Can you show this leads to a contradiction?

 

[Ackbach]

Re: Extra practice help!

Incidentally, these questions are more number theory than discrete math. I will move your thread (and also retitle to be more specific).

 

[nano1]

I think I've figured out the second question I posted at least but the contradiction that I'm trying to find for the one you're helping me with still is stumping me. I apologize, if I'm being irritating since you've already restated the steps twice.

 

[MarkFL]

I think I've figured out the second question I posted at least but the contradiction that I'm trying to find for the one you're helping me with still is stumping me. I apologize, if I'm being irritating since you've already restated the steps twice.

You are not being irritating, especially given that I fouled up the first statement of the steps I gave. :D

In the statement:

$$b-c\ne k_1a$$

Substitute for $b$ and $c$ using the statements:

$$b=k_2a$$

$$c=k_3a$$

What do you get?

 

[nano1]

It wouldn't be k₁a by any chance since it is the contradiction to your statement?

 

[MarkFL]

It wouldn't be k₁a by any chance since it is the contradiction to your statement?

Let's take it one step at a time. What do you get when you make the substitutions I suggested?

 

[nano1]

It would be

k₂a - k₃a cannot equal k₁a

 

[MarkFL]

It would be

k₂a - k₃a cannot equal k₁a

Okay, good! :D

Now factor the left side. What happens if we define:

$$k_1\equiv k_2-k_3$$ ?

 

[nano1]

So that it now looks like

k₂ - k₃ = k₂ - k₃ ?

 

[MarkFL]

So that it now looks like

k₂ - k₃ = k₂ - k₃ ?

No, it would be:

$$a\left(k_2-k_3 \right)\ne a\left(k_2-k_3 \right)$$

 

[nano1]

Oh forgot about including the a. Well, that looks like a contradiction right? Since both sides are identical yet it says it doesn't equal each other?

 

[MarkFL]

Oh forgot about including the a. Well, that looks like a contradiction right? Since both sides are identical yet it says it doesn't equal each other?

Good, yes this is the contradiction for which we were looking. So we now know we must have:

$a$ does not divide $b$ or $a$ does not divide $c$

And this is what we were looking to prove.