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

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Discussion Overview

The discussion revolves around the question of whether there exist two positive integers x and y that satisfy the equation x^2 - 4y^2 = 14. Participants explore methods of proof, particularly indirect proofs, and discuss related number theory concepts.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Some participants suggest using an indirect proof to demonstrate that no two positive integers x and y can satisfy the equation x^2 - 4y^2 = 14.
  • There is a proposal to factor the expression x^2 - 4y^2 and consider the integral factorizations of 14 as part of the proof process.
  • Participants discuss the implications of assuming a certain condition (e.g., a does not divide b - c) and how this leads to contradictions when substituting values for b and c.
  • One participant expresses confusion about the notation used in the proofs, specifically regarding the meaning of k in the context of divisibility.
  • Another participant clarifies that k represents arbitrary natural numbers in the context of the proof.
  • There is a back-and-forth exchange where participants refine their understanding of the proof steps and work towards identifying contradictions.

Areas of Agreement / Disagreement

Participants generally agree on the approach of using indirect proofs and factoring, but the specific steps and clarity of the arguments lead to some confusion and require further discussion. The existence of a contradiction is identified, but the overall question of whether two positive integers x and y exist that satisfy the equation remains unresolved.

Contextual Notes

Participants mention the need for clarity in mathematical notation and the importance of understanding the assumptions behind the proofs. The discussion also highlights the complexity of proving statements in number theory.

nano1
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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 di erent, 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!
 
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Re: Extra practice help!

nano said:
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

$\chi$ $\sigma$
 
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?
 
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
 
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?
 
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).
 
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.
 
nano said:
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?
 
It wouldn't be k1a by any chance since it is the contradiction to your statement?
 
  • #10
nano said:
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?
 
  • #11
It would be

k2a - k3a cannot equal k1a
 
  • #12
nano said:
It would be

k2a - k3a cannot equal k1a

Okay, good! :D

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

$$k_1\equiv k_2-k_3$$ ?
 
  • #13
So that it now looks like

k2 - k3 = k2 - k3 ?
 
  • #14
nano said:
So that it now looks like

k2 - k3 = k2 - k3 ?

No, it would be:

$$a\left(k_2-k_3 \right)\ne a\left(k_2-k_3 \right)$$
 
  • #15
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?
 
  • #16
nano said:
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.
 

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