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

In summary: This won't work, because $k_1e k_2-k_3$.In summary, the theorem is true for all integers a, b, and c if a does not divide b - c, but does not divide b or c.
  • #1
nano1
8
0
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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  • #2
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

$\chi$ $\sigma$
 
  • #3
Re: Extra practice help!

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

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

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:

\(\displaystyle a=k_2b\)

\(\displaystyle a=k_3c\)

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?
 
  • #4
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
 
  • #5
Re: Extra practice help!

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

\(\displaystyle b=ka\)

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:

\(\displaystyle b-c\ne k_1a\)

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:

\(\displaystyle b=k_2a\)

\(\displaystyle c=k_3a\)

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

Can you show this leads to a contradiction?
 
  • #6
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).
 
  • #7
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.
 
  • #8
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:

\(\displaystyle b-c\ne k_1a\)

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

\(\displaystyle b=k_2a\)

\(\displaystyle c=k_3a\)

What do you get?
 
  • #9
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:

\(\displaystyle k_1\equiv k_2-k_3\) ?
 
  • #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:

\(\displaystyle a\left(k_2-k_3 \right)\ne a\left(k_2-k_3 \right)\)
 
  • #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.
 

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

What are divisibility problems?

Divisibility problems are mathematical questions that involve determining if a number can be divided evenly by another number without leaving a remainder. For example, 12 is divisible by 3 because it can be divided into 3 equal groups of 4, with no remainder.

How do you solve divisibility problems?

In order to solve a divisibility problem, you need to know the divisibility rules for different numbers. For example, a number is divisible by 2 if it is even, and a number is divisible by 3 if the sum of its digits is divisible by 3. By applying these rules, you can determine if a number is divisible by another number.

Why are divisibility problems important?

Divisibility problems are important because they help us understand the fundamental principles of division and prime numbers. They also have practical applications in fields such as cryptography, where divisibility rules are used to determine if a large number is prime.

What are some common divisibility problems?

Some common divisibility problems include finding if a number is divisible by 2, 3, 4, 5, 6, 9, or 10. These numbers have simple divisibility rules that make them easier to solve. Other divisibility problems involve finding if a number is divisible by a combination of these numbers, such as finding if a number is divisible by both 3 and 5.

How can you check if a number is divisible by another number?

There are several methods for checking if a number is divisible by another number. One way is to use long division to see if the remainder is 0. Another way is to use the divisibility rules for that number. You can also use a calculator to divide the numbers and see if the result is a whole number.

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