If ## p\geq q\geq 5 ## and ## p ## and ## q ## are both primes ....

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In summary, if ## p\geq q\geq 5## and ## p ## and ## q ## are both prime, then ## 24\mid p^{2}-q^{2} ##.
  • #1
Math100
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Homework Statement
If ## p\geq q\geq 5 ## and ## p ## and ## q ## are both primes, prove that ## 24\mid p^{2}-q^{2} ##.
Relevant Equations
None.
Proof:

Suppose ## p\geq q\geq 5 ## and ## p ## and ## q ## are both primes.
Note that ## p ## and ## q ## are not divisible by ## 3 ##,
so we have ## p^{2}-1\equiv 0 (mod 3) ## and ## q^{2}-1\equiv 0 (mod 3) ##.
This means ## 3\mid((p^{2}-1)-(q^{2}-1)) ##,
and so ## 3\mid p^{2}-q^{2} ##.
Since ## p^{2}-q^{2} ## is divisible by 3, it follows that ## p^{2}-q^{2} ## is also divisible by 24.
Thus, ## 24\mid p^{2}-q^{2} ##.
Therefore, if ## p\geq q\geq 5 ## and ## p ## and ## q ## are both primes,
then ## 24\mid p^{2}-q^{2} ##.
 
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  • #2
Math100 said:
Since ## p^{2}-q^{2} ## is divisible by 3, it follows that ## p^{2}-q^{2} ## is also divisible by 24.
I'd like to see some justification for that! If a number is divisible by ##24##, then it is divisible by ##3##. But not the converse.
 
  • #3
This must be the third or fourth problem now where you have failed to factorise a difference of squares. It should be completely automatic by now that ##x^2 - y^2 = (x-y)(x+y)##. Especially in the context of factorisation problems, this should be the first thing you think about.
 
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  • #4
The other thing I suspect you never do is write down some examples first to see what's going on. The first thing I did was write down ##25, 49, 121, 169, 289##. Just looking at those numbers gave me a big clue on how to solve this.
 
  • #5
PeroK said:
This must be the third or fourth problem now where you have failed to factorise a difference of squares. It should be completely automatic by now that ##x^2 - y^2 = (x-y)(x+y)##. Especially in the context of factorisation problems, this should be the first thing you think about.
I know that ## p^{2}-q^{2}=(p-q)(p+q) ##. But what can we do with this?
 
  • #6
Math100 said:
I know that ## p^{2}-q^{2}=(p-q)(p+q) ##. But what can we do with this?
Looks for factors of ##2##. There must be a factor of ##2^3## there somewhere!

Also, see post #4.
 
  • #7
PeroK said:
The other thing I suspect you never do is write down some examples first to see what's going on. The first thing I did was write down ##25, 49, 121, 169, 289##. Just looking at those numbers gave me a big clue on how to solve this.
I know that ## p^{2}-q^{2}=49-25=24 ##,
also ## p^{2}-q^{2}=121-49=72 ##,
and all of these results are divisible by 24.
But how should I express them in terms of solving this in proof?
 
  • #8
Math100 said:
But how should I express them in terms of solving this in proof?
What's the relationship between one odd prime and the next?
 
  • #9
Math100 said:
I know that ## p^{2}-q^{2}=(p-q)(p+q) ##. But what can we do with this?

For starters, [itex]p[/itex] and [itex]q[/itex] are odd, so [itex]p + q[/itex] and [itex]p - q[/itex] are even. Now you have 4 as a factor of [itex]p^2 - q^2[/itex].
 
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  • #10
pasmith said:
For starters, [itex]p[/itex] and [itex]q[/itex] are odd, so [itex]p + q[/itex] and [itex]p - q[/itex] are even. Now you have 4 as a factor of [itex]p^2 - q^2[/itex].
What do I do from there?
 
  • #11
Since ## p ## and ## q ## are odd, it follows that ## p+q ## and ## p-q ## are even.
Now we have ## p+q=2m ## and ## p-q=2n ## for ## m, n\in\mathbb{Z} ##.
Thus ## p^{2}-q^{2}=(p+q)(p-q) ##
=(2m)(2n)
=4mn
=4k,
where k=mn is an integer.
 
  • #12
Math100 said:
Since ## p ## and ## q ## are odd, it follows that ## p+q ## and ## p-q ## are even.
Now we have ## p+q=2m ## and ## p-q=2n ## for ## m, n\in\mathbb{Z} ##.
You have that already. You need an additional idea. See post #8.
 
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  • #13
PeroK said:
What's the relationship between one odd prime and the next?
Odd primes: ## 5, 7, 11, 13, 17, 19, 23... ##.
Their differences are ## 2, 4, 2, 4, 2, 4... ##.
But note that ## 25 ## isn't a prime, and ## 27 ## isn't a prime either.
 
  • #14
Math100 said:
Odd primes: ## 5, 7, 11, 13, 17, 19, 23... ##.
Their differences are ## 2, 4, 2, 4, 2, 4... ##.
But note that ## 25 ## isn't a prime, and ## 27 ## isn't a prime either.
So, perhaps, let ##p = q + 2k##.
 
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  • #15
If ## p=2k+q ##, then ## p^{2}=(2k+q)^{2}=4k^2+4kq+q^2 ##.
 
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  • #16
Math100 said:
If ## p=2k+q ##, then ## p^{2}=(2k+q)^{2}=4k^2+4kq+q^2 ##.
You already have ##p^2 - q^2 = (p-q)(p+q)##.
 
  • #17
So ## p^{2}-q^{2}=(p-q)(p+q) ##
## =(q+2k-q)(q+2k+q) ##
## =2k(2k+2q) ##
## =4k(k+q) ##.
 
  • #18
Math100 said:
So ## p^{2}-q^{2}=(p-q)(p+q) ##
## =(q+2k-q)(q+2k+q) ##
## =2k(2k+2q) ##
## =4k(k+q) ##.
Nearly there!
 
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  • #19
But what does ## p^{2}-q^{2}=4k(k+q) ## have anything to do with ## p^{2}-q^{2} ## being divisible by 24?
 
  • #20
Math100 said:
But what does ## p^{2}-q^{2}=4k(k+q) ## have anything to do with ## p^{2}-q^{2} ## being divisible by 24?
You only have to show that ##k(k + q)## is even and that gives you the factor of ##8##. The factor of ##3## you have already found.
 
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  • #21
How to show/prove that ## k(k+q) ## is even?
 
  • #22
Math100 said:
How to show/prove that ## k(k+q) ## is even?
If ##k## is even then you're done!
 
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  • #23
PeroK said:
If ##k## is even then you're done!
Let ## p=q+2k ## where ## k=2m ## is an even integer.
 
  • #24
Math100 said:
Let ## p=q+2k ## where ## k=2m ## is an even integer.
No, I mean if ##k## is even the ##k(k + q)## is even. At this level that should be "obvious", in the sense that it doesn't need to be justified further.

And, if ##k## is odd ...
 
  • #25
PeroK said:
No, I mean if ##k## is even the ##k(k + q)## is even. At this level that should be "obvious", in the sense that it doesn't need to be justified further.

And, if ##k## is odd ...
I found out that if ## k ## is odd, then ## k=2m+1 ## for some ## m\in\mathbb{Z} ##.
Thus ## k(k+q)=(2m+1)(2m+1+q) ##
## =4m^2+2m+2mq+2m+1+q ##
## =4m^2+4m+2mq+q+1 ##,
but ## k(k+q) ## is not odd.
 
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  • #26
If ##k## is odd, then ##k + q## is even.
 
  • #27
So that means ## p=2k+q ## where ## q=2m+1 ## for some ## k,m\in\mathbb{Z} ##.
 
  • #28
Math100 said:
So that means ## p=2k+q ## where ## q=2m+1 ## for some ## k,m\in\mathbb{Z} ##.
This is not relevant now.

To get back on track we reached this stage:
Math100 said:
But what does ## p^{2}-q^{2}=4k(k+q) ## have anything to do with ## p^{2}-q^{2} ## being divisible by 24?
We need to show that ##k(k + q)## is even, because then ##p^2 - q^2## is a multiple of ##8##. Note that the product of two numbers is even if either number is even. If ##k## is even, then we are done. And, if ##k## is odd, then ##k + q## is even. Either way, ##p^2 - q^2## is a multiple of ##8##.

That should finish the proof.
 
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  • #29
Alternatively, [tex]k(k+q) \equiv k(k+1) \mod 2[/tex] and a product of successive integers is even.
 
  • #30
Okay, so here's my revised proof:

Suppose ## p\geq q\geq 5 ## and ## p ## and ## q ## are both primes.
Note that ## p ## and ## q ## are not divisible by ## 3 ##,
so we have ## p^{2}-1\equiv 0 (mod 3) ## and ## q^{2}-1\equiv 0 (mod 3) ##.
This means ## 3\mid ((p^{2}-1)-(q^{2}-1)) ##, and so ## 3\mid p^{2}-q^{2} ##.
Let ## p=2k+q ## where ## q=2m+1 ## for some ## k, m\in\mathbb{Z} ##.
Then we have ## p^{2}-q^{2}=(2k+q)^{2}-q^{2} ##
## =4k^2+4kq+q^2-q^2 ##
## =4k^2+4kq ##
## =4k(k+q) ##.
Now we consider two cases.
Case #1: Suppose ## k ## is an odd integer.
Then we have ## k=2n+1 ## for some ## n\in\mathbb{Z} ##.
Thus ## k(k+q)=(2n+1)(2n+1+2m+1) ##
## =(2n+1)(2n+2m+2) ##
## =4n^2+4mn+4n+2n+2m+2 ##
## =2(2n^2+2mn+3n+m+1) ##
## =2t ##,
where ## t=2n^2+2mn+3n+m+1 ## is an integer.
Case #2: Suppose ## k ## is an even integer.
Then we have ## k=2n ## for some ## n\in\mathbb{Z} ##.
Thus ## k(k+q)=2n(2n+2m+1) ##
## =2(2n^2+2mn+n) ##
## =2s ##,
where ## s=2n^2+2mn+n ## is an integer.
Since ## k(k+q) ## is even in both cases,
it follows that ## 8\mid p^2-q^2 ##, so ## 24\mid p^2-q^2 ##.
Therefore, if ## p\geq q\geq 5 ## and ## p ## and ## q ## are both primes,
then ## 24\mid p^2-q^2 ##.
 
  • #31
That's far too long. You only need ##p = q + 2k##. You don't need the rest of those variables. To do something rigorously you don't need to write out every detail.

You seem so focused on the details that you forget the ideas involved and what is important.
 
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Related to If ## p\geq q\geq 5 ## and ## p ## and ## q ## are both primes ....

1. What does the notation "p ≥ q ≥ 5" mean?

The notation "p ≥ q ≥ 5" means that p is greater than or equal to q, and q is greater than or equal to 5. In other words, p and q are both at least 5, and p is equal to or larger than q.

2. How do we know that p and q are both prime numbers?

Since the statement says "p and q are both primes," we can assume that they are prime numbers. A prime number is a positive integer that is only divisible by 1 and itself. Therefore, p and q must both be positive integers and cannot be divided by any other number except 1 and itself.

3. Can p and q be any prime numbers, or are there restrictions?

As long as p and q satisfy the condition of p ≥ q ≥ 5, they can be any prime numbers. However, they must both be greater than or equal to 5 in order for the statement to be true.

4. Is this statement always true for any two prime numbers that satisfy the condition?

Yes, this statement is always true for any two prime numbers that satisfy the condition of p ≥ q ≥ 5. This is because prime numbers can only be divided by 1 and itself, so if p is greater than or equal to q, it cannot be divided by q. And since q is greater than or equal to 5, it cannot be divided by any number less than 5. Therefore, p and q must both be prime numbers.

5. What implications does this statement have in mathematics?

This statement has implications in number theory, as it provides a condition for two prime numbers to satisfy in order for a certain relationship to hold true. It also highlights the properties of prime numbers, such as their inability to be divided by any number except 1 and itself.

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