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

[Math100]

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 $24\mid p^{2}-q^{2}$.

 

[PeroK]

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.

 

[PeroK]

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.

 

[PeroK]

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.

 

[Math100]

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?

 

[PeroK]

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.

 

[Math100]

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?

 

[PeroK]

But how should I express them in terms of solving this in proof?

What's the relationship between one odd prime and the next?

 

[pasmith]

I know that $p^{2}-q^{2}=(p-q)(p+q)$. But what can we do with this?

For starters, $p$ and $q$ are odd, so $p + q$ and $p - q$ are even. Now you have 4 as a factor of $p^2 - q^2$.

 

[Math100]

For starters, $p$ and $q$ are odd, so $p + q$ and $p - q$ are even. Now you have 4 as a factor of $p^2 - q^2$.

What do I do from there?

 

[Math100]

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.

 

[PeroK]

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.

 

[Math100]

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.

 

[PeroK]

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 $p = q + 2k$.

 

[Math100]

If $p=2k+q$, then $p^{2}=(2k+q)^{2}=4k^2+4kq+q^2$.

 

[PeroK]

If $p=2k+q$, then $p^{2}=(2k+q)^{2}=4k^2+4kq+q^2$.

You already have $p^2 - q^2 = (p-q)(p+q)$.

 

[Math100]

So $p^{2}-q^{2}=(p-q)(p+q)$
$=(q+2k-q)(q+2k+q)$
$=2k(2k+2q)$
$=4k(k+q)$.

 

[PeroK]

So $p^{2}-q^{2}=(p-q)(p+q)$
$=(q+2k-q)(q+2k+q)$
$=2k(2k+2q)$
$=4k(k+q)$.

Nearly there!

 

[Math100]

But what does $p^{2}-q^{2}=4k(k+q)$ have anything to do with $p^{2}-q^{2}$ being divisible by 24?

 

[PeroK]

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.

 

[Math100]

How to show/prove that $k(k+q)$ is even?

 

[PeroK]

How to show/prove that $k(k+q)$ is even?

If $k$ is even then you're done!

 

[Math100]

If $k$ is even then you're done!

Let $p=q+2k$ where $k=2m$ is an even integer.

 

[PeroK]

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 ...

 

[Math100]

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.

 

[PeroK]

If $k$ is odd, then $k + q$ is even.

 

[Math100]

So that means $p=2k+q$ where $q=2m+1$ for some $k,m\in\mathbb{Z}$.

 

[PeroK]

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:

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.

 

[pasmith]

Alternatively, $$k(k+q) \equiv k(k+1) \mod 2$$ and a product of successive integers is even.

 

[Math100]

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 $24\mid p^2-q^2$.

 

[PeroK]

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.