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