Prove that if three numbers have no prime factor in common..

1. Jan 29, 2017

Eclair_de_XII

1. The problem statement, all variables and given/known data
"If no prime number $p$ divides a hypothetical solution $(x,y,z)∈ℕ×ℕ×ℕ$ to the equation $x^3+y^3=z^3$, prove that exactly one of x, y and z is even."

2. Relevant equations
Given:
~$∃p:(\frac{x}{p},\frac{y}{p},\frac{z}{p})∈ℕ×ℕ×ℕ$ such that $x^3+y^3=z^3$.
In other words, $(x,y,z)≠(ap,bp,cp)$ for some $a,b,c∈ℕ$

Prove: $x$, $y$, or $z$ can be written as a multiple of $2$.

3. The attempt at a solution
So here I assume that $x, y, z$ do not all have a common prime factor $p$ in common. I also assume that these variables are multiples of certain prime numbers or are prime numbers themselves. Taking these assumptions into mind, I set my variables: $x=ap$, $y=bp$, $z=cp_u$. I assume that $p_u≠p$ is an arbitrary prime number. Now plugging in these values for the equation, I have...

$x^3+y^3=z^3$
$a^3p^3+b^3p^3=c^3p_u^3$
$p^3(a^3+b^3)=p_u^3(c^3)$

Now let $a$ be odd, and $b$ be odd. Then I can rewrite them as $a=2m+1$ and $b=2n+1$. Focusing on the factor of $p^3$, and assuming that it is not $1$...

$(a)^3=(4m^2+4m+1)(2m+1)=(8m^3+8m^2+2m)+(4m^2+4m+1)=8m^3+12m^2+6m+1$
$(b)^3=(4n^2+4n+1)(2n+1)=(8n^3+8n^2+2n)+(4n^2+4n+1)=8n^3+12n^2+6n+1$
$a^3+b^3=8(m^3+n^3)+12(m^2+n^2)+6(m+n)+2=2[4(m^3+n^3)+6(m^2+n^2)+3(m+n)+1]$

Since $[4(m^3+n^3)+6(m^2+n^2)+3(m+n)+1]$ is an integer, then $a^3+b^3$ is even. Moreover, since $p^3$ only has itself and $1$ as a factor, it stands to say that either $p_u^3$ or $c^3$--that is to say, either $p_u$ or $c$ are even. In turn, $z$ is an even integer, when $x$ and $y$ are both odd.

Now I'm a bit ambivalent about submitting this as a solution. Should I provide a case-by-case proof, or is this enough? Should I demonstrate that this does not work when $x$ or $y$ are odd; or perhaps should I make either $x$ or $y$ even, and the remaining of the two odd, with $z$ being odd? Or perhaps I should rework the problem so that only one (or perhaps none) of the three variables is a multiple of the prime number $p$? Maybe if they were all distinct primes...? I don't know how many examples I should give for a case-by-case proof for this problem.

Additionally, I will need to prove that one of these terms is a multiple of $3$ with the same conditions stated above.

Last edited: Jan 29, 2017
2. Jan 29, 2017

PeroK

Hint: look for a simple answer. First, don't worry about the prime divisor condition. What can you say about the oddness/evenness of $x, y, z$?

3. Jan 29, 2017

Eclair_de_XII

I can say that $z$ is even if $x$ and $y$ are of the same parity.

4. Jan 29, 2017

PeroK

Is it possible that $x, y, z$ are all odd?
Is it possible that $x, y, z$ are all even?
Is it possible that two are even and one odd?
Is it possible that one is even and two are odd?
Are there any more options?

5. Jan 29, 2017

Eclair_de_XII

Only if $p_u=2$, I think.

Yes.

I don't think so. If, for example, $x$ were even, then $x^3+y^3=2[4(m^3+n^3)+6n^2+3n]+1$, which is odd.

Yes, as I've just demonstrated.

I don't think so.

6. Jan 29, 2017

Eclair_de_XII

Hold on. If $x$ and $y$ are odd, then $x^3$ and $y^3$ are also odd. So their sum is even and equal to $z^3$. It's basically asking for the parity of $x+y=z$.

$x$ odd and $y$ odd ⇒ $z$ even
$x$ even and $y$ even ⇒ $z$ even
$x$ odd and $y$ even ⇒ $z$ odd
$x$ even and $y$ odd ⇒ $z$ odd

And in the second case, the first condition: ~$∃p:(\frac{x}{p},\frac{y}{p},\frac{z}{p})∈ℕ×ℕ×ℕ$ is violated because now there exists a $p=2$ that makes the second argument irrelevant. And then you have only the three. Now how to prove that one of these variables is also a multiple of 3...

7. Jan 29, 2017

PeroK

You have shown:

For any numbers $a+b = c$ you only have two possibilities. Either they are all even or precisely one of them is even.

Now, if they have no common factors, they can't all be even, leaving just one possibility ...

Note that for this problem you need only the basic logic. No equations are necessary.

8. Jan 29, 2017

Eclair_de_XII

Can I ask, before I go to bed, how I can show that one of the variables is divisible by 3? I tried factoring the expanded polynomial from my original post and rearranging terms, but couldn't express it as a multiple of 3. Thanks.

9. Jan 29, 2017

Eclair_de_XII

In the same vein as before, I tried expressing $z$ as a multiple of 3 by writing the variables as $x=3a+1$ and $y=3b+2$. So $x+y=3(a+b+1)=z$. And I can immediately come up with a counterexample to this; basically, if $x$ and $y$ are of the same parity, it does not work.

10. Jan 29, 2017

PeroK

What are you trying to prove?

11. Jan 29, 2017

Eclair_de_XII

I'm trying to prove the following statement:

"If no prime number $p$ divides a hypothetical solution $(x,y,z)∈ℕ×ℕ×ℕ$ to the equation $x^3+y^3=z^3$, prove that exactly one of $x$, $y$, and $z$ is divisible by $3$."