0
$\begingroup$

Lemma: Let $a,b,x, y \in \Bbb{Z}$. Then $ax^2 = z= by^2 \implies z = \pm \operatorname{lcm}(a,b) \gcd(x,y)^2$.

Proof. If $a,b = \pm 1$ then $z = \pm \operatorname{lcm}(\pm 1, \pm 1) \gcd(\pm x, \pm y)^2 = \pm x^2 = \pm y^2$.

Now assume that it's true for $ab = \pm$ some composition of the first $n\geq 0$ primes. And introduce the next prime $p$. Let $a = p^k u$, $\ b = p^r v$ where $p \nmid u,v$ for some $k, r \geq 0$.

Then $z = p^ku x^2 = p^r v y^2$. Without loss of generality assume that $r \geq k$. Then $\dfrac{z}{p^k} = u x^2 = vp^{r-k} y^2$.

By induction, we have that $\dfrac{z}{p^k} = \pm \operatorname{lcm}(a, vp^{r - k}) \gcd(x,y)^2$. So that $z = \pm \operatorname{lcm}(u, vp^{r - k}) p^k \gcd(x,y)^2 = \pm \operatorname{lcm}(up^k, vp^r) \gcd (x,y)^2$.

Can you verify this proof for me?


If it's true, then that helps prove that $U_a = \{ a x^2 - 1 : x \in \Bbb{Z}\}$ forms a basis for a topology on $\Bbb{Z}$.

$\endgroup$
  • 1
    $\begingroup$ With \text{lcm} rather than \operatorname{lcm} you don't automatically get proper spacing in things like $3\operatorname{lcm}(a,b).$ That's why you saw $\pm\text{lcm}$ instead of $\pm\operatorname{lcm},$ and likewise you would see $3\text{lcm}$ instead of $3\operatorname{lcm}. \qquad$ $\endgroup$ – Michael Hardy Mar 28 '18 at 22:33
1
$\begingroup$

Your inductive step doesn't work as written: if you set $z' = z/p^k, a' = u, b' = vp^{r-k}$, then you do indeed have $z' = a'x^2 = b'y^2$, but you don't necessarily know that $a'b'$ doesn't contain a factor of $p$ (because you might have $r > k$).

But your proof can probably be easily modified to show that, in fact, you must have $r=k$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.