5
$\begingroup$

Can someone check on this? If it is wrong please refrain from telling me the answer.

Suppose $(6,a) = (6,b) = 1$. We wish to prove $24 \mid (a^2-b^2)$.

We can see that $(4,a) = (4,b) = 1$ since $2 \mid 4$ and $2 \mid 6$, so, $(24, a) = (24, b) = 1$. By Bezouts identity, any integer can be represented by a linear combination of two integers that are relatively prime, so let $a^2 - bx_2 = ax_1 + 24y_1$ and $b^2 - ax_1 = bx_2 + 24y_2$ for some $x_1,y_1,x_2,y_2 \in \mathbb{Z}$. Then $a^2 - b^2 = bx_2 - bx_2 + ax_1 - ax_2 + 24y_1 - 24y_2$. So $a^2 - b^2 = 24(y_1 - y_2)$. So $24 \mid (a^2 - b^2)$. $\square$

$\endgroup$
  • 2
    $\begingroup$ You have a slight problem: you never say who $x_2$ is for your first expression (that is, you are trying to express $a^2-bx_2$ as a linear combination of $a$ and $24$... but you don't say who $x_2$ is); it can't be the one from the second expression, because in order to calculate the second expression (when you are using $b$ and $24$) you need to know who $x_1$ is; but in order to now who $x_1$ is, you need to know who $x_2$ is. But in order to know who $x_2$ is, you need to know who $x_1$ is. But... $\endgroup$ – Arturo Magidin Sep 6 '19 at 20:12
  • $\begingroup$ Make a list of a few $x$s for which $(6,x)=1$ Check and see if they all work for your hypothesis. I think this is a great question. $\endgroup$ – steven gregory Sep 6 '19 at 20:17
  • $\begingroup$ @stevengregory: Given that there are infinitely many such $x$s, that's going to take a while... $\endgroup$ – Arturo Magidin Sep 6 '19 at 20:17
  • $\begingroup$ @stevengregory I don't recall seeing "a few" before it was edited.... But then, how will checking a few examples prove that something always holds? $\endgroup$ – Arturo Magidin Sep 6 '19 at 20:20
  • $\begingroup$ 5,1 25-1 = 24 and 24 divides 24. 5,7 25 - 49 = -24 and 24 divides -24... $\endgroup$ – John Mancini Sep 6 '19 at 20:21
1
$\begingroup$

I believe you have a chicken-and-egg problem here.

When you say $a^2-bx_2=ax_1+24y_1$, you are saying that, for every given $x_2$, you can find corresponding $x_1, y_1$. Then you say $b^2-ax_1=bx_2+24y_2$, and this probably means that, for every $x_1$ (including the one we just found!), there exist corresponding $x_2, y_2$.

What does not follow is that $x_2$, found in the second step (using $x_1$ found in the first step from the initially chosen $x_2$), matches the initially chosen $x_2$. (I mean, you do have a freedom to choose $x_2$ to start with, but what if none of those $x_2$'s end up winding back to themselves???)

In fact, I wouldn't even use the same symbol $x_2$ for two different things! But then, from that point on, your proof doesn't work, because it substantially relies on $x_2$ being one and the same thing.

$\endgroup$
  • $\begingroup$ Ah I suspected that was faulty. Hmm... I am not quite sure how to proceed from here. Thank you for the help! $\endgroup$ – John Mancini Sep 6 '19 at 20:18
  • $\begingroup$ @JohnMancini You are welcome, wish you good luck in trying to figure it out! $\endgroup$ – Stinking Bishop Sep 6 '19 at 20:21
1
$\begingroup$

Hints.

  1. The numbers are all of the form $6n \pm 1$

  2. If $n$ is odd, then $3n \pm 1$ is even.

  3. $(6n \pm 1)^2 - 1$ is a multiple of 24.

  4. $(6a \pm 1)^2 - (6b \pm 1)^2 = [(6a \pm 1)^2 - 1] - [(6b \pm 1)^2 - 1]$.

$\endgroup$

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.