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 $24 \mid (a^2 - b^2)$. $\square$

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    $\begingroup$ 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

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.

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


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