# Number Theory Observation

I was doing some number theory problems and while working on a problem I came up with an useful generalization. I know it is not very special but it's worth sharing in my view.

1 year, 2 months ago

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Nice result! However, I think you can greatly improve your writing and formatting; it seems very hard to read and follow, but I would probably chalk that down to your possible lack of experience with mathematical writing.

- 1 year, 2 months ago

I know but I cannot figure out how to use Latex .(I have absolutely 0 coding experience as I had never taken up computer science.) I try to use it but the writing gets all messed up. Maybe someone will type it out on Latex.

- 1 year, 2 months ago

I'm not just talking about LaTeX (I can understand that you may not know how to use it); I found the wording of the proof quite hard to follow. For example, instead of saying lines 2 to 7, you could say the following:

"As $2$ and $a$ are coprime, we would like to find $x \equiv a^n$ (here, the "$\equiv$" sign means "defined to be") such that

$x = 1 \quad (\text{mod } 2)$

and

$x = 0 \quad (\text{mod } a).$

We note that this solution is unique by the Chinese Remainder Theorem."

If you were using LaTeX, you'd also want to take advantage of the Theorem/Proof layout, but even if you don't you would need to separate the theorem from the result. The subsequent six lines could also be rewritten in a similar flavour.

- 1 year, 2 months ago

Okay understood. I will try to better it ................You know after we've kicked our asses in cricket :)

- 1 year, 2 months ago

I hate the Aussie cricket team. National disgraces.

- 1 year, 2 months ago

I prefer rugby myself. I am a huge fan of Dave Pocock and Ben Foley

- 1 year, 2 months ago

Somewhat more directly, it just boils down to $a^2 \equiv a \pmod{2a}$. This follows since $a (a-1) \equiv 0 \pmod{2a}$ as the first term is a multiple of $a$ and the second term is a multiple of 2.

Staff - 1 year, 2 months ago

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