For every integers $n$ and $k$, prove that if $k\mid(2n+1)$ and $k\mid(4n^2+1)$ then $k \in \{-1,1\}$

Question

I thought maybe I could prove this with $k|(2n+1)$ and $k|(4n^2+1)$ so then $k|(4n^2+1-2n-1)=4n^2-2n$ which can be factored as $2n(2n-1)$ which means k has to be the GCD(2n, 2n-1)? I don't think that's true because it can just divide $2n$ or $2n-1$ and still be true but if it were true then the GCD of an odd and even is obviously 1 or -1 and it'd be solved.

Answer 1

If $k$ divides $2n +1$ then $k$ also divides $(2n + 1)(2n - 1) = 4n^2 - 1$.

Since $2n+1$ is odd, so is $k$. I'm sure you know what to do next.

Answer 2

Similar to answer by WhatsUp, $k\mid (2n+1) \Rightarrow k\mid (2n+1)^2 \Rightarrow k\mid (4n^2+4n+1)$

Therefore $k\mid ((4n^2+4n+1)-(4n^2+1)) \Rightarrow k\mid 4n$

Therefore $k\mid 2(2n+1)-(4n) \Rightarrow k\mid 2$

Since $k$ is odd, $k=\pm 1$

Answer 3

Put $\ a=\color{#c00}{2n}\ $ in $\,\overbrace{\gcd(\color{#90f}{a+1},\,\color{#0a0}{a^2\!+\!1}) = \underbrace{\gcd(a+1,\,\color{#0a0}2)}_{\textstyle = 1\ \ {\rm if}\ \ \color{#c00}{2\mid a}\ }}^{\!\!\!\!\!\!\textstyle \color{#90f}{a\equiv -1}\ \Rightarrow\ \color{#0a0}{a^2\!+\!1\,\equiv\, 2_{\phantom |}}}\,\ $ by Euclid

Answer 4

Hint: Take WhatsUp's lead. If $k$| $ 4n^2 + 1$ and $k$| $4n^2 - 1$, while $k$ is odd, what can we conclude? Just reproduce exactly what you did in the first line of your post and you're there :).

Answer 5

? I don't think that's true because it can just divide 2n or 2n−1 and still be true

But can there be any number that divides both $2n$ and $2n-1$.

but if it were true then the GCD of an odd and even is obviously 1 or -1 and it'd be solved.

That'd be true if the even number was exactly one more than the odd number. And it'd be true if the odd number was exactly one more than the even number.

But if all you have is that the even number could be any even number and the odd number could be any odd number possibly quite far away from the even number you can't conclude anything.

..........

This all boils down to: If $a|M$ and $a|N$ then $a|M \pm N$.

So if $k|2n+1$ and $k|4n^2+1$ then $k|4n^2 + 1-(2n+1)$ so $k|4n^2-2n$ or $k|2n(2n-1)$

This, by itself, does not mean $k$ is the $\gcd(2n,2n-1)$ but a few things are worth noting.

1) $2n+1$ is odd. So if any prime $p|k$ than $p|2n+1$ so $p\ne 2$ so $k$ is odd

2) If any prime $p|2n+1$ and if any prime $p|2n$ then $p|(2n+1)-2n$ and $p|1$ but that's not possible if $p>1$. So if $k|2n+1$ then $k$ has no factors in common with $2n$.

3) So if $k$ and $2n$ have no factors in common, and $k|2n(2n-1)$ that means $k|2n-1$.

4) So $k|2n+1$ and $k|2n-1$ so $k|(2n+1)-(2n+1)$ so $k|2$.

.....

But the only integers that divide $2$ are $1,-1, 2$ and $-2$. But we determined $k$ is odd. So $k = 1$ or $-1$.

....

Now as to your question about the $\gcd$ of an even number $2m$ and an odd number $2j + 1$. If $d=\gcd(2m, 2j+1)$ then $d|2m$ and $d|2j+1$. So $d$ divides $(2j+1)-2m = 2(j-m)+1$ and ... well.... now what?

We can contiune that $d|2j+1 -(2(j-m)+1) = 2m-j$ and ... that's true... we can keep doing this but we don't know how it ends.... well, actually we do. We'll eventually get $d|\gcd(2m, 2j+1)=d$ and then we won't be able to go any further.

That's actually Euclids algorithm and that is how we find $\gcd(2m,2j+1)$ in the first place.

Example: Let $d= \gcd(28, 63)$ then $d|28$ and $d|63$ so $d|63-28=35$ so $d|35-28=7$ so $d|28-7=21$ and $d|21-7=14$ and $d|14-7=7$ and $d|7-7 =0$ and .... can't go any further. $\gcd(28,63)=7$.