For positive integers $m$ and $n$, let $m > n$ be such that $gcd(m , n) = 1$. Which of the following statements are true?

Question

For positive integers $m$ and $n$,let $gcd(m, n)$ denote greatest common divisor of $m$ and $n$. Let $m > n$ be such that $gcd(m , n) = 1$. Which of the following statements are true ?

1. $gcd(m-n , m+n) = 1$

2. $gcd(m-n , m+n)$ can have a prime divisor.

3. There exists integers x , y such that $nx-my = 3$

4. $gcd(m-n , m+n)$ can be an odd prime divisor.

My Attempt:

I know that "If $d = gcd(a , b)$, then $\exists x , y \in \Bbb Z$ such that $d = ax + by$" Here given that $gcd(m,n) = 1$ so $\exists p , q \in \Bbb Z$ such that $np + mq = 1$

Multiply both sides by $3$ , we get $n(3p) + m(3q) = 3 \implies ny + m(-x) = 3$ where $ x = - 3q$ and $y = 3p$. Thus there exists integers $x , y$ such that $nx - my = 3$. So option 3 is true.

Also I know that " If $d$ divides $a$ and $b$ then $d$ divides $a + b$ and $a - b$". So let if possible $gcd(m-n , m+n) = d \neq 1$. This implies that $d$ divides $m + n$ and $m - n$. Hence $d/(m+n)+(m-n)$ and $d/(m+n)-(m-n)$. Therefore $d/2m$ and $d/2n$.
So there are two cases,

Case (i) $d/m$ and $d/n$ $\implies d/gcd(m,n) = 1 \implies d = 1$ which is contradiction.

Case (ii) $d/2 \implies d = 2$ So option $2$ is true.

I have no more information about this question. Please help me. Thanks.

Answer 1

$1,2$. Notice Nurator's comment and consider the case where the prime divisor is $2$.

$3.$ Recall Bezout's lemma with $gcd(m,n) = 1$: There exist $x_{1},y_{1}$ such that $nx_{1}+my_{1}=1$. Convert this to the relevant form to show that this can be done.

$4$. Assume this is true with $p$ being some odd prime. Then $p|m-n, p|m+n \implies p |(m+n)-(m-n)=2n$. Since $p \nmid 2, p \mid n$. How does this result in a contradiction to the problem's hypothesis?