if $m$ and $n$ are positive integers ,it means $m=\left\{1,2,3,4,.......\right\}$ and $n=\left\{1,2,3,4,............\right\}$
$A)$ If $m$ and $n$ are co-prime, there exist integers $a$ and $b$ such that $am + bn=1$
Lets take $m=1,n=3$
$\implies am+bn=1$
$\implies a+3b=1$
take $a=-2,b=1$
$\implies -2+3=1$
$\implies 1 = 1$
So, this is true.
Proof:
$B)$ $m^{n-1}=1$ $mod$ $n$
Lets take some values$:m=1,n=1$
$\implies1^{1-1}=1$ $mod$ $1$$\implies 1^{0}=0$
$\implies 1=0$ $($ False because $1\neq 0)$
I took a counterexample and proved that the given statement is false.
So, this is false.
$C)$ The rational number $\frac{n}{m} \cdot \frac{n-1}{m-1} \cdot \frac{n-2}{m-2} \dots \frac{n-(m-2)}{m-(m-2)} \cdot \frac{n-(m-1)}{m-(m-1)}$ is an integer
We can prove this one using proof by contradiction.
Let us assume The rational number $\frac{n}{m} \cdot \frac{n-1}{m-1} \cdot \frac{n-2}{m-2} \dots \frac{n-(m-2)}{m-(m-2)} \cdot \frac{n-(m-1)}{m-(m-1)}$ is not a integer.
lets take $m=4,n=8$
$\implies\frac{8}{4} \cdot \frac{8-1}{4-1} \cdot \frac{8-2}{4-2} \dots \frac{8-(4-2)}{4-(4-2)} \cdot \frac{8-(4-1)}{4-(4-1)}$
$\implies\frac{2}{1}\cdot \frac{7}{3}\cdot \frac{6}{2} \cdot \frac{5}{1}=70$ It is a integer.
Here our assumption is false.
So, the given statement is true.
$D)\ m+1$ is a factor of $m^{n(n+1)}-1$
We can prove this using proof by contradiction.
Let us assume $\ m+1$ is not a factor of $m^{n(n+1)}-1$
Take $m=3,n=2$
$3+1$ is a factor of $3^{2(2+1)}-1$
$4$ is a factor of $3^{6}-1$
$4$ is a factor of $729-1$
$4$ is a factor of $728$
Factor of $728:1,2,4,7,8,13,14,26,28,52,56,91,104,182,364,728$
Here our assumption is false.
So, the given statement is true.
$E)$ If $2^{n}-1$ is prime ,then $n$ is prime
Lets take an example$:n(Prime)=2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,...........$
$P_{n}=2^{n}-1\rightarrow(1)$
Put value one by one and observe the output,
- $P_{2}=2^{2}-1=4-1=3\text{(Prime)}$
- $P_{3}=2^{3}-1=8-1=7\text{(Prime)}$
- $P_{5}=2^{5}-1=32-1=31\text{(Prime)}$
- $P_{7}=2^{7}-1=128-1=127\text{(Prime)}$
- $P_{11}=2^{11}-1=2048-1=2047=23\times 89{\color{Magenta}{\text{(Not Prime)}} }$
- $P_{13}=2^{13}-1=8192-1=8191\text{(Prime)}$
$\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots$
$\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots$
For this one $P_{11}=2^{11}-1=2048-1=2047=23\times 89{\color{Magenta}{\text{(Not Prime)}} }$
If $2^{11}-1$ is prime then $11$ is prime.
we can write like this
$2^{11}-1$ is prime $\implies 11$ is prime.
$\text{False}\implies\text{anything}\equiv \text{True}$
Proof:
So, this is true.
(Reference:https://brilliant.org/wiki/mersenne-prime/)
So, the correct answer is $(B).$