Let $ a$ be a positive integer. Show that $\text{gcd}(a,a-1) = 1$. use the result of par t $ a)$ to solve the Diophantine equation $ a+b=ab$ Not sure if I did part a right, not sure how to complete part $b)$
$a)$ Let $a$ be a positive integer. Show that $\text{gcd}(a,a-1) = 1$.
Proof by contradiction
suppose $\text{gcd}(n, n-1) = p > 1$. 
Then $n$ is a multiple of $p$, so $n = ap$ for some integer $a$. 
Similarly, $n-1 = bp$ for some integer $b$. 
Next multiple of $p$ after $n$ will be $(a-1)p = n-p$, which is greater than $n-1$. We have 
$n=ap < n-1=bp < n-p=(a-1)p$ 
Dividing everything by $p$ we get 
$a < b < a-1$ 
meaning that $b$ is an integer sandwiched between $a$ and $a-1$. This is impossible.
$b)$ Use the result of part $a)$ to solve the Diophantine equation
 $a+b=ab$
     where $(a,b)$ are positive integers.

Let $d = \text{gcd}(a,a-1) \to d\mid a, d\mid (a-1) \to d\mid [a-(a-1)] = 1 \to d = 1$.
$a+b = ab \to a = ab - b = (a-1)b \to b = \dfrac{a}{a-1} = 1 + \dfrac{1}{a-1}$. Can you go from here?
