Algebra problem of Natural numbers

Question

i faced this problem and i am not sure that my solution is right:

i have these 2 equations: Let $a,b \in \Bbb{N} $

1. a+b=n

2. a*b=(n-1)

how can i prove that the only answer is when a=1 and b=(n-1) or vice versa,

many thanks

Answer 1

From 1, $b=n-a$; from 2, $a(n-a)=n-1$, that is $$ a^2-na+n-1=0 $$ Can you solve the quadratic equation?

Answer 2

If you add 1 to equation 2) you get $a+b=n=a\cdot b+1$. The equation $a+b=a\cdot b+1$ is equivalent to $a-a\cdot b+b-1=0$, which is equivalent to $a(1-b)-(1-b)=0$, where you can factor by $1-b$ and get the result.

Answer 3

$$(a+b)^2 - (a-b)^2=4ab$$ $$n^2-(a-b)^2=4(n-1)$$ $$(a-b)^2=n^2-4(n-1)=(n-2)^2$$ $a-b = n-2$ or $2-n$

Now you can solve the two equations.