Hi again a question on if my proof is correct, since the way it is written differs much with how the book proofs it..
The Question given is :
"$r^2+2rs+s^2$ is composite for all positive integer $r$ and $s$"
Here's the proof I have written :
Assume that $r$ and $s$ are particular integers
It is given the number is $r^2+2rs+s^2$
It can be written as $r^2+2rs+s^2 = (r+s)\cdot(r+s)$
Let $m = (r +s)$ and $n=(r+s)$ not that $m,n$ are positive integers
Hence, $\exists m,n \in \mathbb{Z}^+$, such that $r^2+2rs+s^s = m \cdot n$
$\therefore r^2+2rs+s^2$ is composite
This is the answer given in the solution book :
It is given that, if r and s are both positive, i.e., $r>0, s>0$
Then, $r^2+2rs+s^2 = (r+s)^2 =(r+s)\cdot(r+s)$
$\therefore r^2+2rs+s^2$ is composite
Since $r>0$ and $s>0$, then $r+s \neq 1$
By the definition of a composite number, the given number is composite