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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

Diceble
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  • it is literally the same proof, you just forgot to check the (trivial) case that one of the brackets is 1, but otherwise, they are same – Aditya_math Jan 17 '23 at 12:53

1 Answers1

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There are two differences that I can see.

One is that you chose to give two (different) notations $m,n$ for the two factors of $r+s$; the book proof is more concise and does not introduce new notation at all.

The other is you wrote "note that $m,n$ are positive integers", which is not sufficient to complete the proof. You also have to note that, and prove that $m,n \ne 1$; or, as the more concise book proof does, you have to prove that $r+s \ne 1$.

Lee Mosher
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  • does this mean that if I would say; note that $m,n$ are integers and since $m>0$ and $n>0$ , then $m+n\neq1$ would be sufficient to prove that m and n are positive integers? – Diceble Jan 17 '23 at 13:04
  • Yes, that would be fine. – Lee Mosher Jan 17 '23 at 13:05
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    I'll say that your latest comment points out one further difference in your proof and the book's proof: the book leaves it implicit that $r+s$ is a positive integer, whereas you spell that issue out in a bit more detail. – Lee Mosher Jan 17 '23 at 13:07
  • thanks you for the explanation. this will really help me in preparing for my exam :) – Diceble Jan 17 '23 at 13:14