Why is $r^2 + 2rs + s^2$ composite for all positive integer $r$ and $s$?

Question

Why is $r^2 + 2rs + s^2$ composite for all positive integer $r$ and $s$?

By definition,

A integer $n > 1$ is composite if and only if there exist positive integer, $r$ and $s$, so that $n = r \times s$ and $r$ does not equal $1$ and $s$ does not equal $1$.

So from the question above, it says for all positive integer $r$ and $s$.

All positive integers includes $1$, however, by definition, it says $r$ and $s$ must not be equal to $1$.

Therefore, I conclude they are not composite.

Unfortunately, the answer said that they are composite.

How is this so and how do I prove that they are composite even it says "for all positive integer $r$ and $s$"?

Answer 1

This is essentially a rephrasing of the other answers that might make you understand it.

We have that $$r^2 + 2rs + s^2 = (r+s)\cdot(r+s). $$

Hence the expression is composed of two factors, and it is a composite number as long as non of the factors are equal to $1$.

However, since $r$ and $s$ are positive integers they satisfy $r\geq 1$ and $s\geq 1$, and hence the factors satisfy $r+s\geq 2$. Since the factors are larger than $2$, they are in particular not equal to $1$, and hence the given number is composite.

Answer 2

When we write definitions like 'composite', we need to use symbols to denote the things we're discussing. Those symbols are arbitrary, but since the same symbols get used over and over again, things can get confusing. It can often help to replace the symbols in a definition with new symbols to get an equivalent definition that doesn't use symbols for which you already have a prior meaning or use.

If you replace the $r$ and $s$ in the definition of composite, you get this

A integer n > 1 is composite if and only if there exist positive integers, $u$ and $v$, so that $n = u \times v$ and $u \ne 1$ and $v \ne 1$.

This is exactly the same definition! But with this definition, you can read the question about $r^2 + 2rs + s^2$ without confusing the $r$ and $s$ of the question with the $r$ and $s$ of the definition.

With that substitution, the question (slightly rewritten) says this:

Pick any two positive integers $r$ and $s$ (which are allowed to be $1$, since $1$ is a positive integer). Compute $r^2 + 2rs + s^2$. Then there are positive integers $u$ and $v$, neither of which is $1$, with the property that $$ r^2 + 2rs + s^2 = u \times v. $$

Proving this may be tricky for you, but at least you know what you're trying to prove.

Hint: Can you factor $r^2 + 2rs + s^2$?

Post-comment additions

OK, here's the whole proof, in hideous detail.

Claim: For $r, s$ positive integers, the number $n = r^2 + 2rs + s^2$ is composite.

Proof: We must exhibit numbers $u$ and $v$ with the property that $uv = n$ and $u \ge 1$ and $v \ge 1$, and must also show that $n > 1$.

Basic algebra shows that

$$ r^2 + 2r2 + s^2 = (r+s)(r+s). $$

Let $u = r+s$ and $v = r + s$. Then the previous equation shows that $$ n = uv. $$ We now have to show that $u > 1$, $v > 1$, and $n > 0$.

First, since $u = v$, we'll just show that $u > 1$.

Now $r$ is a positive integer, which means it's an integer greater than 0. It is therefore greater than or equal to 1. The same goes for $s$. So we have \begin{align} r &\ge 1 & \text{by the paragraph above}\\ s &\ge 1 & \text{by the paragraph above}\\ r + s &\ge 1 + 1 & \text{addition of inequalities}\\ u &\ge 2 > 1. & \text{because $1+1 =2$.} \end{align} Thus $u > 1$, and we also have $v > 1$ (since $v = u$).

Finally, knowing that $u > 1$ and $v > 1$ tells us that $u\cdot v > 1 \cdot 1$ (by the rule for products of inequalities of positive numbers), so $n > 1 \cdot 1$ so $n > 1$, and we've proved all the things needed in the definition of "composite." Thus $n$ is composite. QED.

Answer 3

$$(r+s)(r+s)=r^2+2rs+s^2$$

So if $r,s\ge1$ then $r+s\geq2$ so $\ldots$