Why is $ab=(a-b)(a+b)$ false when $a,b$ are coprime? I could not find this, and do not have a background in number theory so sorry if this sounds trivial.
If $a$ and $b$ are coprime integers, why is the statement
$$ab=(a-b)(a+b)$$
a contradiction?

Suppose we have equality, then we have that 
$$ab = a^2 - b^2$$
or equivalently:
$$ab + b^2 = a^2.$$ 
Since $b \mid (ab + b^2) = a^2$, this would imply that $b$ divides $a^2$. However, since $a,b$ are coprime, so are $b$ and $a^2$. Indeed: if $b, a^2$ are not coprime, we can look at the prime factorization of their greatest common divisor (which would be different from $1$). Then there is a prime $p$ such that $p \mid b$ and $p \mid a^2$. Therefore $p \mid a$, hence $p \mid \text{gcd}(a,b)$, so $a,b$ would not be coprime.
Since we have shown that $b$ divides $a^2$, whereas they have to be coprime, we need that $b = 1$ (or $-1$). However, applying the same reasoning would show that $a$ divides $b^2$, so then also $a = 1$ or $-1$, but clearly this is not possible (since we would never have $ab = a^2 - b^2$). 
Therefore, we have a contradiction, so $ab \neq (a -b)(a + b)$.
