Given $a + b$ is prime and $a < -1$, show $a^2 + ab + b < 2$ I want to show $a^2 + ab + b < 2$ given $a, b \in \mathbb{Z}$, $a + b$ is prime and finally $a < -1$.
I've reduced the bounds a little. As $a + b$ is prime, $a + b \ge 2$. Since $a \in \mathbb{Z}$ the first integer less than $-1$ is $-2$, and thus $a \le -2$.
Starting with $a + b \ge 2$ I've done the following steps.
$$a + b \ge 2 \implies b \ge 2 - a$$
$$\implies ab \le a(2 - a)$$
$$\implies a^2 + ab + b \le a^2 + a(2 - a) + b$$
$$\implies a^2 + ab + b \le a^2 + 2a - a^2 + b \implies a^2 + ab + b \le 2a + b$$
However this is nowhere near a tight enough upper bound to show the result I am looking for. How should I continue?

Let $a+b = p$. $p$ prime so $p\geq 2$. Then $a^2+ab+b = (p-1)a+p$. What can you conclude now?
