Prove that if the roots of the polynomial $x^2+ ax+ b + 1 $ are positive integers, then a and b are integers.

Question

Prove that if the roots of the polynomial $x^2+ ax+ b + 1 $ are positive integers, then a and b are integers.

I'm having problems with this exercise, I'd like to know how to start it.

Thank you.

Answer 1

Viete theorem says: $-a = x_1+x_2$. This means $a$ is an integer if $x_1,x_2$ are. Also $b+1 = x_1x_2 \implies b = -1 +x_1x_2$. So $b$ is also an integer.

Answer 2

Suppose the integers $p, q$ are distinct roots. Then you have $$p^2+ap+b+1=0$$and $$q^2+aq+b+1=0$$

Now subtract to obtain $$(p^2-q^2)+a(p-q)=0$$ and since $p\neq q$ we can divide by $p-q$ and obtain $a=-p=q$

Now multiply the first of the equations by $q$ and the second by $p$ and subtract to obtain $$pq(p-q)+(b+1)(q-p)=0$$ and similarly divide through by $(p-q)$ to obtain $b+1=pq$.

Note that this is simple linear algebra and uses no theorem about roots or factorisation.

If there is only one root, I don't see how to do the same - you are left with the one equation $p^2+ap+b+1=0$ and you have to know some special property of a double root (or use a property like continuity of the reals or a synthetic form of differentiation). Maybe someone else will see an equivalent solution in the equal roots case.

Answer 3

If $x_1$ and $x_2$ are the integer roots then $x = {x_1 + x_2 \over 2} = {-a \over 2}$ is the vertical axis so $-a = x_1 + x_2$ and $a$ is an integer. Plugging in integer roots for $x$, we see that $$b = -x^2 -ax -1$$ is also an integer.