Real coefficients such that $x^2+bx+c=0$ and $2x^2+(b+1)x+c+1=0$ both have two integer roots

Question

Do there exist real $b,c$ such that each of the equations

$x^2+bx+c=0 \space,\space 2x^2+(b+1)x+c+1=0$ has two integer roots ?

Answer 1

No.

Let $m,n,k,l$ be the roots. Then $x^2+bx+c=(x-m)(x-n)$ and $2x^2+(b+1)x+(c+1)=2(x-k)(x-l)$. Hence $b=-m-n$, $c=mn$, $b+1=-2(k+l)$, $c+1=2kl$. From the last two conditions, $b,c$ are odd integers, hence $m,n$ are both odd, but then $b=-m-n$ is even.

Alternatively, note that the coefficients of $(x-m)(x-n)$ are integers, hence both $b$ and $b+\frac12$ must be integers (as $x^2+bx+c$ and $x^2+(b+\frac12)x+(c+\frac12)$ have integer roots).

Answer 2

If there exist such $b,c\in\Bbb{R}$ then for some $m_1,m_2,n_1,n_2\in\Bbb{Z}$ they satisfy $$x^2+bx+c=(x-m_1)(x-m_2)\qquad\text{ and }\qquad 2x^2+(b+1)x+c+1=2(x-n_1)(x-n_2),$$ or equivalently $$b=-(m_1+m_2)\qquad c=m_1m_2\qquad b+1=-2(n_1+n_2)\qquad c+1=2n_1n_2.$$ This shows that $b$ and $c$ must also be integers. From the last equation it follows that $c$ is odd, so both $m_1$ and $m_2$ are odd by the second. Then $b$ is even by the first, but $b$ is odd by the third, a contradiction. Hence such $b,c\in\Bbb{R}$ do not exist.