Integer roots to $ax^2 + bx + c$ and $(a+1)x^2 + (b+1)x + (c+1)$ Do there exist quadratic functions $ax^2 + bx + c$ and $(a+1)x^2 + (b+1)x + (c+1)$ with integer coefficients such that both have only integer roots?

I was thinking of taking the discriminant in an attempt to try and start off this proof, but I don't quite think it'd be a much use due to the condition that we need integer roots. Than, how exactly should I start this problem?
 A: No there does not (for $a\ne 0,-1$). We shall prove this with a simple parity argument.
One of $a, a+1$ must be even. For simplicity, suppose $a$ is even. The other case is similar.
Since $ax^2+bx+c=0$ has integer roots, it can be written as:
$$0=a(x-\alpha)(x-\beta)=ax^2-a(\alpha+\beta)x+a\alpha\beta=ax^2+bx+c$$
Hence $b$ and $c$ are both divisible by $a$, and hence are both even.
Now consider the other equation, $(a+1)x^2+(b+1)x+(c+1)=0$.
All three coefficients are odd, and in particular, $a+1$ and $c+1$ can only have odd factors.
If this equation indeed has integer roots, it can be factorized.
Considering cross method, or similar methods, we see that b+1 is a sum of two products of two odd numbers, which is even, and a contradiction. I intend to improve the rigour of this statement, if possible.
EDIT: we can make this rigourous using the root argument above, write:
$$0=(a+1)(x-\alpha')(x-\beta')=(a+1)x^2-(a+1)(\alpha'+\beta')x+(a+1)\alpha'\beta'$$
Since $c+1$ is odd, both $\alpha'$ and $\beta'$ are odd. This makes $\alpha'+\beta'$, and thus $b+1$, even.
A: If $c=0$ then the equations are $x(ax+b)=0$ and $x((a+1)x + b+1)=0$, the solution sets of which are $\{0,-b/a\}$ and $\{0,-(b+1)/(a+1)\}$, respectively.
Taking $a=1$ and $b$ an odd integer $2k+1$ makes the solutions all integers since $b$ is divisible by $1$ and $b+1$ is divisible by $2$.
EDIT: Okay, now the question has been changed so this is not a correct solution.
