If $x^2+bx+a=0$ and $x^2+ax+b=0$ have a common root $c$, Then what values of $(a,b)$ would work? Let $a$ and $b$ be distinct integers. If $x^2+bx+a=0$ and $x^2+ax+b=0$ have a common root $c$, Then which of the following statements are true?
1) $c*(a+c)=-b$
2) $a+b=-1$  
3) $a+b+c=0$ 
4) $c=0$ 
Update
I just tried to sub $c$ into both of the equations:
$c^2+cb+a=0$ and $c^2+ac+b=0$ which then gives us the equality
$c^2+cb+a=c^2+ac+b$
$ => cb+a=ac+b$
$=> b(c-1)=a(c-1)$
which gives me then a=b which is contradictory because the integers are supposed to be distinct.
Update #2
Ok it looks like 1) is true, 2) is true, 3) is true, and 4) is false .. right?

Factor Theorem implies $x-c$ is a factor of $x^2+ax+b-(x^2+bx+a)$, and this is equal to $(a-b)(x-1)$, so $c=1$, let $r$ be the another root of $x^2+ax+b$ then $(x-1)(x-r)=x^2-(1+r)x+r=x^2+ax+b$, so $r=b$ and $a=-(1+r)=-1-b$. Therefore $a+b=-1$.
