Find all pair of cubic equations Find all pair of cubic equations $x^3+ax^2+bx+c=0$ and $x^3+bx^2+ax+c=0$, where $a,b$ are positive integers and $c$ not equal to $0$ is an integer, such that both the equations have three integer roots and exactly one of those three roots is common to both the equations.
I tried the sum and product relationships with the coefficients, but I have more variables than the number of equations.All I got was that the common root has to be $x=1$. What do I next?Thanks.
 A: Suppose the roots of the first polynomial are $s,t,u$ and the roots of the second are $s,v,w$, where $s,t,u,v,w$ are integers.  You have
$$ 
\eqalign{ s t u &= svw = -c\cr
          st + su + tu &= -s -v - w = b\cr
          -s - t - u &= sv + sw + vw = a\cr}$$
Since $c \ne 0$ we may divide out $s$ from the first equation, obtaining
$tu = vw$.
The resultant of $st + su + tu + s + v + w$ and $sv + sw + vw + s + t + u$ with respect to $s$ is
$$ -utw-utv+uwv+twv+{u}^{2}+tu+{t}^{2}-{w}^{2}-wv-{v}^{2}+u+t-w-v$$
and the resultant of this and $tu - vw$ with respect to $t$ is
$$  \left( u-w \right)  \left( u-v \right)  \left( uwv+{u}^{2}+uv+uw+wv+u
 \right) 
$$
Now we don't want $u=w$ or $u=v$ because $s$ should be the only root in common, so we must have $$uwv+{u}^{2}+uv+uw+wv+u = u^2 + (w+1)(v+1) u + vw = 0$$
Among the integer solutions of this are
$$ \eqalign{v = 0,& u = 0, w = \text{arbitrary}\cr
            v = 0,& u = -w - 1, w = \text{arbitrary}\cr
            v = -1,& w = u^2, u = \text{arbitrary}\cr
            u = -1, & w = -v, v = \text{arbitrary}\cr
            u = v^2, & w = -v, v = \text{arbitrary}\cr
            u = -2, & v = -3, w = -8\cr
            u = -2, & v = -4, w = -5\cr
            u = -3, & v = -2, w = -12\cr
            u = -3, & v = -3, w = -5\cr
            u = -4, & v = -2, w = -10\cr
            u = -5, & v = -2, w = -10\cr
            u = -5, & v = -3, w = -5\cr
            u = -8, & v = -2, w = -12\cr
            u = -10, & v = -4, w = -5\cr
            u = -12, & v = -3, w = -8\cr}$$
and those obtained from these by interchanging $v$ and $w$.
Substituting these into the first set of equations, I get 
the following integer solutions with $a,b > 0$ and $c \ne 0$:
$$ \eqalign{a &= 13, b = 10, c = -24\cr
            a &= 11, b = 8, c = -20\cr}$$
            I'm not sure these are the only solutions satisfying the requirements, but I wouldn't be surprised.
A: We may assume $a>b\geq1$. An $x$ that solves both equations is $\ne0$ and also solves the equation
$$(a-b)(x^2-x)=0\ .$$
This implies $x=1$ and entails $c=-1-a-b$. Deflating the polynomials  in question by the factor $x-1$ leaves us with the pair of equations
$$\left.\eqalign{x^2+(a+1)x+a+b+1&=0\cr y^2+(b+1)y+a+b+1&=0\cr}\right\}$$
that should have altogether four different pairwise integer solutions $\ne1$. Let $-r$, $-s$ be the solutions of the first of these equations, and $-u$, $-v$ be the solutions of the second. It follows that
$$\eqalign{r+s&=a+1,\qquad rs=a+b+1\geq4\cr
u+v&=b+1,\qquad uv=a+b+1\geq4\ .\cr}\tag{1}$$
It follows that all four quantities $r$, $s$, $u$, $v$ are $>0$. We may assume $r>s$, $\>u>v$.
We now consider $r$ and $s$ as given and have to check which pairs $(r,s)$ are admissible. From the upper line $(1)$ we obtain that
$$a=r+s-1,\qquad b=rs-r-s\ ,$$
so that the second line $(1)$ gives
$$\eqalign{u+v&=rs-r-s+1=(r-1)(s-1)\ ,\cr
uv&=rs\ .\cr}\tag{2}$$
The condition $u+v=b+1\geq2$ then implies $r>s\geq2$.
The equatons $(2)$ imply that $u$ and $v$ are the solutions $t_1$, $t_2$ of the quadratic equation
$$t^2-(rs-r-s+1) t+rs=0\tag{3}$$
with discriminant
$$D:=(rs-r-s+1)^2-4rs=(rs-r-s-1)^2-4(r+s)\ .\tag{4}$$
We want that $D$ is a square. The number $D$ can only be a square if $D\leq(rs-r-s-2)^2$. Now the condition
$$(rs-r-s-1)^2-4(r+s)\leq(rs-r-s-2)^2$$
simplifies to $2(r-3)(s-3)\leq21$, or
$$(r-3)(s-3)\leq10\ .$$
This leaves us with the cases (i) $r>s=2$, (ii) $r>s=3$ and $${\rm (iii):}\qquad 4\leq s<r\leq 3+{10\over s-3}\ .\tag{5}$$ In the case (i) we obtain from $(4)$ that
$D=(r-5)^2-24$. Since $D$ has to be a square this means that we have to represent $24$ as a sum of subsequent odd numbers. There are two such representations, namely $24=3+5+7+9$ and $24=11+13$, leading to the admissible pairs $(10,2)$ and $(12,2)$.
In the case (ii) we obtain from $(4)$ that $D=(2r-5)^2-21$. This  means that we have to represent $21$ as a sum of subsequent odd numbers. There are two such representations, namely $21=5+7+9$ and $21=21$, leading to the admissible pairs $(5,3)$, $(8,3)$.
In the case (iii) the conditions $(5)$ admit $s=4$ and $r\in[5..13]$, and $s=5$ and $r\in[6..8]$. Checking the cases reveils that only $(5,4)$ is an admissible pair.
All in all we have found five admissible pairs $(r,s)$, namely
$$(10,2),\quad(12,2),\quad (5,3),\quad(8,3),\quad (5,4)\ .$$
The corresponding solutions $\{u,v\}$ of $(3)$ are then
$$\{5,4\},\quad\{8,3\},\quad\{5,3\},\quad\{12,2\},\quad\{10,2\}\ .$$
Since we want $r$, $s$, $u$, $v$ all different we have to discard the pair $(r,s)=(5,3)$, and we are left with essentially two solutions to the original problem, corresponding to the pairs $(r,s)=(10,2)$ and $(r,s)=(12,2)$. From $(1)$ and $c=-1-a-b$ we then obtain the coefficient triples 
$$(a,b,c)=(11,8,-20)\quad{\rm and}\quad (a,b,c)=(13,10,-24)\ .$$
