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.

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 $$(a,b,c)=(11,8,-20)\quad{\rm and}\quad (a,b,c)=(13,10,-24)\ .$$

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.