What can we say about b?

This polynomial factorizes as $(x - ab)(ax^2 - 1)$ . Since it only has integer roots, we deduce that $a=1$ , and hence the roots of the cubic are $a=1,-1,b$ . Certainly all positive integer factors of $a=1$ are roots of the cubic. Since all positive integer factors of $b$ are roots of the cubic, we deduce that the only positive integer factors of $b$ are $b$ and $1$ , and hence $b$ is prime.

Let ${ x }_{ 1 }; { x }_{ 2 }; { x }_{ 3 }$ be the whole number roots of the polynomial. Then: ${ x }_{ i }=\frac { p }{ q }$ where $p$ and $q$ are coprime whole numbers. Every root of the polynomial must be a whole number therefore we know: $q=1$ and by the rational root theorem we know that $a|q\rightarrow a|1$ . Because $a$ is a positive whole number $a$ must be $1$ . And then the polynomial simplifies to: ${ x }^{ 3 }-b\cdot { x }^{ 2 }-x+b$ $a$ has only one factor ( $1$ ). Thus $1$ must be a root of the polynomial and by polynomial division( $\left( { x }^{ 3 }-b\cdot { x }^{ 2 }-x+b \right) :\left( x-1 \right)$ ) we get: ${ x }^{ 2 }+{ x }\cdot \left( 1-b \right) -b$ And we can solve this with the quadratic formula: $\begin{aligned} { x }_{ 2|3 } & = \frac { b-1\pm \sqrt { { \left( 1-b \right) }^{ 2 }+4\cdot b } }{ 2 } \\ & = \frac { b-1\pm \sqrt { { 1 }-2\cdot b+{ b }^{ 2 }+4\cdot b } }{ 2 } \\ & = \frac { b-1\pm \sqrt { { 1 }+2\cdot b+{ b }^{ 2 } } }{ 2 } \\ & = \frac { b-1\pm \sqrt { { \left( 1+b \right) }^{ 2 } } }{ 2 } \\ & = \frac { b-1\pm \left( 1+b \right) }{ 2 } \end{aligned}$ Thus our 3 roots are $1$ , $b$ and $-1$ . If $b$ has only one factor than $b$ must be equal $1$ which is by definition false. If $b$ has 2 factors, then it must be a prime and its factors are $1$ and $b$ itself (which are roots of the polynomial).If $b$ has more than 2 factors, than $b$ has at least 3 factors . $1$ , $d$ and $b$ . But $d$ is not a root of the polynomial.

Thus $b$ must have 2 divisiors and therefore $b$ must be a prime.