Problem :
Suppose $p(x)$ is a polynomial with integer coefficients. Show that if $p(a) =1$ for some integer $a$ then $p(x)$ has at most two integer roots.
Let $p(x)$ be a cubic polynomial such that $p(x) =ax^3+bx^2+cx +d =0 $
Let $p(\alpha) =a\alpha^3 +b \alpha^2 + c\alpha +d =1 $ ( as per the question) .
where $a,b,c$ are integers.
Now how to prove the given problem please suggest thanks.....
Hint $\ $ It is the special case $\,\color{#c00}{f(n)=1} $ of the following general
Key Idea $\ $ The possible factorizations of a polynomial $\in\Bbb Z[x]$ are constrained by the factorizations of the integer values that the polynomial takes. For a simple example, if some integer value has few factorizations (e.g. a unit $\,\pm1 $ or prime $p$) then the polynomial must also have few factors, asssuming that that the factors are distinct at the evaluation point. More precisely
If $\, f(x) = f_1(x)\cdots f_k(x)\,$ and $\,f_i\in\Bbb Z[x]\,$ satisfy $\color{#0a0}{f_i(n) \ne f_j(n)}\,$ for $\,i\ne j,$ all $\,n\in \Bbb Z$
$\quad \color{#c00}{f(n) =\pm1}\,\Rightarrow\, k\le 2\ $ else $1$ would have $\rm\,3\,\ \color{#0a0}{distinct}$ factors $\,f_1(n),f_2(n),f_3(n)$
$\quad f(n) = \pm p\,\Rightarrow\, k\le \color{#c0f}4\ $ since a prime $p$ has at most $\,\color{#c0f}4\,$ distinct factors $\,\pm1,\pm p$
Yours is a special case of the first (unit) case, where the $f_i$ are linear.
Remark $\ $ One can push the key idea to the hilt to obtain a simple algorithm for polynomial factorization using factorization of its integer values and Lagrange interpolation. The ideas behind this algorithm are due in part to Bernoulli, Schubert, Kronecker. See this answer for references.
Hints:
Suppose your cubic has three integer roots $\;u_1,u_2,u_3\;$ , then
$$p(x)=a(x-u_1)(x-u_2)(x-u_3)$$
It is given though that
$$1=p(\alpha)=a(\alpha-u_1)(\alpha-u_2)(\alpha-u_3)\;,\;\alpha\in\Bbb Z$$
But all the factors above are integers, so either all are $\;1\;$ or two exactly are $\;-1\;$ or...
