We are given a polynomial $f$ with integer coefficients such that for 4 distinct integers $a_1,a_2,a_3$ and $ a_4$, $f(a_1)=f(a_2)=f(a_3)=f(a_4)=1991$. Show that there exists no integer $b$ such that $f(b)=1993$.

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    $\begingroup$ You really need to explain what you have tried and where you are stuck, rather than just giving the question. $\endgroup$ – Mark Bennet Sep 27 '14 at 8:11
  • $\begingroup$ First impulse: have a look at polynomial $f(x)-1991$. I am not sure, but it might help. $\endgroup$ – drhab Sep 27 '14 at 8:16

There must be a polynomial with integer coefficients say $q(x)$ with$$f(x)=(x-a_1)(x-a_2)(x-a_3)(x-a_4)q(x)+1991$$
If $f(b)=1993$ then $$1993=(b-a_1)(b-a_2)(b-a_3)(b-a_4)q(b)+1991$$
$$\Rightarrow (b-a_1)(b-a_2)(b-a_3)(b-a_4)q(b)=2$$.
Since $b,q(b),a_1 ,a_2,a_3,a_4$ are distinct integers this means that 2 can be written as a product of at least 4 different integers which is a contradiction.


Hint: $f(x)=(x-a_1)(x-a_2)(x-a_3)(x-a_4)g(x)+1991$.


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