2021/052) Let f(x) be a polynomial in x with integer coefficients and suppose that for 5 distinct integers $a_1,a_2,a_3,a_4,a_5$ one has $f(a_1)= f(a_2)= f(a-3) = f(a_4) = f(a_5) = 2$ Show that there is no integer b such that $f(b) = 9$

Because  $f(a_1)= f(a_2)= f(a-3) = f(a_4) = f(a_5) = 2$

So defining  $g(x) = f(x) - 2$  

We get $g(a_1)= g(a_2)= g(a-3) = g(a_4) = g(a_5) = 0$

So $g(x) =  (x-a_1)(x-a_2)(x-a_3)x-a_4)(x-a_5)h(x)$ for some function h(x) of x

So g(x) is product of atleast 5 different numbers 

Because $f(b) = 9$ so $g(b) = 7$ 

For g(x) to be 7 at some b 7 should be product of 5 different numbers but is is not product of have at least 5 different factors but 7  has at most 3(as -7 * -1 * 1) so it is not possible for any b such that $g(b) = 7$ or $f(b)  = 9$