This is a Test of Mathematics Solution of Subjective 42 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.
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Let f(x) be a polynomial with integer coefficients. Suppose that there exist distinct integers , such that . Show that there does not exist any integer b with f(b) = 14.
Consider the auxiliary polynomial g(x) = f(x) – 3. Clearly, according to the problem, g(x) has four distinct integer roots . Hence we may write where Q(x) is an integer coefficient polynomial. (Since by factor theorem if ‘t’ is a root of a polynomial f(x) then (x-t) is it’s factor).
Suppose there exists an integer b such that f(b) = 14; then g(b) = 14-3 =11. Hence . are distinct so are . Therefore the equation indicates that 11 can be written as the product of at least 4 different integers which is impossible (as 11 is a prime). Indeed even if we consider -1, 1 and 11 , we have only three integers whose product is 11.
Thus we have a contradiction and proving that there does not exist an integer b such that f(b) 14.
Key Idea: Factor Theorem