Let px be a polynomial with integer coefficients satisfying

Let p(x) be a polynomial with integer coefficients satisfying p(0) = p(1) = 2017. Show that p has no integer zeros. I really wanted to write this question with 2016 instead of 2017. but couldn\'t. Why? Now, generalize this result and prove it.

Solution

Sufficient to show that, Let p(x) be a polynomial with integer coefficients satisfying that p(0) and p(1) are odd. Since 2017 is an odd number.Show that p has no integer zeros.

We know f(0)1(mod2) This means that the constant term of f is an odd integer.

We also know f(1)1(mod2) This means that the sum of the coefficients of f is odd.

Consider f(x) for a general xZ

.Case 1: x is even.If x is even, all of the powers of x must also be even, and since the product of even numbers with any integer coefficients is also even, everything but the constant term of f is even. Thus, f(x) must be odd.

Case 2: xis odd.If x is odd, all of the powers of x must also be odd. We have a bunch of odd numbers xi (including x0), and we are adding multiples of them together. We know that the sum of the coefficients is odd; this means we have, in total, an odd number of xis. The sum of an odd number of odd numbers must be odd, and so f(x) is odd.

We\'ve proven that f(x) is odd for integer x. It follows, a fortiori, that f(x) is nonzero.