Let f(x)=ax^5+bx^4+cx^3+dx^2+ex+1 be a polynomial. Also assume the f(1)=4 and f(2)=11. Additionally, all the c

Question

Let f(x)=ax^5+bx^4+cx^3+dx^2+ex+1 be a polynomial. Also assume the f(1)=4 and f(2)=11. Additionally, all the coefficients are integers. Prove that the equation does not have a solution that is an integer. here is the full question yahoo cut some of it off. I need all the work with it to

Answer 1

By Gauss's Lemma or the Rational Root theorem, or whatever you call it, for the polynomial f to have an integer zero, we would have to have a = +1 or a= -1, and the only possible zeroes would be x = +1 or x = -1.

We are told that 4=f(1)=a+b+c+d+e+1, so a+b+c+d+e = 3;
call this equation A. Note that it rules out the possibility that x=1 is a zero of f.

We are also told that 11 = f(2) = 32a + 16b + 8c + 4d + 2e + 1, so 32a + 16b + 8c + 4d + 2e = 10 and thus
16a+8b+4c+2d+e=5. Call this equation B.

Is it possible that f(-1)=0? Well,
f(-1) = -a+b-c+d-e+1, so if f(-1)=0, we must have
-a+b-c+d-e = -1. Call this equation C.

If we add equations A and C, we get 2(b+d)=2, so b+d = 1. If we subtract equation C from equation A, we get 2(a+c+e) = 4, so a+c+e=2. Now substitute into equation B:
5 = 16a+8b+4c+2d+e = 16a+8b+4c+2(1-b)+e
= 15a +8b +3c + 2(1-b) + (a+c+e) = 15a +8b +3c + 2(1-b) + 2
= 15a + 6b +3c + 4.

Hence 15a+6b+3c = 5-4=1. But the left hand side of this equation is divisible by three, and the right hand side is not, which is a contradiction. Thus f(-1) cannot be zero, and hence f has no integer zero.