Question

The polynomial equation with integer coefficient of 1, -2, -5

Answer 1

Answer:

x Let P(x) be a polynomial with integer coefficients, and suppose that there are five distinct

integers a1, a2, a3, a4 and a5 such that P(a1) = P(a2) = P(a3) = P(a4) = P(a5) = 5. Find the

number of solutions in integers of the equation P(x) = 0.

Solution. Let Q(x) = P(x)−5. Then the polynomial Q has distinct roots a1, a2, a3, a4 and a5, so we can write

Q(x) = R(x)S(x), where S(x) = (x − a1)(x − a2)(x − a3)(x − a4)(x − a5), and R(x) is a polynomial with

rational coefficients. Since S(x) is monic with integer coefficients, and R(x) is the result of long division of

Q(x) by R(x), it follows that R(x) also has integer coefficients. Therefore we have

P(x) = Q(x) + 5 = R(x)S(x) + 5

If α is an integer root of P(x), we have

R(α)(α − a1)(α − a2). . .(α − a5) = −5.

Recall that the integers a1, a2, a3, a4 and a5 are distinct, so that −5 has at least 5 distinct integer divisors.

But this is absurd, since the only integers dividing 5 are 1, −1, 5 and −5. It follows that there are no integer

solutions of P(x) = 0.