Multiple choice

Let $P(x)=x^{2}+ax+b$ be a quadratic polynomial where $a$ and $b$ are real numbers. Suppose $P(-1)^{2}.P(0)^{2}.P(1)^{2})$ is an arithmetic progression of integers , then $a$ and $b$ are integers.

  1. True

  2. False
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A Correct answer
Explanation

The condition that P(-1)^2, P(0)^2, P(1)^2 form an arithmetic progression forces b = -(a^2 + 1)/2. Requiring all three squared values to be integers further constrains a^2 + 1 to be even and a^2 to be an odd integer, which makes a and b integers. No non-integer real values of a can satisfy the integer AP condition.

AI explanation

Let the terms of the arithmetic progression of squares be u^2, v^2 and w^2 corresponding to P(-1), P(0) and P(1). Since they form an arithmetic progression, we have 2v^2 = u^2 + w^2. Multiplying by 2 gives the equation (u - w)^2 + (u + w)^2 = (2v)^2, and since these are integer squares, this forms a Pythagorean triple where (2v)^2 is the hypotenuse. Any primitive Pythagorean triple has an odd hypotenuse, meaning 2v must be odd, which is impossible for an integer v unless the progression is a trivial constant sequence. This implies u = v = w, meaning the polynomial values P(-1) = P(0) = P(1) are all the same integer. From P(-1) = P(0) we get 1 - a + b = b, yielding a = 0. Substituting a = 0 into the polynomial shows b = P(0), which must be an integer. Therefore, the statement that a and b are integers is true.

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