Question

The polynomial $p(x) = x^{2} + ax + b$ has positive integer coefficients $a$ and $b.$ If $p(60)$ is a perfect square and the equation $p(x) = 0$ has two distinct integer solutions, what is the least possible value of $b\,?$

Answer 1

To find the least possible value of b, we need to consider the conditions given in the problem. The polynomial p(x) = x² + ax + b has positive integer coefficients a and b. We are told that p(60) is a perfect square and that the equation p(x) = 0 has two distinct integer solutions.

Step-by-step explanation:

To find the least possible value of b, we need to consider the conditions given in the problem. The polynomial p(x) = x² + ax + b has positive integer coefficients a and b. We are told that p(60) is a perfect square and that the equation p(x) = 0 has two distinct integer solutions.

We can start by substituting x = 60 into the polynomial: p(60) = 60² + 60a + b. Since we want p(60) to be a perfect square, it must be expressed as n², where n is an integer. Therefore, we can write the equation as 60² + 60a + b = n².

Next, we need to consider the solutions of p(x) = 0. Since the equation has two distinct integer solutions, it means that the discriminant of the quadratic formula, b² - 4ac, must be a perfect square. In this case, the discriminant is a² - 4b.

By analyzing the given conditions and equations, we can solve for the least possible value of b that satisfies the requirements.