Understanding the Polynomial Property

We are given a polynomial \(P(x) = x^3 + px^2 + qx + r\). The core condition is that this polynomial must evaluate to an integer value for every integer value of \(x\) we substitute into it. Our goal is to determine if the coefficients \(p\), \(q\), and \(r\) must also be integers based on this condition.

Analyzing Statements using Specific Values

Let's test the polynomial's behavior with a few specific integer inputs for \(x\). These specific choices will help us uncover necessary conditions for the coefficients.

• Case \(x = 0\): Substituting \(x=0\) into the polynomial gives: \(P(0) = (0)^3 + p(0)^2 + q(0) + r = r\) Since \(P(x)\) must be an integer for all integer \(x\), \(P(0)\) must be an integer. Therefore, \(r\) must be an integer. This means Statement III is correct.
• Case \(x = 1\): Substituting \(x=1\) gives: \(P(1) = (1)^3 + p(1)^2 + q(1) + r = 1 + p + q + r\) We know \(P(1)\) must be an integer, and we've established that \(r\) is an integer. Therefore, \(1 + p + q\) must also be an integer. Let's call this integer \(K_1\). So, \(p+q = K_1 - 1\), which means \(p+q\) must be an integer.
• Case \(x = -1\): Substituting \(x=-1\) gives: \(P(-1) = (-1)^3 + p(-1)^2 + q(-1) + r = -1 + p - q + r\) Similarly, \(P(-1)\) must be an integer. Since \(r\) is an integer, \(-1 + p - q\) must also be an integer. Let's call this integer \(K_2\). So, \(p-q = K_2 + 1\), which means \(p-q\) must be an integer.

Analyzing Statements using Finite Differences

Another powerful technique is to look at the differences between consecutive values of the polynomial. If \(P(x)\) is always an integer, then its differences must also always be integers.

First Difference: Define \(\Delta P(x) = P(x+1) - P(x)\).

\( \Delta P(x) = [ (x+1)^3 + p(x+1)^2 + q(x+1) + r ] - [ x^3 + px^2 + qx + r ] \) Expanding and simplifying, we find: \( \Delta P(x) = (3x^2 + 3x + 1) + p(2x + 1) + q \) \( \Delta P(x) = 3x^2 + (3 + 2p)x + (1 + p + q) \) Since \(P(x)\) and \(P(x+1)\) are integers for all integers \(x\), their difference, \(\Delta P(x)\), must also be an integer for all integers \(x\).

Second Difference: Define \(\Delta^2 P(x) = \Delta P(x+1) - \Delta P(x)\).

Calculating the second difference: \( \Delta^2 P(x) = [ 3(x+1)^2 + (3+2p)(x+1) + (1+p+q) ] - [ 3x^2 + (3+2p)x + (1+p+q) ] \) Expanding and simplifying this expression leads to: \( \Delta^2 P(x) = 3(2x+1) + (3+2p) = 6x + 3 + 3 + 2p = 6x + (6+p) \) Since \(\Delta P(x)\) must be an integer for all integers \(x\), the second difference, \(\Delta^2 P(x)\), must also be an integer for all integers \(x\).

Now, let's analyze \(\Delta^2 P(x) = 6x + (6+p)\). For this expression to yield an integer value for every integer \(x\), the term \((6+p)\) must itself allow this. Consider \(x=0\): \(\Delta^2 P(0) = 6(0) + (6+p) = 6+p\). Since \(\Delta^2 P(0)\) must be an integer, \(6+p\) must be an integer. Because 6 is an integer, \(p\) must be an integer. This confirms that Statement I is correct.

Now that we know \(p\) is an integer, let's revisit the first difference: \(\Delta P(x) = 3x^2 + (3+2p)x + (1+p+q)\). We know \(\Delta P(x)\) must be an integer for all integer \(x\). Since \(p\) is an integer, \(3x^2\), \(3\), \(2p\), and \(x\) are all integers or involve integers. Therefore, \(3x^2\) is an integer, and \((3+2p)x\) is an integer. For the entire expression \(\Delta P(x)\) to be an integer, the term \((1+p+q)\) must also contribute in a way that results in an integer. As \(1\) and \(p\) are integers, this requires \(q\) to be an integer as well. This confirms that Statement II is correct.

Conclusion

By analyzing the polynomial with specific integer values (\(x=0, 1, -1\)) and using the method of finite differences, we have concluded the following:

• \(r\) must be an integer (from \(P(0)\)).
• \(p\) must be an integer (from \(\Delta^2 P(x)\)).
• \(q\) must be an integer (from \(\Delta P(x)\) and knowing \(p\) is integer).

Therefore, all three statements I, II, and III are correct.

The correct option is the one that includes all three statements.