Question

F(x) is a fourth-order polynomial with integer coefficients and no common factor among the coefficients. The roots of F(x) are -2, -1, 1 and 2. If p is a prime number greater than 97, then the largest integer that divides F(p) for all such values of p is:

• A. 72
• B. 120
• C. 240
• D. 360

Hint:

Write F(p) as a product of four factors around p, then check what power of 2, 3 and 5 is guaranteed every time.

Answer 1

The Correct Option is D

Step 1: Write down F(x).
Since F(x) has roots -2, -1, 1, 2 and integer coefficients with no common factor, the simplest such polynomial is:
\[ F(x) = (x+2)(x+1)(x-1)(x-2) \]
Any other valid F(x) would just be an integer multiple of this one, and multiplying by anything bigger than 1 would add back a common factor, which is not allowed. So this is the F(x) meant here.

Step 2: Write F(p) as a product of four numbers around p.
\[ F(p) = (p-2)(p-1)(p+1)(p+2) \]
Since p is a prime bigger than 97, p is odd and is not divisible by 2, 3 or 5.

Step 3: Find the guaranteed power of 2.
Since p is odd, \(p-1\) and \(p+1\) are both even, and they are two consecutive even numbers, so one of them is divisible by 4. That means \((p-1)(p+1)\) always contributes at least \(2 \times 4 = 8\), that is \(2^3\), to the product. \(p-2\) and \(p+2\) stay odd, so they add nothing more to this count.

Step 4: Find the guaranteed power of 3.
Since p is not divisible by 3, p leaves remainder 1 or 2 when divided by 3. If \(p \equiv 1 \pmod 3\), then \(p-1\) and \(p+2\) are both divisible by 3. If \(p \equiv 2 \pmod 3\), then \(p+1\) and \(p-2\) are both divisible by 3. Either way, exactly two of the four factors carry a factor of 3, so the product always has at least \(3^2 = 9\).

Step 5: Find the guaranteed power of 5.
Since p is not divisible by 5, p leaves remainder 1, 2, 3 or 4 when divided by 5, and exactly one of \(p-2, p-1, p+1, p+2\) matches the residue that is divisible by 5. So the product is always divisible by 5, but there is no guarantee of a second factor of 5 for every such p.

Step 6: Combine and check tightness.
Multiplying the guaranteed pieces: \(2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360\). Testing this on actual primes such as p = 101 and p = 103 shows the product is always a multiple of 360, but not always a multiple of anything bigger, since the extra power of 3 or 2 seen for one prime drops back down for another, so 360 is the true largest guaranteed divisor.

Final Answer:
\[ \boxed{360} \]