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Let P(x) = x2 +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. Prove that a and b are integers. |
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Answer» Observe that P(-1) = 1 - a + b, P(0) = b; P(1) = 1 + a + b. The given condition gives 2b2 = (1 - a + b)2 + (1 + a + b)2 = 2(1 + b)2 + 2a2 = 2 + 4b + 2b2 + 2a2. Hence a2 + 2b + 1 = 0. Observe 1 + a2 + b2 + 2a + 2b + 2ab = (1 + a + b)2 ∈ Z. But 1, b2, 2a2 +4b are all integers. Hence 4a+4ab ∈ Z. This gives 16a2(1+b)2 is an integer. But a2 = -(2b + 1). Hence 16(2b + 1)(1 + b)2 is an integer. But 16(2b + 1)(1 + b)2 = 16(1 + 4b + 5b2 + 2b3). Hence 16b(4 + 2b2) is an integer. If b = 0, then b is an integer. Otherwise, this shows that b is a rational number. Because b2 ∈ Z, it follows that b is an integer. Since a2 = -(2b + 1), we get that a2 is an integer. Now 4a(1 + b) ∈ Z. If b ≠ -1, then a is rational and hence a is an integer. If b = -1, then we see that P(-1) = -a, P(0) = b = -1 and P(1) = a. Hence a2, 1, a2 is an AP. This implies that a2 = 1 and hence a = ±1. |
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