Problem: Let n be the number of integer values of x such that P=x4+6x3+11x2+3x+31 is the square of an integer. Then n is:
Answer Choices:
A. 4
B. 3
C. 2
D. 1
E. 0
Solution:
Let P=x4+6x3+11x2+3x+31=(x2+3x+1)2−3(x−10)=y2
∴(x2+3x+1)2−y2=3(x−10). When x=10,P=(x2+3x+1)2
=1312=y2. To prove that 10 is the only possible value we use the following lemma: If ∣N∣>∣M∣,N,M integers, then N2−M2≧2∣N∣−1 (This lemma is easy to prove, try it)
Case I If x>10, then 3(x−10)=(x2+3x+1)2−y2≧2∣∣∣x2+3x+1∣∣∣−1, an impossibility
Case II If x<10, then 3(10−x)=y2−(x2+3x+1)2≧2∣y∣−1>2∣∣∣x2+3x+1∣∣∣−1. This inequality holds for the integers x=2,1,0,−1,−2,−3,−4,−5,−6, but none of these values makes P the square of an integer.