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Solutions to 46(2) HEXA Challenge Problems

Challenge 1:

The sequence n 1 , n 2 , n 3 … is defined as the positive integers such that

is a perfect square. Find the first five terms in this sequence.

SOLUTION: From the definition of the factorial, ( n +1)!= n !( n +1), and so we may write n !( n +1)!/3 = ( n !) 2 ( n +1)/3 = M 2 , where M is a positive integer. Since ( n !) 2 and M 2 are both perfect squares, the equality may only hold if ( n +1)/3 = k 2 for some positive integer k . That is, n =3 k 2 −1. Therefore, the first five terms would be n 1 =2, n 2 =11, n 3 =26, n 4 =47, and n 5 =74. This grows very fast! (74!75!)/3 ≈ 2.736×10 216 . Challenge 2: Depending on the value of a , the parabola y = x 2 + a can intersect the circle x 2 + y 2 =4 at zero, one, two, three, or four points. Determine the number of points of intersection as a function of a .

SOLUTION: Substituting the equation of the parabola into the equation of the circle yields

x 2 +( x 2 + a ) 2 =4

x 4 +(1+2 a ) x 2 + a 2 −4=0

This equation is biquadratic. That is, a quadratic equation for x 2 , and so its four solutions satisfy

These may be real or complex depending on the value of a. A necessary condition for any of them to be real is 17+4 a ≥0, meaning that if a <−17/4, there are no points of intersection. (The entire circle is above the parab ola.) If a =−17/4, there are two distinct solutions x =± √(15 )/2, each with multiplicity 2 and with corresponding y - values of −1/2. For a > −17/4, all four solutions are real as long as a <−2, giving four points of intersection. When a =−2, two of the points collide at ( x,y )=(0,−2). For this value, the bottom of the circle intersects the bottom of the parab ola, and two other intersection points are at (± √(3),1) for a total of three. For a > −2, the vertex of the parabola moves into the interior of the circle, leaving only two intersection points with

Virginia Mathematics Teacher vol. 47, no. 1

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