Pairing
How many pairs , where and are real numbers, are there such that whenever is a root of the equation , is also a root of the equation?
The answer is 6.
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1 solution
Consider the equation x+ax+b=0. It has two roots(not necessairly real), say 'c' and 'd'. Either c=d or c≠d. CASE 1: Suppose c=d, so that c is a double root. Since c-2 is also a root, the only possibility is c=c²-2 This reduce to (c+1)(c-2)=0. Hence c=-1 or c=2. Observe that a=-2c and b=c² thus ab=(2,1) or (-4,4). CASE 2 : Suppose c≠d. There are four possibilities; (1)c=c²-2 and d=d²-2 (2)c=d²-2 and d=c²-2 (3)c=c²-2=d²-2 and c≠d (4)d=c²-2=d²-2 and c≠d. (1)Here (c, d) =(2,-1)or(-1,2). Hence (a, b) =(-(c+d), cd) =(-1,-2). (2)suppose c=d²-2and d=c²-2. Then c-d=d²-c²=(d-c)(d+c). Since c≠d, we get d+c=-1. However, we also have c+d =d²+c²-4=(c+d)²-2cd-4. Thus -1=1-2cd-4, which implies that cd=-1. Therefore (a, b)=(-(c+d), cd)=(1,-1). (3)If c=c²-2=d²-2 and c≠d, then c=-d. Thus c=2, d=-2 or c=-1,d=1. In such case (a, b)=(0,-4) and (0,-1). (4)Note that d=c²-2=d²-2 and c≠d is identical to (3),so that we get exactly same pairs (a, b). Thus we get 6 pairs; (a,b)=(-4,4), (2,1),(-1,-2),(1,-1),(0,-4),(0,-1).