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Algebra Difficulty 4.1 AIME Prove it Canada

A sequence is created in such a way that

a real number is chosen as the first number in the sequence,
and
each of the following numbers in the sequence is generated by
applying a function to the previous number in the sequence.

For example, if the first number in a sequence is 1 and the following
numbers are generated by the function $x^2-5$ , then the first three numbers in the sequence are $1, -4$ and $11$ since $1^2 - 5 = -4$ and $(-4)^2 - 5 = 11$ .

The first number in a sequence is $3$ and the sequence is generated by the function $x^2-3x+1$ . What are the first four numbers in the sequence? The number 7 is the third number in a sequence generated by the function $x^2-4x+7$ . What are all possible first numbers in the sequence? The first number in a sequence is $c$ and the sequence is generated by the function $x^2 - 7x - 48$ . If all numbers in the sequence are equal to $c$ , determine all possible values of $c$ . A sequence generated by the function $x^2 - 12x + 39$ alternates between two different numbers. That is, the sequence is $a,b,a,b,a,b,\dots$ , with $a \neq b$ . Determine all possible values of $a$ .

Solution

If the first number in a sequence is 3 and the sequence is
generated by the function $x^2-3x+1$ , then the second number in the sequence is $3^2-3(3)+1=1,$ and the third number in the sequence is $1^2-3(1)+1=-1,$ and the fourth number in the sequence is $(-1)^2-3(-1)+1=5.$ The first four numbers in the sequence are $3, 1, -1,
5. Let the first and second numbers in the sequence generated by the function

Figure for this problem

Figure for this problem x^2-4x+7 $be$ f $and$ s, respectively. Then, the first three numbers in the sequence are

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Figure for this problem f,s,7. Since the third number in the sequence is 7, then

Figure for this problem

Figure for this problem s^2-4s+7=7 $. Solving this equation, we get$ s^2-4s=0 $or$ s(s-4)=0 $, which has solutions$ s=0 $and$ s=4, and so the first three numbers in the sequence could be

Figure for this problem

Figure for this problem f,0,7 $or$ f,4,7. If the second number in the sequence is 0, then

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Figure for this problem f^2-4f+7=0. The discriminant of this equation is

Figure for this problem

Figure for this problem (-4)^2-4(1)(7)=-12 (less than zero) and so there are no real solutions. Thus, there is no first number in this sequence for which the second number is 0. If the second number in the sequence is 4, then

Figure for this problem

Figure for this problem f^2-4f+7=4 $, or$ f^2-4f+3=0 $and so$ (f-1)(f-3)=0 $, which has solutions$ f=1 $and$ f=3. Therefore, if 7 is the third number in a sequence generated by the function

Figure for this problem

Figure for this problem x^2-4x+7, then the first three numbers in the sequence could be

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Figure for this problem 1,4,7 $or$ 3,4,7, and so the possible first numbers in the sequence are 1 and 3. The first two numbers in the sequence are

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Figure for this problem c,c $, and so$ c^2-7c-48=c $. Solving this equation, we get$ c^2-8c-48=0 $or$ (c+4)(c-12)=0 $, which has solutions$ c=-4 $and$ c=12. The first number in the sequence is

Figure for this problem

Figure for this problem a and the second number is

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Figure for this problem b $, and so$ $a^2-12a+39=b.$ The second number in the sequence is

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Figure for this problem b and the third number is

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Figure for this problem a $, and so$ $b^2-12b+39=a.$ Subtracting the second equation from the first and simplifying, we get

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Figure for this problem $\text{(a 2-12a+39)-(b 2-12b+39) =b-a a 2-b 2-12a+12b =b-a a 2-b 2-11a+11b =0 (a-b)(a+b)-11(a-b) =0 (a-b)(a+b-11) =0}$ $Since$ $a\neq$ b $, then$ $a-b\neq$ 0 $and so$ a+b-11=0 $or$ b=11-a. Substituting into the first equation, we get

Figure for this problem

Figure for this problem a^2-12a+39=11-a $or$ a^2-11a+28=0 $. Factoring gives$ (a-4)(a-7)=0 $and so the possible values of$ a are 4 and 7. (Note that the two possible sequences are

Figure for this problem

Figure for this problem 4,7,4, $\dots$ $and$ $7,4,7,\dots\$.)$

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