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Difference between revisions of "1985 AHSME Problems/Problem 21"

(Created page with "==Problem== How many integers <math> x </math> satisfy the equation <math> (x^2-x-1)^{x+2}=1 </math> <math> \mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 4 \q...")
 
(Improved solution and formatting)
 
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==Problem==
 
==Problem==
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How many integers <math> x </math> satisfy the equation <math> (x^2-x-1)^{x+2}=1 </math>
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How many integers <math>x</math> satisfy the equation <cmath>\left(x^2-x-1\right)^{x+2} = 1?</cmath>
  
 
<math> \mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \  } 4 \qquad \mathrm{(D) \  } 5 \qquad \mathrm{(E) \  }\text{none of these} </math>
 
<math> \mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \  } 4 \qquad \mathrm{(D) \  } 5 \qquad \mathrm{(E) \  }\text{none of these} </math>
  
 
==Solution==
 
==Solution==
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Notice that any power of <math> 1 </math> is <math> 1 </math>, so <math> x^2-x-1=1 </math> would give valid solutions.  
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We recall that for real numbers <math>a</math> and <math>b</math>, there are exactly <math>3</math> ways in which we can have <math>a^b = 1</math>, namely <math>a = 1</math>; <math>b = 0</math> and <math>a \neq 0</math>; or <math>a = -1</math> and <math>b</math> is an even integer.
  
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<math> x^2-x-2=0 </math>
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The first case therefore gives <cmath>\begin{align*}x^2-x-1 = 1 &\iff x^2-x-2 = 0 \\&\iff (x-2)(x+1) = 0 \\&\iff x = 2 \text{ or } x = -1.\end{align*}</cmath>
  
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<math> (x-2)(x+1)=0 </math>
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Similarly, the second case gives <math>x+2 = 0</math>, i.e. <math>x = -2</math>, and this indeed gives <math>x^2-x-1 = 4+2-1 = 5 \neq 0</math>, so <math>x = -2</math> is a further valid solution.
  
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<math> x=2, -1 </math>
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Lastly, for the third case, we have <cmath>\begin{align*}x^2-x-1 = -1 &\iff x^2-x = 0 \\&\iff x(x-1) = 0 \\&\iff x = 0 \text{ or } x = 1,\end{align*}</cmath> but <math>x = 1</math> would give <math>x+2 = 3</math>, which is odd, whereas <math>x = 0</math> gives <math>x+2 = 2</math>, which is even. Therefore, this case gives only one further solution, namely <math>x = 0</math>.
  
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Also, <math> -1 </math> to an even power also gives <math> 1 </math>, so we check <math> x^2-x-1=-1 </math>
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Accordingly, the possible values of <math>x = -2</math>, <math>-1</math>, <math>0</math>, or <math>2</math>, yielding a total of <math>\boxed{\text{(C)} \ 4}</math> solutions.
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<math> x^2-x=0 </math>
 
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<math> x(x-1)=0 </math>
 
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<math> x=0, 1 </math>
 
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However, <math> x=1 </math> gives an odd power of <math> -1 </math>, so this is discarded. Finally, notice that anything to the <math> 0\text{th} </math> power (except for <math> 0 </math>, which is debatable) gives <math> 1 </math>.
 
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<math> x+2=0 </math>
 
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<math> x=-2 </math>
 
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This doesn't make <math> x^2-x-1=0 </math>, so this is also valid.
 
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Overall, our valid solutions are <math> x=-2, -1, 0, 2 </math> for a grand total of <math> 4, \boxed{\text{C}} </math>.
 
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=20|num-a=22}}
 
{{AHSME box|year=1985|num-b=20|num-a=22}}
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{{MAA Notice}}

Latest revision as of 22:29, 19 March 2024

Problem

How many integers $x$ satisfy the equation \[\left(x^2-x-1\right)^{x+2} = 1?\]

$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ } 5 \qquad \mathrm{(E) \ }\text{none of these}$

Solution

We recall that for real numbers $a$ and $b$, there are exactly $3$ ways in which we can have $a^b = 1$, namely $a = 1$; $b = 0$ and $a \neq 0$; or $a = -1$ and $b$ is an even integer.

The first case therefore gives \begin{align*}x^2-x-1 = 1 &\iff x^2-x-2 = 0 \\&\iff (x-2)(x+1) = 0 \\&\iff x = 2 \text{ or } x = -1.\end{align*}

Similarly, the second case gives $x+2 = 0$, i.e. $x = -2$, and this indeed gives $x^2-x-1 = 4+2-1 = 5 \neq 0$, so $x = -2$ is a further valid solution.

Lastly, for the third case, we have \begin{align*}x^2-x-1 = -1 &\iff x^2-x = 0 \\&\iff x(x-1) = 0 \\&\iff x = 0 \text{ or } x = 1,\end{align*} but $x = 1$ would give $x+2 = 3$, which is odd, whereas $x = 0$ gives $x+2 = 2$, which is even. Therefore, this case gives only one further solution, namely $x = 0$.

Accordingly, the possible values of $x = -2$, $-1$, $0$, or $2$, yielding a total of $\boxed{\text{(C)} \ 4}$ solutions.

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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