Difference between revisions of "1967 AHSME Problems/Problem 31"
(Created page with "== Problem == Let <math>D=a^2+b^2+c^2</math>, where <math>a</math>, <math>b</math>, are consecutive integers and <math>c=ab</math>. Then <math>\sqrt{D}</math> is: <math>\textbf...") |
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== Solution == | == Solution == | ||
− | <math>\fbox{C}</math> | + | Let <math>a=x, b=x+1, c = x(x+1)</math>. Then <math>D = x^2 + (x+1)^2 + x^2(x+1)^2</math>, which simplifies to <math>x^4 + 2x^3 + 3x^2 + 2x + 1</math>. |
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+ | From the options, we want to test if <math>D</math> is always a perfect square. Because the polynomial expression for <math>D</math> is quartic, if it is a perfect square, it would be the square of a quadratic expression. Thus, <math>D</math> could be written in the form <math>(Ax^2 + Bx + C)^2</math> for some <math>(A, B, C)</math>. | ||
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+ | Setting <math>(Ax^2 + Bx + C)^2 = x^4 + 2x^3 + 3x^2 + 2x + 1</math>, we can compare coefficients. From the <math>x^4</math> coefficient, we get <math>A = \pm 1</math>. Note that if <math>(Ax^2 + Bx^2 + C)^2</math> works, so does <math>(-Ax^2 - Bx - C)^2</math>, so we can arbitrarily pick <math>A=1</math>. | ||
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+ | We now have <math>(x^2 + Bx + C)^2= x^4 + 2x^3 + 3x^2 + 2x + 1</math>. Setting the cubic terms equal gives <math>2B = 2</math>, or <math>B = 1</math>. This leaves <math>(x^2 + x + C)^2 = x^4 + 2x^3 + 3x^2 + 2x + 1</math>. We can quickly inspect the constant term to determine that <math>C = \pm 1</math>. We reject <math>C = -1</math>, since the quadratic and linear terms won't match up, which leaves <math>(x^2 + x + 1)^2</math> as the only possibility - and, in fact, it works. | ||
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+ | Thus, <math>D</math> is always the square of an integer - namely <math>x^2 + x + 1</math>. This in turn means that <math>\sqrt{D}</math> is always rational, which leaves choices <math>A, B, C</math> as the only possible correct answers. | ||
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+ | The question now is whether <math>\sqrt{D}</math>, or <math>x^2 + x + 1</math>, is odd, even, or could be both. We have two cases for <math>x</math>: | ||
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+ | If <math>x \equiv 0 \pmod 2</math>, then <math>x^2 \equiv 0^2 \pmod 2</math>. This means <math>x^2 + x + 1 \equiv 0 + 0 + 1 \equiv 1 \pmod 2</math>. | ||
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+ | If <math>x \equiv 1 \pmod 2</math>, then <math>x^2 \equiv 1^2 \pmod 2</math>, and <math>x^2 + x + 1 \equiv 1 + 1 + 1 \equiv 1 \pmod 2</math>. | ||
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+ | Either way, <math>x^2 + x + 1</math> is an odd integer, and the answer is <math>\fbox{C}</math> | ||
== See also == | == See also == |
Latest revision as of 10:40, 12 July 2019
Problem
Let , where , , are consecutive integers and . Then is:
Solution
Let . Then , which simplifies to .
From the options, we want to test if is always a perfect square. Because the polynomial expression for is quartic, if it is a perfect square, it would be the square of a quadratic expression. Thus, could be written in the form for some .
Setting , we can compare coefficients. From the coefficient, we get . Note that if works, so does , so we can arbitrarily pick .
We now have . Setting the cubic terms equal gives , or . This leaves . We can quickly inspect the constant term to determine that . We reject , since the quadratic and linear terms won't match up, which leaves as the only possibility - and, in fact, it works.
Thus, is always the square of an integer - namely . This in turn means that is always rational, which leaves choices as the only possible correct answers.
The question now is whether , or , is odd, even, or could be both. We have two cases for :
If , then . This means .
If , then , and .
Either way, is an odd integer, and the answer is
See also
1967 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 30 |
Followed by Problem 32 | |
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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