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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...")
 
(→‎Solution)
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== Solution ==
 
== Solution ==
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<math>\fbox{C}</math>
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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>.
 +
 
 +
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>.
 +
 
 +
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>.
 +
 
 +
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>:
 +
 
 +
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>. 
 +
 
 +
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 ==

Revision as of 10:40, 12 July 2019

Problem

Let $D=a^2+b^2+c^2$, where $a$, $b$, are consecutive integers and $c=ab$. Then $\sqrt{D}$ is:

$\textbf{(A)}\ \text{always an even integer}\qquad \textbf{(B)}\ \text{sometimes an odd integer, sometimes not}\\ \textbf{(C)}\ \text{always an odd integer}\qquad \textbf{(D)}\ \text{sometimes rational, sometimes not}\\ \textbf{(E)}\ \text{always irrational}$

Solution

Let $a=x, b=x+1, c = x(x+1)$. Then $D = x^2 + (x+1)^2 + x^2(x+1)^2$, which simplifies to $x^4 + 2x^3 + 3x^2 + 2x + 1$.

From the options, we want to test if $D$ is always a perfect square. Because the polynomial expression for $D$ is quartic, if it is a perfect square, it would be the square of a quadratic expression. Thus, $D$ could be written in the form $(Ax^2 + Bx + C)^2$ for some $(A, B, C)$.

Setting $(Ax^2 + Bx + C)^2 = x^4 + 2x^3 + 3x^2 + 2x + 1$, we can compare coefficients. From the $x^4$ coefficient, we get $A = \pm 1$. Note that if $(Ax^2 + Bx^2 + C)^2$ works, so does $(-Ax^2 - Bx - C)^2$, so we can arbitrarily pick $A=1$.

We now have $(x^2 + Bx + C)^2= x^4 + 2x^3 + 3x^2 + 2x + 1$. Setting the cubic terms equal gives $2B = 2$, or $B = 1$. This leaves $(x^2 + x + C)^2 = x^4 + 2x^3 + 3x^2 + 2x + 1$. We can quickly inspect the constant term to determine that $C = \pm 1$. We reject $C = -1$, since the quadratic and linear terms won't match up, which leaves $(x^2 + x + 1)^2$ as the only possibility - and, in fact, it works.

Thus, $D$ is always the square of an integer - namely $x^2 + x + 1$. This in turn means that $\sqrt{D}$ is always rational, which leaves choices $A, B, C$ as the only possible correct answers.

The question now is whether $\sqrt{D}$, or $x^2 + x + 1$, is odd, even, or could be both. We have two cases for $x$:

If $x \equiv 0 \pmod 2$, then $x^2 \equiv 0^2 \pmod 2$. This means $x^2 + x + 1 \equiv 0 + 0 + 1 \equiv 1 \pmod 2$.

If $x \equiv 1 \pmod 2$, then $x^2 \equiv 1^2 \pmod 2$, and $x^2 + x + 1 \equiv 1 + 1 + 1 \equiv 1 \pmod 2$.

Either way, $x^2 + x + 1$ is an odd integer, and the answer is $\fbox{C}$

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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