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Latest revision as of 00:40, 16 August 2023

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 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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