Question

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

always an even integer

sometimes an odd integer, sometimes not

always an odd integer

sometimes rational, sometimes not

always irrational

Answer: C

Concepts:algebraic manipulation parity perfect square

Difficulty rating: 1500

Small Hint:

Set $b=a+1$ and expand $D$

Big Hint:

Try to recognize $D$ as the square of $a^2+a+1$

Solution:

With $b=a+1$ and $c=a(a+1),$ $\begin{aligned} D&=a^2+(a+1)^2+a^2(a+1)^2\\ &=(a^2+a+1)^2. \end{aligned}$ Thus $\sqrt D=a^2+a+1.$ Since $a(a+1)$ is even, this integer is always odd.

Therefore, the correct answer is C.

1967 AMC 12 Problem 31

Answer: C

Difficulty rating: 1500

Small Hint:

Big Hint:

Solution:

Problem 31 in Other Years