1967 AMC 12 Problem 31

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

Set

b=a+1

and expand

D

Try to recognize

D

as the square of

a^2+a+1

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.

Problem 31 in other years