1962 AMC 12 Problem 36

Question

If both

x

and

y

are integers, how many solutions are there to the equation

(x-8)(x-10)=2^y?

0

1

2

3

more than

3

Rewrite the left side as

(x-9)^2-1

If

n=x-9,

then the consecutive even factors

n-1

and

n+1

must both be powers of

2

Put

n=x-9.

Then

2^y=n^2-1=(n-1)(n+1).

Since the product is a power of

2,

both factors must have no odd prime divisor. The only consecutive even integers differing by

2

that are both signed powers of

2

are

(-4,-2)

and

(2,4).

Thus

n=\pm3

and

y=3,

giving

x=6

or

12.

There are

2

ordered pairs

(x,y).

Therefore, the correct answer is C.

Problem 36 in other years