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If a, b are integers, and (a−b ) 2 +8 b 2 =108 (a−b)2+8

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by Jay@ManhattanReview » Thu May 18, 2017 9:02 am
ziyuenlau wrote:If a, b are integers, and (a−b)^2+8b^2=108, what is the number of the ordered pairs (a,b)?

A. 2
B. 4
C. 6
D. 8
E. 10

Source : Math Revolution
OA=D
We have (a−b)^2+8b^2=108.

Since a and b are integers, b^2 must be 0, 1, 4, 9, etc.

Let's plug-in the values of b^2 and find out whether (a-b)^2 is a perfect square.

1. b^2 = 0

=> (a-b)^2 = 108 - 0 = 108. Not probable as 108 is not a perfect square

2. b^2 = 1

=> (a-b)^2 = 108 - 8 = 100. Probable as 100 is a perfect square.

=> b = +/-1 and a = +/-9 and +/-11. There are four ordered pairs: (11, 1), (9, -1), (-9, 1), and (-11, -1)

3. b^2 = 4

=> (a-b)^2 = 108 - 32 = 76. Not probable as 76 is not a perfect square

4. b^2 = 9

=> (a-b)^2 = 108 - 72 = 36. Probable as 36 is a perfect square.

=> b = +/-3 and a = +/-3 and +/-9. There are four ordered pairs: (9, 3), (3, -3), (-3, 3), and (-9, -3)

5. b^2 = 16

=> (a-b)^2 = 108 - 128 = -20. Not a probable one as a perfect square cannot be nagative.

There are eight ordered pairs.

The correct answer: D

Hope this helps!

-Jay
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by [email protected] » Thu May 18, 2017 2:06 pm
Hi ziyuenlau,

While this question 'looks' complex, it can be organized into some simple ideas; those ideas - along with a bit of BRUTE FORCE arithmetic - can get you to the solution without too much trouble.

We're told that (A - B)^2 + 8(B^2) = 108 and that A and B are both INTEGERS.

In simple terms:

(A perfect square) + 8(A perfect square) = 108

Since perfect squares cannot be negative, and that second perfect square is multiplied by EIGHT, there can't be that many solutions to this equation. Let's focus on that second perfect square....If the first perfect square was 0, then we'd have....

0 + 8(a perfect square) = 108
(a perfect square) = 108/8 = 13.5

Since B is an INTEGER, we know that this equation is NOT possible - but we can use it to determine that that perfect square is less than 13.5. Thus, we really only have to think about 0, 1, 4 and 9 for that second perfect square. Let's start 'big' and work our way down...

IF.... B = 3 or -3, then we have...
(A-B)^2 + 72 = 108
(A-B)^2 = 36
(A-B) = +6 or -6

Thus, we could have FOUR solutions here:
A = 9, B = 3
A = -3, B = 3
A = -9, B = -3
A = 3, B = -3

IF.... B = 2 or -2, then we have...
(A-B)^2 + 32 = 108
(A-B)^2 = 76

76 is NOT a perfect square though, so there are no solutions here.

IF.... B = 1 or -1, then we have...
(A-B)^2 + 8 = 108
(A-B)^2 = 100
(A-B) = +10 or -10

Thus, we could have FOUR solutions here:
A = 11, B = 1
A = -9, B = 1
A = 9, B = -1
A = -11, B = -1

IF.... B = 0, then we have...
(A-0)^2 + 0 = 108
(A)^2 = 108

108 is NOT a perfect square though, so there are no solutions here.

Total solutions = four + four = 8

Final Answer: D

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