Verified answer

3x² - 8y² + 3x²y² = 2008

3x² (1 + y²) - 8y² = 2008

3x² (1 + y²) - 8(1 + y²)= 2000

(3x² - 8)(1 + y²) = 2000

2000 = (2000)(1) = (1000)(2) = (500)(4) = (400)(5) = (200)(10) = (100)(20) = (50)(40) = (25)(80) = (5)(400)

There are some choices we can easily rule out.

(500)(4), (100)(20)

We are left with (2000)(1), (1000)(2), (400)(5), (200)(10), (50)(4) (25)(80).

3x² - 8 = 2000 doesn't work

3x² - 8 = 1000 is impossible hence (1000)(2) doesn't work.

3x² - 8 = 400

3x² = 408

x² = 136

This is impossible

3x² - 8 = 200

3x² = 208

This is also impossible.

3x² - 8 = 40

3x² = 48

x² = 16

x = 4

Of course, y² = 49, y = 7.

x = 4, y = 7, xy = 28 is the answer.

3x² - 8 = 25

3x² = 33

x² = 11

11 is not a perfect square hence (25)(80) is not correct.

ANSWER: 28

This may look tedious, but the guessing and checking requires little time. Some people may be able to rule out even more choices just by experience.

If you look at the AIME, there are similar problems except the guessing and checking is less tedious.

A similar problem, but much simpler is

Find all solutions (x,y) such that xy + x + y = 70.

(x + 1)(y + 1) = 71

x + 1 = 1

y + 1 = 71

x = 0, y = 70

The reverse also works.

x = 70, y = 0

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EDIT:

Approach 2:

3x² - 8y² + 3x²y² = 2008

x² (3 + 3y²) - (8y² + 2008) = 0

x = (√ (16(3 + 3y²)(2y² + 502))) / 2

6(y² + 1)(y² + 251) is a perfect square

Keep checking and you'll find it.

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EDIT:

AIME 1987 - Problem # 5:

Find 3x²y² if x and y are integers such that y² + 3x²y² = 3x² + 517.

3x² - 8y² + 3x²y²= 2008

=> 3x²(1 + y²) - 8(1 + y²) = 2008

=> (1 + y²)(3x² - 8) = 2000

=> 3x² - 8 = 2000/(1 + y²)

For LHS to be an integer, y = 2, 3 or 7

[Note : Trying all possible y upto [ √(2000) ], i.e., 1 to 44]

y = 2 => 3x² - 8 = 2000/5 = 400 which does not give interger value of x

y = 3 also does not give x as an integer

y = 7 => 3x² - 8 = 2000/50 = 40 => x = 4

=> xy = 4*7 = 28.

3x² - 8y² + 3x²y² = 2008

3x²y² + 3x² = 8y² + 2008

3x²(y² + 1) = 8(y² + 251)

3x²(y² + 1) = 8(y² + 1 + 250)

(3/8)x²[y² + 1] = y² + 1 + 250

(3/8)x²[y² + 1] - (y² + 1) = 250

[(3/8)x² - 1][y² + 1] = 250

y² + 1 is one of the integer factors of 250, so (3/8)x² - 1 will be the other.

trying case by case, we find

y² + 1 = 50, y² = 49, y = 7 and

(3/8)x² - 1 = 5, (3/8)x² = 6, x² = 16, x = 4 works.

so xy = 4(7) = 28

Rewrite it as:

3x²(1 + y²) = 8(251 + y²)

Now y² ≡ 0 or 1 (mod 4) so 1+y² ≡ 1 or 2 (mod 4).

This means that x² is an even number in order for the left side to be divisible by 8. Since this also means x is even, write x = 2k.

3k²(1 + y²) = 2(251 + y²) = 500 + 2(1 + y²)

3k² = 500 / (1+y²) + 2

shows that 1 + y² divides 500 = 4*125.

Divisors of 500 are:

1,2,4,5,10,20,25,50,100,125,250,500

so y² equals one of

0,1,3,4,9,19,24,49,99,124,249,499

y is then one of:

1, 2, 3, 7

Case 1: y=1

3k² = 250 + 2 gives k² = 84 =><=

Case 2: y=2

3k² = 100 + 2 gives k² = 34 =><=

Case 3: y=3

3k² = 50 + 2 gives k² = 52/3 =><=.

Case 4: y=7

3k² = 10 + 2 gives k² = 4 and k = 2.

So x = 2k = 4 and y = 7.

This gives xy = 28.

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Coincidentally, if we remove the restriction that x and y be positive (of course, they still need to be integers for the argument to work) then the only additional answer is -28.