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Question: If x, y are integral solutions of $2x^2 - 3xy - 2y^2 = 7$, then the value of |x + y| is...

If x, y are integral solutions of 2x2−3xy−2y2=72x^2 - 3xy - 2y^2 = 7, then the value of |x + y| is

Answer

4

Explanation

Solution

The given equation is 2x2−3xy−2y2=72x^2 - 3xy - 2y^2 = 7. We seek integer solutions (x,y)(x, y).

The left side of the equation is a homogeneous quadratic expression in xx and yy. We can factor it: 2x2−3xy−2y2=(x−2y)(2x+y)2x^2 - 3xy - 2y^2 = (x - 2y)(2x + y).

The given equation becomes (x−2y)(2x+y)=7(x - 2y)(2x + y) = 7. Since xx and yy are integers, (x−2y)(x - 2y) and (2x+y)(2x + y) must also be integers. The integer factors of 7 are (1,7),(7,1),(−1,−7),(−7,−1)(1, 7), (7, 1), (-1, -7), (-7, -1). We set up systems of linear equations based on these factor pairs:

Case 1: x−2y=1x - 2y = 1 and 2x+y=72x + y = 7. Multiply the second equation by 2: 4x+2y=144x + 2y = 14. Add the first equation to this: (x−2y)+(4x+2y)=1+14  ⟹  5x=15  ⟹  x=3(x - 2y) + (4x + 2y) = 1 + 14 \implies 5x = 15 \implies x = 3. Substitute x=3x=3 into 2x+y=7  ⟹  2(3)+y=7  ⟹  6+y=7  ⟹  y=12x + y = 7 \implies 2(3) + y = 7 \implies 6 + y = 7 \implies y = 1. Solution: (x,y)=(3,1)(x, y) = (3, 1). For this solution, x+y=3+1=4x+y = 3+1 = 4. ∣x+y∣=∣4∣=4|x+y| = |4| = 4.

Case 2: x−2y=7x - 2y = 7 and 2x+y=12x + y = 1. Multiply the second equation by 2: 4x+2y=24x + 2y = 2. Add the first equation to this: (x−2y)+(4x+2y)=7+2  ⟹  5x=9  ⟹  x=9/5(x - 2y) + (4x + 2y) = 7 + 2 \implies 5x = 9 \implies x = 9/5. This is not an integer, so there is no integer solution in this case.

Case 3: x−2y=−1x - 2y = -1 and 2x+y=−72x + y = -7. Multiply the second equation by 2: 4x+2y=−144x + 2y = -14. Add the first equation to this: (x−2y)+(4x+2y)=−1+(−14)  ⟹  5x=−15  ⟹  x=−3(x - 2y) + (4x + 2y) = -1 + (-14) \implies 5x = -15 \implies x = -3. Substitute x=−3x=-3 into 2x+y=−7  ⟹  2(−3)+y=−7  ⟹  −6+y=−7  ⟹  y=−12x + y = -7 \implies 2(-3) + y = -7 \implies -6 + y = -7 \implies y = -1. Solution: (x,y)=(−3,−1)(x, y) = (-3, -1). For this solution, x+y=−3+(−1)=−4x+y = -3 + (-1) = -4. ∣x+y∣=∣−4∣=4|x+y| = |-4| = 4.

Case 4: x−2y=−7x - 2y = -7 and 2x+y=−12x + y = -1. Multiply the second equation by 2: 4x+2y=−24x + 2y = -2. Add the first equation to this: (x−2y)+(4x+2y)=−7+(−2)  ⟹  5x=−9  ⟹  x=−9/5(x - 2y) + (4x + 2y) = -7 + (-2) \implies 5x = -9 \implies x = -9/5. This is not an integer, so there is no integer solution in this case.

The integer solutions are (3,1)(3, 1) and (−3,−1)(-3, -1). For (3,1)(3, 1), x+y=4x+y = 4, and ∣x+y∣=4|x+y| = 4. For (−3,−1)(-3, -1), x+y=−4x+y = -4, and ∣x+y∣=4|x+y| = 4. In both cases, the value of ∣x+y∣|x+y| is 4.