MHB How Can I Solve This Bilinear Integer Equation?

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The discussion centers on solving the bilinear integer equation 2x^2 - 3xy - 2y^2 = 7. The original poster attempts to factor the equation but makes an error in their factorization. A participant points out the mistake and suggests that the correct factorization should lead to integer solutions. They note that since the product of two integers equals 7, one must be ±1 and the other ±7. The conversation concludes with the original poster acknowledging the correction and seeking further assistance.
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If $x,y$ are integer ordered pair of $2x^2-3xy-2y^2 = 7,$ Then $\left|x+y \right| = $

My Try:: Given $2x^2-3xy-2y^2 = 7\Rightarrow 2x^2-4xy+xy-2y^2 = 7$

So $(2x-y)\cdot (x-2y) = 7.$

Now How can I solve after that

Help me

Thanks
 
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jacks said:
If $x,y$ are integer ordered pair of $2x^2-3xy-2y^2 = 7,$ Then $\left|x+y \right| = $

My Try:: Given $2x^2-3xy-2y^2 = 7\Rightarrow 2x^2-4xy+xy-2y^2 = 7$

So $(2x-y)\cdot (x-2y) = 7.$

Now How can I solve after that

Help me

Thanks


You are just a little bit off. You can check your work. Let's try that here.

$(2x-y)\cdot (x-2y) = 2x^2 - 4xy -xy + 2y^2 = 2x^2 -5xy + 2y^2 \neq 2x^2 - 3xy - 2y^2$.

Knowing this, what do you think the correct factorization is?
 
Once you have got the correct factorisation sorted out, notice that if the product of two integers is $7$, then one of them must be $\pm1$ and the other one must be $\pm7$.
 
Thanks Aryth, opalg Got it.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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