Proof of No Solution for x^2 - 3xy + 2y^2 = 10 Conjecture | Polynomial Homework

[fishturtle1]

Homework Statement

Prove or refute the following conjecture: There are no positive integers x and y such that $x^2 - 3xy + 2y^2 = 10$

Homework Equations

$10 = 5*2$
$10 = 10*1$

The Attempt at a Solution

I graphed it using a graphing calculator, so I know this is true.

Proof: This will be a proof by contradiction. Suppose $x$ and $y$ are positive integers and $x^2 - 3xy + 2y^2 = 10$. By factoring, we have $(x-2y)(x-y) = 10$.

im not sure how to get further..
Like I did in the previous problem, I tried to set $x-2y = 10$, then $x-y = 10 + y$ but I don't think I can get a contradiction on this path

 

[Merlin3189]

Well, I'm not very good at maths, but haven't you proved there are integer values for x and y?
If x=8 and y=3, $x^2 -3xy + 2x^2 = 64 - 72 + 18 = 10$

 

[SammyS]

Homework Statement

Prove or refute the following conjecture: There are no positive integers x and y such that $x^2 - 3xy + 2y^2 = 10$

Homework Equations

$10 = 5*2$
$10 = 10*1$

The Attempt at a Solution

I graphed it using a graphing calculator, so I know this is true.

Proof: This will be a proof by contradiction. Suppose $x$ and $y$ are positive integers and $x^2 - 3xy + 2y^2 = 10$. By factoring, we have $(x-2y)(x-y) = 10$.

im not sure how to get further..
Like I did in the previous problem, I tried to set $x-2y = 10$, then $x-y = 10 + y$ but I don't think I can get a contradiction on this path

If x and y are both positive, which is larger, x − y, or x − 2y ?

 

[fishturtle1]

Well, I'm not very good at maths, but haven't you proved there are integer values for x and y?
If x=8 and y=3, $x^2 -3xy + 2x^2 = 64 - 72 + 18 = 10$

Thanks for the response, I think you did the proving when made x=8 and y=3.

I made a mistake.. I'm not sure what the x intercepts represented when I graphed it..

I think since I couldn't just see it, I could have set $x-2y = 1,2,5,$ or $10$ and then set $x-y = 1,2,5$ or $10$ depending on the $x-2y$ value.. I did this and solved for x and y and just discarded the negative values, and got your answer

 

[fishturtle1]

If x and y are both positive, which is larger, x − y, or x − 2y ?

then x-y is larger.. so from my above post, I would automatically know x-y =5 or x-y = 10

Ok so if x-y = 10 then x-2y = 1. Solving these equations gives x = 19 and y = 9. We confirm this by (19-9)(19-2(9)) = 10*1 = 10. So this is one solution.

If we let x-y = 5 then x-2y = 5. Solving these equations gives x = 8 and y = 3. We confirm this by (8-3)(8-(3(2)) = 5*2 = 10.

So there are two solutions to this equation where x and y are positive integers.

 

[SammyS]

then x-y is larger.. so from my above post, I would automatically know x-y =5 or x-y = 10

Ok so if x-y = 10 then x-2y = 1. Solving these equations gives x = 19 and y = 9. We confirm this by (19-9)(19-2(9)) = 10*1 = 10. So this is one solution.

If we let x-y = 5 then x-2y = 5. Solving these equations gives x = 8 and y = 3. We confirm this by (8-3)(8-(3(2)) = 5*2 = 10.

So there are two solutions to this equation where x and y are positive integers.

Could there possibly be one or two more set of solutions?

The factor pairs could also be −5, −2, and − 10, − 1 . Can either of these occur with both x and y being positive ?

 

[fishturtle1]

Could there possibly be one or two more set of solutions?

The factor pairs could also be −5, −2, and − 10, − 1 . Can either of these occur with both x and y being positive ?

I hadn't thought about that,

if we let x-2y = -5 and x-y = -2 then x = 1 and y = 3 which is a solution to the problem. (1-6)(1-3) = (-5)(-2) = 10

if we let x-2y = -10 and x-y = -1, then x = 8 and y = 9 which is a solution. (8-18)(8-9) = (-10)(-1) = 10

I forgot to add my actual answer.. since there are 4 solutions, I could use any of them to refute this..

Proof: Let x = 1 and y = 3. Then x and y are positive integers. Then $x^2 -3xy + 2y^2 = (1)^2 -3(1)(3) + 2(3)^2 = 1 - 9 + 18 = -8 + 18 = 10$. Therefore there does exist positive integers x and y such that $x^2 -3xy + 2y^2 = 10$. []

 

[SammyS]

Excellent !