# Polynomial Proof

## Homework Statement

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

##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
Homework Helper
Gold Member
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
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

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

##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 ?

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

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
Staff Emeritus
Homework Helper
Gold Member
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 ?

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
SammyS
Staff Emeritus