Number theory: finding integer solution to an equation

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mtayab1994
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Homework Statement



(E): x^2+y^2=6+2xy+3x

The Attempt at a Solution



[tex]x^{2}+y^{2}=6+2xy+3x\Longleftrightarrow x^{2}-2xy-3x+y^{2}=6\Longleftrightarrow x^{2}+x(-2y-3)+y^{2}=6[/tex]

Any further help to find the answer??
 
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You might try looking at ##x^2 - 2xy+y^2=6+3x##
 
Joffan said:
You might try looking at ##x^2 - 2xy+y^2=6+3x##

That's [tex](x-y)^{2}-3x=6[/tex]
 
mtayab1994 said:
That's [tex](x-y)^{2}-3x=6[/tex]
Or ##(x-y)^2=3(x+2)## - which should be more interesting.
 
Joffan said:
Or ##(x-y)^2=3(x+2)## - which should be more interesting.

Can we use substitution and say that x+2=n?
 
mtayab1994 said:
Can we use substitution and say that x+2=n?
Sure, although we will find a better substitution... what can you tell me about ##n##?
 
Joffan said:
Sure, although we will find a better substitution... what can you tell me about ##n##?

On the question before I proved that x^2 Ξ 0(mod3) and that means that x^2=3n.
 
mtayab1994 said:
On the question before I proved that x^2 Ξ 0(mod3) and that means that x^2=3n.
Hmm, not really. Let's define ##m:=(x-y)## - what can you tell me about ##m##?
 
Joffan said:
Hmm, not really. Let's define ##m:=(x-y)## - what can you tell me about ##m##?

That means that m is the difference of x and y.
 
mtayab1994 said:
That means that m is the difference of x and y.

My question is, what does ##m^2=3n## (using the new definitions) tell you about ##m##?
 
Joffan said:
My question is, what does ##m^2=3n## (using the new definitions) tell you about ##m##?

That means that m^2 is divisible by 3 hence divisible by all multiples of 3.
 
mtayab1994 said:
That means that m^2 is divisible by 3
Yes... what does that mean for ##m##?
mtayab1994 said:
... hence divisible by all multiples of 3.
What? No. For example, 36 is divisible by 3 but not by 15.
 
vela said:
Is that what you meant to say? 81 is divisible by 3, but it's not divisible by 6, which is a multiple of 3.

If we said that m=x-y then m^2=(x-y)^2=3n .
So then we get that (x-y)^2=3n right?
 
vela said:
Yes, but it doesn't follow that m2 is divisible by all multiples of 3, which is what you claimed.


But how are we supposed to find integer solutions out of that? I found that y=n^2-3n-2 and that's wrong I think.
 
Joffan said:
My question is, what does ##m^2=3n## (using the new definitions) tell you about ##m##?

Which new definitions m=x-y?
 
mtayab1994 said:
Which new definitions m=x-y?
Don't backtrack. We have defined new variables ##m## and ##n##; the formula translates into those variables as shown; now you need to understand what ##m^2=3n## tells you about ##m##.
 
Joffan said:
Don't backtrack. We have defined new variables ##m## and ##n##; the formula translates into those variables as shown; now you need to understand what ##m^2=3n## tells you about ##m##.

Well there are 3 cases:

Case 1: if m Ξ 0(mod3) then m^2 Ξ 0(mod3)

Case 2: if m Ξ 1(mod3) then m^2 Ξ 1(mod3)

Case 3: if m Ξ 2(mod3) then m^2 Ξ 4(mod3) with is m^2 Ξ 1(mod3)

So that means that m^2=3n . So that means the when m is divided by 3 you get either a remainder of 0,1, or 2 am i right?
 
mtayab1994 said:
Well there are 3 cases:

Case 1: if m Ξ 0(mod3) then m^2 Ξ 0(mod3)

Case 2: if m Ξ 1(mod3) then m^2 Ξ 1(mod3)

Case 3: if m Ξ 2(mod3) then m^2 Ξ 4(mod3) with is m^2 Ξ 1(mod3)
Looks good.

So that means that m^2=3n . So that means the when m is divided by 3 you get either a remainder of 0,1, or 2 am i right?
How did you come up with that conclusion based on what you wrote above?
 
What is the value of ##3n## mod 3?
 
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Joffan said:
What is the value of ##3n## mod 3?

3n mod 3 means that 3n=3k so that means 3k equals the multiples of 3 which are 3n.