# Number theory: finding integer solution to an equation

## Homework Statement

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

## The Attempt at a Solution

$$x^{2}+y^{2}=6+2xy+3x\Longleftrightarrow x^{2}-2xy-3x+y^{2}=6\Longleftrightarrow x^{2}+x(-2y-3)+y^{2}=6$$

Any further help to find the answer??

## Answers and Replies

You might try looking at ##x^2 - 2xy+y^2=6+3x##

You might try looking at ##x^2 - 2xy+y^2=6+3x##

That's $$(x-y)^{2}-3x=6$$

That's $$(x-y)^{2}-3x=6$$
Or ##(x-y)^2=3(x+2)## - which should be more interesting.

Or ##(x-y)^2=3(x+2)## - which should be more interesting.

Can we use substitution and say that x+2=n?

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##?

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.

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##?

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.

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##?

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.

vela
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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.

That means that m^2 is divisible by 3
Yes... what does that mean for ##m##???
... hence divisible by all multiples of 3.
What??? No. For example, 36 is divisible by 3 but not by 15.

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
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Yes, but it doesn't follow that m2 is divisible by all multiples of 3, which is what you claimed.

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.

vela
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At this point, you're going to have to think about it a bit on your own. You've got all the pieces. You just need that last little insight, which is what Joffan's been trying to get you to see.

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

Which new definitions m=x-y???

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##.

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?

vela
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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?

Last edited:
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.