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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??
Joffan said:You might try looking at ##x^2 - 2xy+y^2=6+3x##
Or ##(x-y)^2=3(x+2)## - which should be more interesting.mtayab1994 said:That's [tex](x-y)^{2}-3x=6[/tex]
Joffan said:Or ##(x-y)^2=3(x+2)## - which should be more interesting.
Sure, although we will find a better substitution... what can you tell me about ##n##?mtayab1994 said:Can we use substitution and say that x+2=n?
Joffan said:Sure, although we will find a better substitution... what can you tell me about ##n##?
Hmm, not really. Let's define ##m:=(x-y)## - what can you tell me about ##m##?mtayab1994 said:On the question before I proved that x^2 Ξ 0(mod3) and that means that x^2=3n.
Joffan said:Hmm, not really. Let's define ##m:=(x-y)## - what can you tell me about ##m##?
mtayab1994 said:That means that m is the difference of x and y.
Joffan said:My question is, what does ##m^2=3n## (using the new definitions) tell you about ##m##?
Yes... what does that mean for ##m##?mtayab1994 said:That means that m^2 is divisible by 3
What? No. For example, 36 is divisible by 3 but not by 15.mtayab1994 said:... hence divisible by all multiples of 3.
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.
vela said:Yes, but it doesn't follow that m2 is divisible by all multiples of 3, which is what you claimed.
Joffan said:My question is, what does ##m^2=3n## (using the new definitions) tell 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##.mtayab1994 said:Which new definitions m=x-y?
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##.
Looks good.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)
How did you come up with that conclusion based on what you wrote above?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?
Joffan said:What is the value of ##3n## mod 3?
The first step is to check if the equation is in the form of a linear Diophantine equation, where all variables are raised to the power of 1. If so, you can use the Extended Euclidean Algorithm to find the greatest common divisor (GCD) of the coefficients. If the GCD divides the constant term, then there are integer solutions. If not, then there are no integer solutions.
No, not all equations can be solved using number theory. In fact, there is no general algorithm to solve all Diophantine equations. Some equations may require more advanced techniques or may not have any integer solutions at all.
The simplest method is to use the Extended Euclidean Algorithm to find the GCD of the coefficients, and then use this GCD to find the smallest positive solution. Alternatively, you can use other techniques such as modular arithmetic or linear congruences to find the smallest positive solution.
Yes, negative integers can be considered as solutions to a Diophantine equation. In fact, equations with negative solutions are often easier to solve than those without. However, it is important to specify whether you are looking for only positive solutions or all integer solutions.
Integer solutions play a crucial role in number theory, as they can provide insights into the properties of numbers and their relationships. They can also be used to find patterns and make conjectures about more complex equations. Additionally, integer solutions have practical applications in fields such as cryptography and coding theory.