x² - xy + y² = x + y ==> (x + y)² - 3xy = x + y Making x + y = k ==> y = k - x (eq. 1) Then k² - 3x(k - x) = k ==> k² - (3x +1) + 3x² = 0 Solving the quadratic equation in k: k = [3x + 1 ± √((3x + 1)² - 12x²)] / 2 k = [3x + 1 ± √(-3x² + 6x + 1)] / 2 (eq.2) For k to be real, it is necessary that - 3x² + 6x + 1 ≥ 0 The only integer solutions for x are x = 0, x = 1 and x = 2 For x = 0 (from eq.2): k = [1 ± √1] / 2 => k = 0 or k = 1 From eq.1: y = 0 or y = 1 ==> (0 , 0) , (0 , 1) For x = 1: k = [4 ± √4] / 2 => k = 1 or k = 3 From eq.1: y = 0 or y = 2 ==> (1 , 0) , (1 , 2) For x = 2: k = [7 ± √1] / 2 => k = 3 or k = 4 From eq.1: y = 1 or y = 2 ==> (2 , 1) , (2 , 2)
I used the following method: i turned it to the quadratic equation of x with y as a parameter and calculated the determinant as a function of y (which turned to be quadratic function). SInce we need real sol. the determinant must be greater or equal to 0, that gave me 3 possible integer values for y (0, 1, 2). Then I found x in each of those cases. Imo much simpler and easier to come up with method
x(x-1) + y(y-1-x) = 0 i.e (x-(y+1)/2)^2= ((y+1)/2)^2 +y(1-y) Term at the RHS is rewritten as (y*y + 2y +1 +4y -4y*y )/4 = (1 +6y -3y*y)/4 = (4 - 3(y-1)^2 )/4 This essentially means 4(x-(y+1)/2)^2 + 3(y-1)^2 = 4 or (2x- y-1)^2 + 3(y-1)^2= 4 Here x, y being integer each of these two square term is square number case I [ (2x- y-1)^2 = 1 and (y-1)^2=1] Hereby 2x-y-1 = 1,. y -1 = 1 i.e y = 2, x= 2 2x-y-1 = 1, y-1 = -1 i.e. y = 0, x=1 2x-y-1 = -1,. y -1 = 1 i.e y = 2, x= 1 2x-y-1 = -1, y-1 = -1 i.e. y = 0, x=0 The solution set of (x,y) for case I is { (0,0), (1,0), (1,2), (2,2)} case II [(2x- y-1)^2 = 4 and (y-1)^2=0] Hereby 2x-y-1 = 2, y-1 = 0 i.e. y = 1, x=2 2x-y-1 = -2,. y -1 = 0 i.e y = 1, x= 0 The solution set of (x,y) for case II is { (0,1), (2,1)} Hereby the solution set for (x,y).is { (0,0), (1,0), (0,1),(1,2), (2,1), (2,2)}
How I did it: Make x the parameter and complete the square. With a bit of cleanup you can rewrite the original equation as (2x - y - 1)^2 = -(3y^2 - 6y - 1) Because the LHS is a square, 3y^2 - 6y - 1
Diophantine Equations are a fun and exciting topic in Number Theory because you have more variables than equations and solution method is not always very straightforward. In these equations, we are looking for integer solutions. Some equations have no solutions or no non-trivial solutions. Some have infinitely many solutions such as the Pythagorean Triples in a^2+b^2=c^2. What if a^3+b^3=b^3 or a^4+b^4=c^4? This brings us to Fermat's Last Theorem whose proof is way too complicated! This particular problem can be considered somewhat easy but it still involves some manipulations. There are many methods that can be used to solve Diophantine Equations one of which is Modular Arithmetic. Factoring also plays an important role. Any thoughts?
A fairly simple solution is to express the original equation as a quadratic in x. Then we use the quadratic formula to get x in terms of y. The determinant has to be a perfect square, which is only possible for y = 0,1,2. Then for each case, you can find corresponding x values.
@@soraiya3053 first, arrange the figures so you can use the quadratic equation x^2 + (-y - 1)x + (y^2 - y) = 0 this means: a = 1 b = -y - 1 c = y^2 - y now plug these in to the quadratic formula, and you get: 2x = y + 1 +- sqrt( -3y^2 + 6y + 1 ) and put the radical by itself: 2x - y - 1 = +- sqrt( -3y^2 + 6y + 1 ) it's clear the LHS is an integer because x, y, and 1 are integers that means RHS must be an integer. That's only possible when the contents of the radical are a non-negative perfect square number. define a new function f(y) = -3y^2 + 6y + 1 f(y) must be greater than or equal to zero for all possible solutions (and of course, y must be an integer) you can solve f(y) = 0 using the quadratic formula to get: 6y = -6 +- sqrt(48) it is relatively easy to solve that to find what all possible values of y are. You will get get that only integer values of y for which f(y) >= 0 are y=0, y=1, and y=2 that's how you can be sure that all solutions have y= 0, 1, or 2 from there it is relatively easy to solve the problem.
x²-xy+y²=x+y is a cyclic quadratic equation. Written as x²-(y+1)x+(y²-y)=0 it is a quadratic equation in x with a=1, b=-(y+1), c=(y²-y). Therefore x=½[-(y+1)±sqrt{(y+1)²-4(y²-y)}] =½[-(y+1)±sqrt(-3y²+6y+1)] =½[-(y+1)±sqrt{-3(y-1)²+4}] Let D=-3(y-1)²+4. As x and y are integers D>=0 and a perfect square. It is so if: • y=2 --> D=1, then x=½[-(2+1)±1] --> x=-1 or x=-2 • y=1 --> D=4, then x=½[-(1+1)±2] --> x=0 or x=-2 • y=0 --> D=1, then x=½[-(0+1)±1] --> x=0 or x=-1 When the original equation is viewed as a quadratic equation in y similar results will be obtained with reverse value of y for x. Therefore the solutions are (x,y)={(-2,2),(-1,2),(0,1),(-2,1),(-1,0),(0,0),(0,-1),(1,-2),(1,0),(2,-1),(2,-2)} (x,y)={(-2,2),(-2,1),(-1,0),(-1,2),(0,1),(0,0),(
My approach :first i let x =y then the only solutions is (0,0)(2,2) Then i realised an important thing x^2+y^2 is always biger or equal to x+y (for x,y integers) if they are equal the only way is when (x,y)(0,0)(1,1)(1,0)and vise versa beacaus the symetrie and any combination except these force the (-xy) to be negative beacaus if its positive it will be for sure biger than (x+y) So x and y must positive or negative to guarantee (-xy) is negative And if you write this equation in the following form :(x-y)^2+xy=x+y Then you will notice (xy) is always positive by my proof above and (x-y)^2 is always positive (square) Therefor the only way (xy) to be less then (x+y) is to x=1or y=1 (or 0) Assum x =1 1^2-y +y^2=1+y y^2-2y=0 (y=0 or y=2) by symetrie we are done !!! We can also switch the answers (0,0)(1,0)(1,2)(2,2)
Can you make a video on how to approach Diophantine equations ? The thing is , I know some of the basic methods , but I don't know which one to apply when :(
You always choose interesting and instructive approaches to solving these problems. Looking at it I would have just seen it as a quadratic in say x and used the formula. By requiring the expression under the root to be an integer the solution just drop out. But I think your method is more instructive,
The method sometimes depends on the problem. I'm glad you brought up the idea of writing this as a quadratic equation in x and then solving it that way! This never crossed my mind... and thank you for the kind words! 🥰
8：45 Since we need to find integer solutions only, we can conclude that: WLOG, assume x>=y. Each of x-y, x-1, y-1 is =-√2. Then only things we need to check is (x,y)=(2,2),(2,1),(1,1),(1,0),(0,0).
Excuse me SyberMath, I'm stuck in a question. My teacher asked me the this question: "Find all two digit numbers such that each is divisible by the product of its two digits" So, I framed the Eqn : 10x+y = xy*K => (Kx-1)(Ky-10)= 10 By making cases I found the following solutions : (Kx,Ky)={ (2,20) , (3,15) , (6,12) , (11,11) } But I don't know how to proceed...It would be nice of you if you can help me in the further steps....
As for me there're too many words describing the solution in this video. Actually the full solution can be found using the simple substitution x=u+1 and y=w+1 the equation is then converted to u^2+w^2+(u-w)^2=2 And there are only three options in terms of integer values for u and w: u^2=0 w^2=1 (u-w)^2=1 u=0;w=±1 Similarly u^2=1 w^2=0 (u-w)^2=1 u=±1;w=0 And the last option u^2=1 w^2=1 (u-w)^2=0 u=w=±1 Getting back to x and y x=1;y={0,2} x={0,2};y=1 x=y={0,2} So there are six options in the solution. For better understanding the solutions may be shown as points on the 2-dimensional plane. Six points in the square 2×2.
The method of the video is too tricky. A more straightforward way is like this: let y = x + a, a, x, y are all integers x^2 - (x+a)*x + (x+a)^2 = x + x+a => x^2 + (a-2)x + a^2 - a = 0 => x = (2 - a +/- sqrt(4 - 3a^2) )/2 4-3a^2 >= 0, so the integer a can only be 0, 1 or -1 if a = 0, x = 0 or 2, the possible combinations of (x, y) are (0, 0), (2, 2) if a = 1, x = 0 or 1, the possible combinations of (x, y) are (0, 1), (1, 2) if a = -1, x = 1 or 2, the possible combinations of (x, y) are (1, 0), (2, 1) Those are all the 6 possible solutions.
x² - xy + y² = x + y ==> (x + y)² - 3xy = x + y
Making x + y = k ==> y = k - x (eq. 1)
Then k² - 3x(k - x) = k ==> k² - (3x +1) + 3x² = 0
Solving the quadratic equation in k:
k = [3x + 1 ± √((3x + 1)² - 12x²)] / 2
k = [3x + 1 ± √(-3x² + 6x + 1)] / 2 (eq.2)
For k to be real, it is necessary that - 3x² + 6x + 1 ≥ 0
The only integer solutions for x are x = 0, x = 1 and x = 2
For x = 0 (from eq.2): k = [1 ± √1] / 2 => k = 0 or k = 1
From eq.1: y = 0 or y = 1 ==> (0 , 0) , (0 , 1)
For x = 1: k = [4 ± √4] / 2 => k = 1 or k = 3
From eq.1: y = 0 or y = 2 ==> (1 , 0) , (1 , 2)
For x = 2: k = [7 ± √1] / 2 => k = 3 or k = 4
From eq.1: y = 1 or y = 2 ==> (2 , 1) , (2 , 2)
I used the following method: i turned it to the quadratic equation of x with y as a parameter and calculated the determinant as a function of y (which turned to be quadratic function). SInce we need real sol. the determinant must be greater or equal to 0, that gave me 3 possible integer values for y (0, 1, 2). Then I found x in each of those cases. Imo much simpler and easier to come up with method
Nice!
I used this method but can you explain to me how you get integer values for y. I get y=3+2√3/3 and I dont know how to continue
@@soraiya3053 he guessed and checked within the range, but 2 years
x(x-1) + y(y-1-x) = 0
i.e
(x-(y+1)/2)^2= ((y+1)/2)^2 +y(1-y)
Term at the RHS is rewritten as
(y*y + 2y +1 +4y -4y*y )/4
= (1 +6y -3y*y)/4
= (4 - 3(y-1)^2 )/4
This essentially means
4(x-(y+1)/2)^2 + 3(y-1)^2 = 4
or (2x- y-1)^2 + 3(y-1)^2= 4
Here x, y being integer each of these two square term is square number
case I [ (2x- y-1)^2 = 1 and (y-1)^2=1]
Hereby
2x-y-1 = 1,. y -1 = 1 i.e y = 2, x= 2
2x-y-1 = 1, y-1 = -1 i.e. y = 0, x=1
2x-y-1 = -1,. y -1 = 1 i.e y = 2, x= 1
2x-y-1 = -1, y-1 = -1 i.e. y = 0, x=0
The solution set of (x,y) for case I is
{ (0,0), (1,0), (1,2), (2,2)}
case II [(2x- y-1)^2 = 4 and (y-1)^2=0]
Hereby
2x-y-1 = 2, y-1 = 0 i.e. y = 1, x=2
2x-y-1 = -2,. y -1 = 0 i.e y = 1, x= 0
The solution set of (x,y) for case II is
{ (0,1), (2,1)}
Hereby the solution set for (x,y).is
{ (0,0), (1,0), (0,1),(1,2), (2,1), (2,2)}
How I did it:
Make x the parameter and complete the square. With a bit of cleanup you can rewrite the original equation as
(2x - y - 1)^2 = -(3y^2 - 6y - 1)
Because the LHS is a square, 3y^2 - 6y - 1
Very good!
Diophantine Equations are a fun and exciting topic in Number Theory because you have more variables than equations and solution method is not always very straightforward. In these equations, we are looking for integer solutions. Some equations have no solutions or no non-trivial solutions. Some have infinitely many solutions such as the Pythagorean Triples in a^2+b^2=c^2. What if a^3+b^3=b^3 or a^4+b^4=c^4? This brings us to Fermat's Last Theorem whose proof is way too complicated! This particular problem can be considered somewhat easy but it still involves some manipulations. There are many methods that can be used to solve Diophantine Equations one of which is Modular Arithmetic. Factoring also plays an important role. Any thoughts?
Thoughts about what?
(x+y)^2- 3xy =x+y
(x+y)^2- (x+y)=3xy
(x+y)(x+y- 1)=3xy
x+y=3,x+y- 1=3
x+y=4
If x+y=3, x-y=1
x,y=(2,1)
x+y=4 then x=y
x,y=(2,2)
A fairly simple solution is to express the original equation as a quadratic in x. Then we use the quadratic formula to get x in terms of y. The determinant has to be a perfect square, which is only possible for y = 0,1,2. Then for each case, you can find corresponding x values.
Good thinking!
Can you explain to me how do you get that y=0, 1, 2. I dont get it
@@soraiya3053 first, arrange the figures so you can use the quadratic equation
x^2 + (-y - 1)x + (y^2 - y) = 0
this means:
a = 1
b = -y - 1
c = y^2 - y
now plug these in to the quadratic formula, and you get:
2x = y + 1 +- sqrt( -3y^2 + 6y + 1 )
and put the radical by itself:
2x - y - 1 = +- sqrt( -3y^2 + 6y + 1 )
it's clear the LHS is an integer because x, y, and 1 are integers
that means RHS must be an integer. That's only possible when the contents of the radical are a non-negative perfect square number.
define a new function f(y) = -3y^2 + 6y + 1
f(y) must be greater than or equal to zero for all possible solutions (and of course, y must be an integer)
you can solve f(y) = 0 using the quadratic formula to get:
6y = -6 +- sqrt(48)
it is relatively easy to solve that to find what all possible values of y are. You will get get that only integer values of y for which f(y) >= 0 are y=0, y=1, and y=2
that's how you can be sure that all solutions have y= 0, 1, or 2
from there it is relatively easy to solve the problem.
x²-xy+y²=x+y is a cyclic quadratic equation. Written as x²-(y+1)x+(y²-y)=0
it is a quadratic equation in x with a=1, b=-(y+1), c=(y²-y). Therefore
x=½[-(y+1)±sqrt{(y+1)²-4(y²-y)}]
=½[-(y+1)±sqrt(-3y²+6y+1)]
=½[-(y+1)±sqrt{-3(y-1)²+4}]
Let D=-3(y-1)²+4. As x and y are integers D>=0 and a perfect square. It is so if:
• y=2 --> D=1, then x=½[-(2+1)±1]
--> x=-1 or x=-2
• y=1 --> D=4, then x=½[-(1+1)±2]
--> x=0 or x=-2
• y=0 --> D=1, then x=½[-(0+1)±1]
--> x=0 or x=-1
When the original equation is viewed as a quadratic equation in y similar results will be obtained with reverse value of y for x.
Therefore the solutions are
(x,y)={(-2,2),(-1,2),(0,1),(-2,1),(-1,0),(0,0),(0,-1),(1,-2),(1,0),(2,-1),(2,-2)}
(x,y)={(-2,2),(-2,1),(-1,0),(-1,2),(0,1),(0,0),(
My approach :first i let x =y then the only solutions is (0,0)(2,2)
Then i realised an important thing
x^2+y^2 is always biger or equal to x+y (for x,y integers) if they are equal the only way is when (x,y)(0,0)(1,1)(1,0)and vise versa beacaus the symetrie and any combination except these force the (-xy) to be negative beacaus if its positive it will be for sure biger than (x+y)
So x and y must positive or negative to guarantee (-xy) is negative
And if you write this equation in the following form :(x-y)^2+xy=x+y
Then you will notice (xy) is always positive by my proof above and (x-y)^2 is always positive (square)
Therefor the only way (xy) to be less then (x+y) is to x=1or y=1 (or 0)
Assum x =1
1^2-y +y^2=1+y
y^2-2y=0 (y=0 or y=2) by symetrie we are done !!! We can also switch the answers (0,0)(1,0)(1,2)(2,2)
Nice!
Can you make a video on how to approach Diophantine equations ? The thing is , I know some of the basic methods , but I don't know which one to apply when :(
That's a great idea! Let me put together some notes and a set of good problems. Thank you for the suggestion! 😊
@@SyberMath You are welcome ! 😀
You always choose interesting and instructive approaches to solving these problems. Looking at it I would have just seen it as a quadratic in say x and used the formula. By requiring the expression under the root to be an integer the solution just drop out. But I think your method is more instructive,
The method sometimes depends on the problem. I'm glad you brought up the idea of writing this as a quadratic equation in x and then solving it that way! This never crossed my mind...
and thank you for the kind words! 🥰
8：45 Since we need to find integer solutions only, we can conclude that: WLOG, assume x>=y. Each of x-y, x-1, y-1 is =-√2. Then only things we need to check is (x,y)=(2,2),(2,1),(1,1),(1,0),(0,0).
You could make x=y+k, k integer, solve for y and find out that k can only be 0 or +/-1, so x=y or x=y+/-1.
What if x+y=0? You only considered the case where x+y≠0
There are no solutions if x+y=0, not even the real ones!
@@SyberMath (x,y) = (0,0)
Finaly i have solved 🎊🎊🎊🎉🎉🎉🎉
Awesome!
Excuse me SyberMath,
I'm stuck in a question.
My teacher asked me the this question:
"Find all two digit numbers such that each is divisible by the product of its two digits"
So, I framed the Eqn :
10x+y = xy*K
=> (Kx-1)(Ky-10)= 10
By making cases I found the following solutions :
(Kx,Ky)={ (2,20) , (3,15) , (6,12) , (11,11) }
But I don't know how to proceed...It would be nice of you if you can help me in the further steps....
As for me there're too many words describing the solution in this video.
Actually the full solution can be found using the simple substitution x=u+1 and y=w+1
the equation is then converted to u^2+w^2+(u-w)^2=2
And there are only three options in terms of integer values for u and w:
u^2=0
w^2=1
(u-w)^2=1
u=0;w=±1
Similarly
u^2=1
w^2=0
(u-w)^2=1
u=±1;w=0
And the last option
u^2=1
w^2=1
(u-w)^2=0
u=w=±1
Getting back to x and y
x=1;y={0,2}
x={0,2};y=1
x=y={0,2}
So there are six options in the solution. For better understanding the solutions may be shown as points on the 2-dimensional plane. Six points in the square 2×2.
Very nice!
Nice problem, nice method. Congratulation.
Thank you! Cheers!
@@SyberMath but that doesnt work 2 cubed plus 3 cubed does not equal 2 plus 3 cubed..so it doesnt hold uo in general..
complex solutions??
This is a diophantine equation
Wow i like when we are asked for integers solution
Until now i didnt try to solve it beacaus im busy but later ofcorse i will try .
I'm glad you like it! Cheers!
It's pretty cool
You rock!
Thanks!
So fun!
Glad you like it!
The method of the video is too tricky.
A more straightforward way is like this:
let y = x + a, a, x, y are all integers
x^2 - (x+a)*x + (x+a)^2 = x + x+a
=> x^2 + (a-2)x + a^2 - a = 0
=> x = (2 - a +/- sqrt(4 - 3a^2) )/2
4-3a^2 >= 0, so the integer a can only be 0, 1 or -1
if a = 0, x = 0 or 2, the possible combinations of (x, y) are (0, 0), (2, 2)
if a = 1, x = 0 or 1, the possible combinations of (x, y) are (0, 1), (1, 2)
if a = -1, x = 1 or 2, the possible combinations of (x, y) are (1, 0), (2, 1)
Those are all the 6 possible solutions.
The original equation is symmetric with respect to x and y, you can use this to justify your last two answers.
are you turkish?
The answer is x=y=2
That's it?
x²-xy+y²=2²-2.2+2²=4-4+4=4=2+2=x+y
So it's still valid