What a video. It's really funny when you realize your mistakes, but at least you solve them. Also, good job at finding the solutions; they were quite surprising.
We can use the two formulas:- x³-y³=(x-y)(x²+y²+xy) and x³+y³=(x+y)(x²+y²-xy) when (x-y) and (x+y) are not 0. We can calculate separately for when they are 0. Substituting values of x³ and y³ in the above formulas, we get 2 = x²+y²+xy and 4 = x²+y²-xy which gives x²+y²=3 Now for the (x-y)=0 and (x+y)=0 cases, we can substitute x=y and x=-y respectively and get x²+y² = 4 and x²+y² = 8 respectively. So total three possible values, and their sum is 3+4+8= 15
Although not technically required to answer the problem, we can solve this equation system completely. Let's do it. Let S be the solution set (as a subset of R^2 with pairs (x, y) as elements). We have 4 disjoint cases: 1) x = y = 0: Observe the following: If (x, y) is in S, then x = 0 iff y = 0. Also note that (0, 0) is in S. Therefore, in this case we have 1 solution (0, 0) and a corresponding x^2 + y^2 = 0. 2) 0 != x = y: The equation system reduces to x^3 = 4x and we have: ((x, x) in S) iff (x^2 = 4) iff (x = 2 or x = -2). Therefore, we have two distinct solutions in this case: (2, 2) and (-2, -2), and the corresponding value of x^2 + y^2 is 8. 3) 0 != x = -y: The system reduces to x^3 = 2x and we have: ((x, -x) in S) iff (x^2 = 2) iff (x = sqrt(2) or x = -sqrt(2)). Therefore, we also have two distinct solutions in this case: (sqrt(2), -sqrt(2)) and (-sqrt(2), sqrt(2)), and the corresponding value of x^2 + y^2 is 4. 4) x != 0 and y != 0 and |x| != |y|: This is the main case. It follows that x + y != 0 and x - y != 0. By adding and subtracting the two equations, we can conclude that x^3 + y^3 = 4(x + y) and x^3 - y^3 = 2(x - y) which is equivalent to x^2 - xy + y^2 = 4 and x^2 + xy + y^2 = 2 in this case (by dividing by (x + y) or (x - y) respectively). Adding and subtracting the last two equations to/from each other yields: 2(x^2 + y^2) = 6 and -2xy = 2 which is equivalent to x^2 + y^2 = 3 and xy = -1. So we already know the value of x^2 + y^2 in this case. But we can derive more: (x + y)^2 = 3 + 2(-1) = 1 and (x - y)^2 = 3 - 2(-1) = 5. This is equivalent to |x + y| = 1 and |x - y| = sqrt(5). Now, whenever (x, y) is in S, we also have (y, x), (-x, -y), and (-y, -x) in S due to the symmetries of the original equation system. The important thing to note here is that in our case (4) all of these 4 solutions are pairwise distinct! Thus, all solutions that fall under this case come in groups of 4 distinct pairs. Therefore, without loss of generality, we can add the additional assumptions (0 < |x| < |y|) and (y > 0) to this case. But are there any solutions in this case and, if yes, how many? Let's continue and see. First, observe that -- under the above additional assumptions -- (x + y) and (x - y) cannot have the same sign, because (x + y)(x - y) = |x|^2 - |y|^2 < 0. Second, note that x + y = x + |y| >= -|x| + |y| > 0 (under the above assumptions). Therefore we have: (x + y > 0) and (x - y < 0), and further: (|x + y| = x + y = 1) and (|x - y| = y - x = sqrt(5)). By adding and subtracting the last two equations to/from each other, we arrive at: (2y = 1 + sqrt(5)) and (2x = 1 - sqrt(5)) which is equivalent to (x = -1/Phi) and (y = Phi) where Phi is the "golden ratio". It's easy to check that (x, y) = (-1/Phi, Phi) is indeed a solution. Therefore, in this case we have 4 distinct solutions: (-1/Phi, Phi), (Phi, -1/Phi), (1/Phi, -Phi), and (-Phi, 1/Phi), and the corresponding value of x^2 + y^2 is 3. -- In summary, the total number of distinct solutions (x, y) in S is: 1 + 2 + 2 + 4 = 9. To answer the posed problem: If we sum over all distinct values of (x^2 + y^2) for solutions (x, y) in S we get: 0 + 8 + 4 + 3 = 15 (this sum was most likely asked for in the problem). If we sum the value of (x^2 + y^2) over all distinct solutions (x, y) in S we get: 1*0 + 2*8 + 2*4 + 4*3 = 36.
Thank you for your wonderful videos. Your love of maths and the natural way your body and personality express this are beautiful and your and our blessing. For this problem, I suggest the last step of adding the two equations you did to get the final value of 3 for x^2 + y^2 needs some justification / expansion. This is because they were only cases - the alternatives to x=y and x=-y respectively - which do not have to be true simultaneously. To avoid the need for this, rather than also factorise x^3 + y^3, I prefer to stick with the factorisation of x^3 - y^3 you have and solve the alternative case to x=y, ie the equation x^2 + y^2 + xy -2 = 0, directly. Dividing equation 1 by x and equation 2 by y and subbing for x^2 + y^2 gives 6 + y/x -+ x/y = 2 - xy. Adding the two fractions on LHS and again subbing for x^2 + y^2 leads to a quadratic equation in xy which factorises and has solutions -1 and -2. The first of these values for xy gives the final value of 3 (= 2--1) for x^2 + y^2. The second value for xy gives the value of 4 (2--2) for x^2 + y^2 you found from the x=~y case. This is because when y=~2/x is subbed into equations 1 and 2, it leads to two identical quadratic equations in x^2 and in y^2 which only have solutions for which x=-y. So the first of the two equations you used in the last step can hold when x=~y as well as when the second equation you used holds. Nonetheless it is possible for both equations to be true simultaneously, leading to the final value of 3 for x^2 + y^2. So with your approach, by justification / expansion, I guess I just mean this. Spelling out that because you have shown that A (x=y) or B (your first equation) is true, and also that C (x=-y) or D (your second equation) is true, 4 cases arise: A and C, A and D, B and C and B and D. And that by adding your two equations you are considering the last of these cases, the first three having already been dealt with, being those leading to the values of 0, 8 and 4 for x^2 + y^2 respectively. Again, thank you for your lovely and inspirational videos and look forward to seeing the next one ❤ so much for your wonderful forward to your next video
I appreciate this derailed feedback. I must say, sometimes I change my strategy when I start recording because I become concerned about making the video too long. I never stop learning.
for x 0 and y 0 we can substitute t=x/y x^2=3+1/ t y^2=3+t So: sum=x^2+y^2=6+t+1/t x^2/y^2=t^2=(3+1/ t)/(3+t) so t^3*(3+t)=3t+1 t^4+3t^3-3t-1=0 (t^4-1)+3t(t^2-1)=0 (t^2+1)(t^2-1)+3t(t^2-1)=0 (t^2+3t+1)(t^2-1)=(t^2+3t+1)(t-1)(t+1)=0 t=1 gives 6+1+1=8 t=-1 gives 6-1-1=4 (t^2+3t+1)=0 gives to solution witch sum is - 3 an product of roots is 1 so t+1/t is sum of roots gives us sum= 6-3=3 0+2+4+8=15
Respect to you Sir 🫡 The videos show thy love for the subject. Today is 'Guru Purnima' in India, a day to respect thy teachers 😇 So, once again, respect to you Sir for the informative videos, Thank you 😇😊
BEFORE WATCHING: I already suspect something with sum and difference of cubes. Let's add and subtract both equations and see what we get x^3 + y^3 = 4x + 4y x^3 - y^3 = 2x - 2y Pull out some factors (x+y)(x^2 - xy + y^2) = 4(x+y) (x-y)(x^2 + xy + y^2) = 2(x-y) Consider for the moment that x ≠ y and x ≠ -y. x^2 - xy + y^2 = 4 x^2 + xy + y^2 = 2 Now add both equations 2x^2 + 2y^2 = 6 x^2 + y^2 = 3 Can this happen? Subtract the equations instead. 2xy = -2 x^2 + 2xy + y^2 = 1 (x+y)^2 = 1 x^2 - 2xy + y^2 = 5 (x-y)^2 = 5 x and y can be real, this checks out What about when x = y? x^3 = 3x + x x^3 = 4x x^3 - 4x = 0 x(x-2)(x+2) = 0 x = 0, 2, -2 -> y = 0, 2, -2 x^2 + y^2 = 0, 8 Or when x = -y? x^3 = 3x - x x^3 = 2x x^3 - 2x = 0 x(x-sqrt(2))(x+sqrt(2)) = 0 x = 0, sqrt(2), -sqrt(2) y = 0, -sqrt(2), sqrt(2) x^2 + y^2 = 0, 4 (0, 0) is a duplicate solution Total: 3 + 0 + 8 + 4 = 15
Muito bonita a questão! Parabéns pela escolha. Essa eu fiz facilmente. Brasil - agosto 2024. Very beautiful question! Congratulations on your choice. I did this easily. Brazil - August 2024.
Thanks for the video and the solution. Maybe it is worth quickly checking that or arguing why x^2 + y^2 = 3 also leads to real solutions for x and y? (this is anyways the case, and 3 is therefore a legitimate value for x^2 + y^2)
Thanks for your interersing problem, that I solved as below. Please tell me if you like my solution (although quite detailed). Of course, I didn't look at your solutions. Keep up the good work Greetings Recall of the system x,y € R (i) x^3=3x+y (ii) y^3=3y+x Recall of some usefull equalities (a) x^2+y^2=(x+y)^2-2xy (b) (x+y)^3=x^3+y^3+3xy(x+y) (c) x^3-y^3=(x-y)(x^2+xy+y^2) (d) x^3+y^3=(x+y)(x^2-xy+y^2) From (c), the operation (i)-(ii) gives us x^3-y^3=(x-y)(x^2+xy+y^2)=3(x-y)+(y-x)=2(x-y) So that (x-y)[(x^2+xy+y^2)-2]=0 Then two cases Case A1 (x-y)=0 or Case B1 xy and [(x^2+xy+y^2)-2]=0 Then Case A1 gives x=y so that (i) and (ii) give x^3=3x+x so x^3-4x=0 so x^2(x-4)=0 So for Case A1 we have x=y=0 so x^2+y^2=0 or x=y=4 so x^2+y^2=32 Or Case B1 xy and (x^2+xy+y^2)-2=0 so xy and x^2+y^2=2-xy From (d), the operation (i)+(ii) gives us x^3+y^3=(x+y)(x^2-xy+y^2)=3(x+y)+(y+x)=4(x+y) So that (x+y)[(x^2-xy+y^2)-4]=0 Then two cases Case A2 (x+y)=0 so that from (i) x=-y gives x^3=3x-x so x^2(x-2)=0 or Case B2 x-y and [(x^2-xy+y^2)-4]=0 So for Case A2 we have x=-y=0 so x^2+y^2=0 or x=-y=2 so x^2+y^2=4 Or Case B2 x-y and (x^2-xy+y^2)-4=0 so x-y and x^2+y^2=4+xy TO SUM UP, we have the following solution with below cases Case A1 with x^2+y^2=0 or x^2+y^2=32 or Case A2 with x^2+y^2=0 or x^2+y^2=4 or Case B (B1 and B2 are gathered) with xy and x-y and x^2+y^2=2-xy and x^2+y^2=4+xy So that by suming the 2 last equations 2(x^2+y^2)=(2-xy)+(4+xy) so x^2+y^2=3 Case B x^2+y^2=3 Let's verify for case B solutions that the conditions xy and x-y are fulfilled. We know that x^2+y^2=3 so 3=2-xy and 3=4+xy then xy=-1 So from (a) x^2+y^2=(x+y)^2-2xy we have 3=(x+y)^2-2.(-1) => (x+y)^2=3-2=1 Then we have case B3 (x+y=1 and xy=-1) or and case B4 (x+y=-1 and xy=-1) From case B3 (x;y) are solutions of t^2-s3.t+p3=0 (s3=1 the sum of solutions, p3=-1 product of solutions) t^2-t-1=0 Then [x3;y3]=[(1+sqrt(5))/2;(1-sqrt(5))/2] then x3y3 and x3-y3 then case B3 solutions fulfills the conditions From case B4 (x;y) are solutions of t^2-s4.t+p4=0 (s4=-1 the sum of solutions, p4=-1 product of solutions) t^2+t-1=0 Then [x4;y4]=[(-1+sqrt(5))/2;(-1-sqrt(5))/2] then x4y4 and x4-y4 then case B4 solutions fulfills the conditions TO CONCLUDE the real solutions to the following system (i) x^3=3x+y (ii) y^3=3y+x give the following values for x^2+y^2 as below (from cases A1, A2 and B) S={(x^2+y^2)}={0;3;4;32} END
My approach was to get the equations into a form involving x^2 + y^2 from the beginning, so I realized that if I multiplied the first equation by x and the second equation by y and then subtracted the second from the first, I get x^4 - y^4 = 3x^2 + 3y^2. Then moving everything to one side and factoring, I get (x + y)(x - y)(x^2 + y^2 - 3) = 0. Then I solved x and y for when x = y and x = -y and also knew that x^2 + y^2 = 3 is also a solution. Is this a valid approach?
It would be appropriate to also ensure that the solution to the equations for which the condition x2+y2=3 is satisfied is real solution, (not complex).
I suppose the question on the thumbnail is kinda ambiguous (it's all I looked at before I attempted the problem). Depending on how you interpret it, the answer could be 15 or 30. You get 15 if you plug in only distinct values of x^2+y^2 (those are 0, 3, 4 and 8). But if you decide to find *every* solution to the system, you will find that there are 7 solutions, where one is (0; 0), but after you have 6 solutions where 3 are just the other 3, but flipped, because the system is symmetric. If you count those, you get to 30. Anyways Imma go watch the video now, I know it's gonna be juicy!
We can also multiply the 1st equation with x and the second one with y. So we get x⁴ - 3x² = y⁴ - 3y² solving this we get all the possibilities we need one with x²-y² = 0 and one with x²+y²=3. Rest is same
Sorry, something doesn't add up to me... Suppose we have (among others) two solutions to the original system: (x1, y1) such that it satisfies the relation x1^2 + y1^2 = 4 + x1y1 (x2, y2) such that it satisfies the relation x2^2 + y2^2 = 2 - x2y2 How can we be sure that x1y1 = x2y2? That doesn't seem at all obvious. Otherwise, the sum (x1^2 + y1^2) + (x2^2 + y2^2) that we want to find as part of our needed quantity is not equal to 6, but is instead equal to 6 + x1y1 - x2y2, which we technically still need to compute. And even if we somehow show that x1y1 = x2y2, why do we divide by two after summing? We need to sum ALL possible values of (x^2 + y^2) to get our answer, and two of those values sum to 6. There are also values x^2 + y^2 = 4 and x^2 + y^2 = 8, we don't divide them by 2 when we sum them, do we? We didn't find ONE solution (x,y) that yields x^2 + y^2 = 3, we found TWO solutions that, together, yield (x1^2 + y1^2) + (x2^2 + y2^2) = 6. I don't think it is correct to divide by 2 here, unless we also show that x1=x2 and y1=y2, thus making it actually only a single (distinct) solution. Cheers!
@@robertpearce8394 Hello! After looking over the problem one more time, I realized the flaw in my reasoning. After transforming our original system into a system of two zero-products, we should never consider any of the consequent equations "in a vacuum", we actually create 4 smaller-degree sub-systems: {x-y=0, x+y=0}; {x-y=0, x^2 + y^2 - 4 - xy=0}; {x+y=0, x^2 + y^2 - 2 + xy=0} and {x^2 + y^2 - 2 + xy = 0, x^2 + y^2 - 4 - xy=0}. These sub-systems give us all our solutions, and both solutions to the last sub-system yield x^2+y^2=3, which we only add to the resulting sum once. I was lost in the sauce, as they say, sorry for the confusion. 😔
@@sujitsivadanam Yes, that's exactly what I was confused about. And we can be sure they are both satisfied because that's one of the cases/subsystems we are considering when we transition from a bigger system: {(x-y)(x^2 + y^2 - 2 + xy) = 0, (x+y)(x^2 + y^2 - 4 - xy)=0} to 4 smaller systems that result from it: {x-y=0, x+y=0} OR {x-y=0, x^2 + y^2 - 4 - xy=0} OR {x+y=0, x^2 + y^2 - 2 + xy=0} OR {x^2 + y^2 - 2 + xy = 0, x^2 + y^2 - 4 - xy=0}. I forgot that you can't just take x,y so that x^2 + y^2 - 2 + xy = 0, you also need x+y=0 or x^2 + y^2 - 4 - xy=0 to hold, otherwise it's not a solution to the original system.
Sorry but I don'get something. If x=y=2,then x^2+y^2=4, then when x=y=-2 we get oneother sum of x^2+y^2=4. Shund't you consider all the values of x and y?
What a video. It's really funny when you realize your mistakes, but at least you solve them. Also, good job at finding the solutions; they were quite surprising.
Yes. It keeps us all alert.
Best math teacher i have ever seen, most think i love is your smile 😊
I love how you catch your mistakes and leave your lessons organic. Reminds me of my university professors!
We can use the two formulas:-
x³-y³=(x-y)(x²+y²+xy) and
x³+y³=(x+y)(x²+y²-xy)
when (x-y) and (x+y) are not 0. We can calculate separately for when they are 0.
Substituting values of x³ and y³ in the above formulas, we get
2 = x²+y²+xy and
4 = x²+y²-xy
which gives x²+y²=3
Now for the (x-y)=0 and (x+y)=0 cases, we can substitute x=y and x=-y respectively and get
x²+y² = 4 and
x²+y² = 8 respectively.
So total three possible values, and their sum is 3+4+8= 15
Muito bom. Eu fiz do mesmo jeito que o seu! Brazil - Agosto 2024. Very good. I did it the same way as yours! Brazil - August 2024.
Although not technically required to answer the problem, we can solve this equation system completely. Let's do it. Let S be the solution set (as a subset of R^2 with pairs (x, y) as elements).
We have 4 disjoint cases:
1) x = y = 0: Observe the following: If (x, y) is in S, then x = 0 iff y = 0. Also note that (0, 0) is in S. Therefore, in this case we have 1 solution (0, 0) and a corresponding x^2 + y^2 = 0.
2) 0 != x = y: The equation system reduces to x^3 = 4x and we have: ((x, x) in S) iff (x^2 = 4) iff (x = 2 or x = -2). Therefore, we have two distinct solutions in this case: (2, 2) and (-2, -2), and the corresponding value of x^2 + y^2 is 8.
3) 0 != x = -y: The system reduces to x^3 = 2x and we have: ((x, -x) in S) iff (x^2 = 2) iff (x = sqrt(2) or x = -sqrt(2)). Therefore, we also have two distinct solutions in this case: (sqrt(2), -sqrt(2)) and (-sqrt(2), sqrt(2)), and the corresponding value of x^2 + y^2 is 4.
4) x != 0 and y != 0 and |x| != |y|: This is the main case. It follows that x + y != 0 and x - y != 0. By adding and subtracting the two equations, we can conclude that x^3 + y^3 = 4(x + y) and x^3 - y^3 = 2(x - y) which is equivalent to x^2 - xy + y^2 = 4 and x^2 + xy + y^2 = 2 in this case (by dividing by (x + y) or (x - y) respectively). Adding and subtracting the last two equations to/from each other yields: 2(x^2 + y^2) = 6 and -2xy = 2 which is equivalent to x^2 + y^2 = 3 and xy = -1. So we already know the value of x^2 + y^2 in this case. But we can derive more: (x + y)^2 = 3 + 2(-1) = 1 and (x - y)^2 = 3 - 2(-1) = 5. This is equivalent to |x + y| = 1 and |x - y| = sqrt(5).
Now, whenever (x, y) is in S, we also have (y, x), (-x, -y), and (-y, -x) in S due to the symmetries of the original equation system. The important thing to note here is that in our case (4) all of these 4 solutions are pairwise distinct! Thus, all solutions that fall under this case come in groups of 4 distinct pairs. Therefore, without loss of generality, we can add the additional assumptions (0 < |x| < |y|) and (y > 0) to this case.
But are there any solutions in this case and, if yes, how many? Let's continue and see. First, observe that -- under the above additional assumptions -- (x + y) and (x - y) cannot have the same sign, because (x + y)(x - y) = |x|^2 - |y|^2 < 0. Second, note that x + y = x + |y| >= -|x| + |y| > 0 (under the above assumptions). Therefore we have: (x + y > 0) and (x - y < 0), and further: (|x + y| = x + y = 1) and (|x - y| = y - x = sqrt(5)).
By adding and subtracting the last two equations to/from each other, we arrive at: (2y = 1 + sqrt(5)) and (2x = 1 - sqrt(5)) which is equivalent to (x = -1/Phi) and (y = Phi) where Phi is the "golden ratio". It's easy to check that (x, y) = (-1/Phi, Phi) is indeed a solution.
Therefore, in this case we have 4 distinct solutions: (-1/Phi, Phi), (Phi, -1/Phi), (1/Phi, -Phi), and (-Phi, 1/Phi), and the corresponding value of x^2 + y^2 is 3.
--
In summary, the total number of distinct solutions (x, y) in S is: 1 + 2 + 2 + 4 = 9.
To answer the posed problem:
If we sum over all distinct values of (x^2 + y^2) for solutions (x, y) in S we get: 0 + 8 + 4 + 3 = 15 (this sum was most likely asked for in the problem).
If we sum the value of (x^2 + y^2) over all distinct solutions (x, y) in S we get: 1*0 + 2*8 + 2*4 + 4*3 = 36.
Thank you for your wonderful videos. Your love of maths and the natural way your body and personality express this are beautiful and your and our blessing.
For this problem, I suggest the last step of adding the two equations you did to get the final value of 3 for x^2 + y^2 needs some justification / expansion. This is because they were only cases - the alternatives to x=y and x=-y respectively - which do not have to be true simultaneously.
To avoid the need for this, rather than also factorise x^3 + y^3, I prefer to stick with the factorisation of x^3 - y^3 you have and solve the alternative case to x=y, ie the equation x^2 + y^2 + xy -2 = 0, directly. Dividing equation 1 by x and equation 2 by y and subbing for x^2 + y^2 gives 6 + y/x -+ x/y = 2 - xy. Adding the two fractions on LHS and again subbing for x^2 + y^2 leads to a quadratic equation in xy which factorises and has solutions -1 and -2.
The first of these values for xy gives the final value of 3 (= 2--1) for x^2 + y^2.
The second value for xy gives the value of 4 (2--2) for x^2 + y^2 you found from the x=~y case. This is because when y=~2/x is subbed into equations 1 and 2, it leads to two identical quadratic equations in x^2 and in y^2 which only have solutions for which x=-y.
So the first of the two equations you used in the last step can hold when x=~y as well as when the second equation you used holds. Nonetheless it is possible for both equations to be true simultaneously, leading to the final value of 3 for x^2 + y^2.
So with your approach, by justification / expansion, I guess I just mean this. Spelling out that because you have shown that A (x=y) or B (your first equation) is true, and also that C (x=-y) or D (your second equation) is true, 4 cases arise: A and C, A and D, B and C and B and D. And that by adding your two equations you are considering the last of these cases, the first three having already been dealt with, being those leading to the values of 0, 8 and 4 for x^2 + y^2 respectively.
Again, thank you for your lovely and inspirational videos and look forward to seeing the next one ❤
so much for your wonderful forward to your next video
I appreciate this derailed feedback. I must say, sometimes I change my strategy when I start recording because I become concerned about making the video too long. I never stop learning.
for x 0 and y 0 we can substitute t=x/y
x^2=3+1/ t
y^2=3+t
So:
sum=x^2+y^2=6+t+1/t
x^2/y^2=t^2=(3+1/ t)/(3+t)
so
t^3*(3+t)=3t+1
t^4+3t^3-3t-1=0
(t^4-1)+3t(t^2-1)=0
(t^2+1)(t^2-1)+3t(t^2-1)=0
(t^2+3t+1)(t^2-1)=(t^2+3t+1)(t-1)(t+1)=0
t=1 gives 6+1+1=8
t=-1 gives 6-1-1=4
(t^2+3t+1)=0 gives to solution witch sum is - 3 an product of roots is 1 so
t+1/t is sum of roots
gives us
sum= 6-3=3
0+2+4+8=15
Very interesting problem. A variation on the solution:
x^3 = 3x + y [eq.1]
y^3 = x + 3y [eq.2]
Either x=y=0 or neither x nor y are 0 (x=0 => y=x^3-3x=0 etc). So in the case where both x and y are non-zero:
[eq.1] / x:
=> x^2 = 3 + y/x [eq.3]
[eq.2] / y:
=> y^2 = 3 + x/y [eq.4]
Let R = x^2 + y^2.
[eq.3] + [eq.4]:
=> R = 6 + y/x + x/y
=> R - 6 = (x^2 + y^2) / xy
=> (R-6)xy = R [eq.5]
[eq.2] - [eq.1]:
=> y^3-x^3 = 2(y - x)
=> (y-x)(x^2 + xy + y^2) - 2(y-x) = 0
=> (y-x)(x^2 + xy + y^2 - 2) = 0
Case 1: x = y
[eq.1] => x^3 = 4x
=> x = y = 0, ±2
=> R = 0, 8
Case 2: x^2 + xy + y^2 - 2 = 0
=> xy = 2 - R [eq.6]
[eq.6] in [eq.5]:
=> (R-6)(2-R) = R
=> 2R - R^2 - 12 + 6R = R
=> R^2 - 7R + 12 = 0
=> (R-4)(R-3) = 0
=> R = 3, 4
∴ R = x^2 + y^2 ∈ {0, 3, 4, 8}
13:20 - "and this, is what we've been looking for" with a look that gives goosebumps.😅
A Mathematican who is a good teacher. A rare combination!!
Respect to you Sir 🫡 The videos show thy love for the subject. Today is 'Guru Purnima' in India, a day to respect thy teachers 😇 So, once again, respect to you Sir for the informative videos, Thank you 😇😊
BEFORE WATCHING:
I already suspect something with sum and difference of cubes. Let's add and subtract both equations and see what we get
x^3 + y^3 = 4x + 4y
x^3 - y^3 = 2x - 2y
Pull out some factors
(x+y)(x^2 - xy + y^2) = 4(x+y)
(x-y)(x^2 + xy + y^2) = 2(x-y)
Consider for the moment that x ≠ y and x ≠ -y.
x^2 - xy + y^2 = 4
x^2 + xy + y^2 = 2
Now add both equations
2x^2 + 2y^2 = 6
x^2 + y^2 = 3
Can this happen?
Subtract the equations instead.
2xy = -2
x^2 + 2xy + y^2 = 1
(x+y)^2 = 1
x^2 - 2xy + y^2 = 5
(x-y)^2 = 5
x and y can be real, this checks out
What about when x = y?
x^3 = 3x + x
x^3 = 4x
x^3 - 4x = 0
x(x-2)(x+2) = 0
x = 0, 2, -2
-> y = 0, 2, -2
x^2 + y^2 = 0, 8
Or when x = -y?
x^3 = 3x - x
x^3 = 2x
x^3 - 2x = 0
x(x-sqrt(2))(x+sqrt(2)) = 0
x = 0, sqrt(2), -sqrt(2)
y = 0, -sqrt(2), sqrt(2)
x^2 + y^2 = 0, 4
(0, 0) is a duplicate solution
Total: 3 + 0 + 8 + 4 = 15
Muito bonita a questão! Parabéns pela escolha. Essa eu fiz facilmente. Brasil - agosto 2024. Very beautiful question! Congratulations on your choice. I did this easily. Brazil - August 2024.
Sir, starting the time of 6:53, all the work is incorrect. You should have been solving x^2-xy+y^2-4 as equal to zero.
Thanks for the video and the solution. Maybe it is worth quickly checking that or arguing why x^2 + y^2 = 3 also leads to real solutions for x and y? (this is anyways the case, and 3 is therefore a legitimate value for x^2 + y^2)
X^3=3X+Y Y^3=3Y+X X^2+Y^2=8 I don’t have to think about it.
Thanks for your interersing problem, that I solved as below.
Please tell me if you like my solution (although quite detailed).
Of course, I didn't look at your solutions.
Keep up the good work
Greetings
Recall of the system
x,y € R
(i) x^3=3x+y
(ii) y^3=3y+x
Recall of some usefull equalities
(a) x^2+y^2=(x+y)^2-2xy
(b) (x+y)^3=x^3+y^3+3xy(x+y)
(c) x^3-y^3=(x-y)(x^2+xy+y^2)
(d) x^3+y^3=(x+y)(x^2-xy+y^2)
From (c), the operation (i)-(ii) gives us
x^3-y^3=(x-y)(x^2+xy+y^2)=3(x-y)+(y-x)=2(x-y)
So that (x-y)[(x^2+xy+y^2)-2]=0
Then two cases
Case A1 (x-y)=0
or
Case B1 xy and [(x^2+xy+y^2)-2]=0
Then
Case A1 gives x=y so that (i) and (ii) give x^3=3x+x so x^3-4x=0 so x^2(x-4)=0
So for Case A1 we have x=y=0 so x^2+y^2=0 or x=y=4 so x^2+y^2=32
Or
Case B1 xy and (x^2+xy+y^2)-2=0 so xy and x^2+y^2=2-xy
From (d), the operation (i)+(ii) gives us
x^3+y^3=(x+y)(x^2-xy+y^2)=3(x+y)+(y+x)=4(x+y)
So that (x+y)[(x^2-xy+y^2)-4]=0
Then two cases
Case A2 (x+y)=0 so that from (i) x=-y gives x^3=3x-x so x^2(x-2)=0
or
Case B2 x-y and [(x^2-xy+y^2)-4]=0
So for Case A2 we have x=-y=0 so x^2+y^2=0 or x=-y=2 so x^2+y^2=4
Or
Case B2 x-y and (x^2-xy+y^2)-4=0 so x-y and x^2+y^2=4+xy
TO SUM UP, we have the following solution with below cases
Case A1 with x^2+y^2=0 or x^2+y^2=32
or
Case A2 with x^2+y^2=0 or x^2+y^2=4
or
Case B (B1 and B2 are gathered) with xy and x-y and x^2+y^2=2-xy and x^2+y^2=4+xy
So that by suming the 2 last equations 2(x^2+y^2)=(2-xy)+(4+xy) so x^2+y^2=3
Case B x^2+y^2=3
Let's verify for case B solutions that the conditions xy and x-y are fulfilled.
We know that x^2+y^2=3 so 3=2-xy and 3=4+xy then xy=-1
So from (a) x^2+y^2=(x+y)^2-2xy we have 3=(x+y)^2-2.(-1) => (x+y)^2=3-2=1
Then we have case B3 (x+y=1 and xy=-1) or and case B4 (x+y=-1 and xy=-1)
From case B3 (x;y) are solutions of t^2-s3.t+p3=0
(s3=1 the sum of solutions, p3=-1 product of solutions) t^2-t-1=0
Then [x3;y3]=[(1+sqrt(5))/2;(1-sqrt(5))/2] then x3y3 and x3-y3 then case B3 solutions fulfills the conditions
From case B4 (x;y) are solutions of t^2-s4.t+p4=0
(s4=-1 the sum of solutions, p4=-1 product of solutions) t^2+t-1=0
Then [x4;y4]=[(-1+sqrt(5))/2;(-1-sqrt(5))/2] then x4y4 and x4-y4 then case B4 solutions fulfills the conditions
TO CONCLUDE the real solutions to the following system
(i) x^3=3x+y
(ii) y^3=3y+x
give the following values for x^2+y^2 as below (from cases A1, A2 and B)
S={(x^2+y^2)}={0;3;4;32}
END
My approach was to get the equations into a form involving x^2 + y^2 from the beginning, so I realized that if I multiplied the first equation by x and the second equation by y and then subtracted the second from the first, I get x^4 - y^4 = 3x^2 + 3y^2. Then moving everything to one side and factoring, I get (x + y)(x - y)(x^2 + y^2 - 3) = 0. Then I solved x and y for when x = y and x = -y and also knew that x^2 + y^2 = 3 is also a solution. Is this a valid approach?
Oops, I meant x^4 - y^4 = 3x^2 - 3y^2 in my comment above. We all seem to be having problems keeping signs straight 🙂
Indeed never stop learning!
Never stop learning 😄🤗🥰...complete
It would be appropriate to also ensure that the solution to the equations for which the condition x2+y2=3 is satisfied is real solution, (not complex).
12:25 you can't conclude that x²+y²=3 is a solition without check x and y are both real.
X^2+Y^2=3
I plot the grafics and should have quite interesting if you have done the same to show the geometric meaning
I suppose the question on the thumbnail is kinda ambiguous (it's all I looked at before I attempted the problem). Depending on how you interpret it, the answer could be 15 or 30. You get 15 if you plug in only distinct values of x^2+y^2 (those are 0, 3, 4 and 8). But if you decide to find *every* solution to the system, you will find that there are 7 solutions, where one is (0; 0), but after you have 6 solutions where 3 are just the other 3, but flipped, because the system is symmetric. If you count those, you get to 30. Anyways Imma go watch the video now, I know it's gonna be juicy!
Have you considered just converting to polar coordinates
X^2+Y^2=4+XY
alternative? g1= x+3*y-y^3 (=0), g2=x^3-3*x-y (=0), g3=x^2+y^2 (=?). remove dependence on 'x' g3-> g3/g1 =r1= y^6 - 6*y^4 + 10*y^2
g2/g1= r2= y^9 - 9*y^7 + 27*y^5 - 30*y^3 + 8*y=0 = y*(y^2 - 2)*(y - 2)*(y + 2)*(- y^2 + y + 1)*(y^2 + y - 1). y=0 => r1=0;
r1/(y+2) = 8, r1/(y-2)=8, r1/(y^2-2) = 4, r1/(y^2 + y - 1)=3, r1/(y^2 - y - 1)=3 =>x^2+y^2 = { 0, 3, 4, 8}
In my understanding the geometric meaning is more interesting than the calculation by itself….plot and see it.
These olympiad type problems are so hard for me
This is one of the easiest questions so far
We have a problem, integrate cos (10x^2)
Been struggling for days now
Isn't it the same like cos(x²)? It's non elementary
LOVE IT ❤
I make some calculus and another solutions are (φ,1/φ), (1/φ,φ), (-φ,-1/φ) and (-1/φ,-φ), since φ=golden ratio((1+sqrt(5))/2)
for which the sum of squares = 3
@@echandler I never said opposite, I only post for curiosity
In the factorization of x^3+y^3 there is a sign error. The result is (x+y)(x^2-xy+y^2) and not (x+y)(x^2+xy+y^2)
That is corrected later, as noted in the caption.
Thanks Sir
I get answer by looking the equations. x^2+y^2=3
Actually x³+y³=(x+y)(x²-xy+y)
We can also multiply the 1st equation with x and the second one with y. So we get x⁴ - 3x² = y⁴ - 3y² solving this we get all the possibilities we need one with x²-y² = 0 and one with x²+y²=3. Rest is same
two kasus, x = y and x =\ y
I tried to solve it and i got no solution
Sorry, something doesn't add up to me...
Suppose we have (among others) two solutions to the original system:
(x1, y1) such that it satisfies the relation x1^2 + y1^2 = 4 + x1y1
(x2, y2) such that it satisfies the relation x2^2 + y2^2 = 2 - x2y2
How can we be sure that x1y1 = x2y2? That doesn't seem at all obvious. Otherwise, the sum (x1^2 + y1^2) + (x2^2 + y2^2) that we want to find as part of our needed quantity is not equal to 6, but is instead equal to 6 + x1y1 - x2y2, which we technically still need to compute.
And even if we somehow show that x1y1 = x2y2, why do we divide by two after summing? We need to sum ALL possible values of (x^2 + y^2) to get our answer, and two of those values sum to 6. There are also values x^2 + y^2 = 4 and x^2 + y^2 = 8, we don't divide them by 2 when we sum them, do we?
We didn't find ONE solution (x,y) that yields x^2 + y^2 = 3, we found TWO solutions that, together, yield (x1^2 + y1^2) + (x2^2 + y2^2) = 6. I don't think it is correct to divide by 2 here, unless we also show that x1=x2 and y1=y2, thus making it actually only a single (distinct) solution.
Cheers!
I am not sure that I am following your argument, but maybe the point is that we are finding solutions where the xy terms can be eliminated.
@@robertpearce8394I think what he's trying to say is how we can be certain that both x^2+y^2 = 4+xy and x^2+y^2 = 2-xy are simultaneously satisfied.
@@robertpearce8394 Hello! After looking over the problem one more time, I realized the flaw in my reasoning. After transforming our original system into a system of two zero-products, we should never consider any of the consequent equations "in a vacuum", we actually create 4 smaller-degree sub-systems:
{x-y=0, x+y=0}; {x-y=0, x^2 + y^2 - 4 - xy=0}; {x+y=0, x^2 + y^2 - 2 + xy=0} and {x^2 + y^2 - 2 + xy = 0, x^2 + y^2 - 4 - xy=0}.
These sub-systems give us all our solutions, and both solutions to the last sub-system yield x^2+y^2=3, which we only add to the resulting sum once.
I was lost in the sauce, as they say, sorry for the confusion. 😔
@@sujitsivadanam Yes, that's exactly what I was confused about. And we can be sure they are both satisfied because that's one of the cases/subsystems we are considering when we transition from a bigger system:
{(x-y)(x^2 + y^2 - 2 + xy) = 0, (x+y)(x^2 + y^2 - 4 - xy)=0}
to 4 smaller systems that result from it:
{x-y=0, x+y=0} OR {x-y=0, x^2 + y^2 - 4 - xy=0} OR {x+y=0, x^2 + y^2 - 2 + xy=0} OR {x^2 + y^2 - 2 + xy = 0, x^2 + y^2 - 4 - xy=0}.
I forgot that you can't just take x,y so that x^2 + y^2 - 2 + xy = 0, you also need x+y=0 or x^2 + y^2 - 4 - xy=0 to hold, otherwise it's not a solution to the original system.
Sorry but I don'get something.
If x=y=2,then x^2+y^2=4, then when x=y=-2 we get oneother sum of x^2+y^2=4.
Shund't you consider all the values of x and y?
what about the integral i sent you ?
0+4+8+3=15 final answer
Prime Newtons AKA *Maths Beyond The Obvious* .
X^2+Y^2=4