nice to see a general solution method!...solved by just repeating Si*S1 (1) xyz (R)& xy+yz+zx (Q) in terms of S2 (2)solved S2 then S5...kinda neat, 2 integer solutions if M unbounded
If the last equation had been x^4 + y^4 + z^4 = 33, the solution would have been so much easier. Simple trial and error reveals x=y=2 and z=-1 satisfies all of the stipulations. Then the answer is simply 32 + 32 -1 = 63. Interesting how such a small change on the one equation makes such a difference.
That was a nice one. I solved it. I used elementary symmetric functions to find the value of symmetric functions of degree 1,2,3 of three variables then put them in vieta's formula to find x,y and z. Then put them in the S5. 😇
Einstein is always Einstein. By the way, you haven't taken Ifter meals from our house. I think you like our hospitability and food. We are now sad about your conduct because you have not taken Iftari foods.
i solved it totally different (and in my opinion easier) way. I directly solved this system of equations and got solutions (x,y,z)=(1,sqrt(2)+1,1-sqrt(2)) and permutations. Then i just find x^5+y^5+z^5 easily
I think that I’ve solved the problem in the same your way. I’ve just used two algebraic tricks: x+y = 3-z, x^3 + y^3 = (x+ y)(x^2 -xy + y^2) = 15 - z^3 and (x^2 + y^2)^2 = 35 - z^4 +2(xy)^2. After these passages solve the new system and you obtain an 6th degree equation in z but really easy to solve using Ruffini for example or some decomposition, you find z = 1 and so substitute this in the previous equations and you get a second degree equation in xy solving that equation you obtain xy =-1 or xy = 15 and always by substitution you’ve done (i.e. your solutions). In this way, more computational, the system can be solved also by one who doesn't know much “advanced” math, in the spirit of IMO (tested for the high school guys).
Yes, you can simplify the system in this two equation where the solutions are x, y and z: λ³−3λ²+λ+1=0 and λ³−3λ−λ+7=0 The first one is easy to solve by factoring the second one you can prove that the solutions aren’t all real (or you can see that S₂>0) But this is his method with a more complicated final (if you want to find x⁵+y⁵+z⁵)
@@nirmankhan2134 Einstein is always Einstein. By the way, you haven't taken Ifter meals from our house. I think you like our hospitability and food. We are now sad about your conduct because you have not taken Iftari foods.
This is the type of Math video i find worth watching, beautifully done!!
The solution is critical to keep in mind!
nice to see a general solution method!...solved by just repeating Si*S1 (1) xyz (R)& xy+yz+zx (Q) in terms of S2 (2)solved S2 then S5...kinda neat, 2 integer solutions if M unbounded
Nice problem and of course nice solution! Would be great if you can upload more combinatorial problems.
If the last equation had been x^4 + y^4 + z^4 = 33, the solution would have been so much easier. Simple trial and error reveals x=y=2 and z=-1 satisfies all of the stipulations. Then the answer is simply 32 + 32 -1 = 63. Interesting how such a small change on the one equation makes such a difference.
That was a nice one. I solved it. I used elementary symmetric functions to find the value of symmetric functions of degree 1,2,3 of three variables then put them in vieta's formula to find x,y and z. Then put them in the S5. 😇
Einstein is always Einstein. By the way, you haven't taken Ifter meals from our house. I think you like our hospitability and food. We are now sad about your conduct because you have not taken Iftari foods.
i solved it totally different (and in my opinion easier) way. I directly solved this system of equations and got solutions (x,y,z)=(1,sqrt(2)+1,1-sqrt(2)) and permutations. Then i just find x^5+y^5+z^5 easily
Futhermore, my method proves that without condition x^2+y^2+z^2
I think that I’ve solved the problem in the same your way. I’ve just used two algebraic tricks: x+y = 3-z, x^3 + y^3 = (x+ y)(x^2 -xy + y^2) = 15 - z^3 and (x^2 + y^2)^2 = 35 - z^4 +2(xy)^2. After these passages solve the new system and you obtain an 6th degree equation in z but really easy to solve using Ruffini for example or some decomposition, you find z = 1 and so substitute this in the previous equations and you get a second degree equation in xy solving that equation you obtain xy =-1 or xy = 15 and always by substitution you’ve done (i.e. your solutions). In this way, more computational, the system can be solved also by one who doesn't know much “advanced” math, in the spirit of IMO (tested for the high school guys).
The two method are the same in reality but they seems different
gained a new appreciation for vieta's formula
I was wondering can you upload solutions to IMO 1969 Problems 1 and 6. Thank you!
so cool!
The image at the start of the video should have the entire problem
I thought you are going to solving for x,y,z. Is this possible in this problem?
Yes, you can simplify the system in this two equation where the solutions are x, y and z:
λ³−3λ²+λ+1=0
and
λ³−3λ−λ+7=0
The first one is easy to solve by factoring the second one you can prove that the solutions aren’t all real (or you can see that S₂>0)
But this is his method with a more complicated final (if you want to find x⁵+y⁵+z⁵)
Class!
👍
I Solved With Girard-Newton Identity,It's Take 5 Minutes
How did you get the value of x²+y²+z² I started solving by this method also and I need the value of only x²+y²+z²
@@brinzanalexandru2150 After Reach X^4 + Y^4 + Z^4,You Just Solve For e2,p2,And e3
@@iceblaze8180 can u explain it further
😍😍😍😍🔥🔥🔥🔥
Where are you from?
@@mustafizrahman2822 hello
@@mustafizrahman2822 nice to meet you. I am from ghe Earth?
@@tianqilong8366 Ghe earth! What is this? Are you from China?
@@nirmankhan2134 Einstein is always Einstein. By the way, you haven't taken Ifter meals from our house. I think you like our hospitability and food. We are now sad about your conduct because you have not taken Iftari foods.
Hi letsthinkcritically! I send you an email about Shef's Scholars which could be of interest to your viewer who are competing in math competitions !