I was totally expecting the first technique. By contrast, that second technique blew my mind. It is so simple, yet so elegant. I can’t believe it wasn’t part of my equation solving arsenal. Well, it is now!
I did it another way. Factorised the second eqn (x+y)(x^4-..........+y ^4) = 82 First brackets is 2. So x^4 - x^3.y + x^2 y^2 - x.y^3 + y^4 = 41 Subtract from this (x+y)^4, expanded, which is just 16, and we get after factorising out -5xy -5xy(x^2 + xy + y^2) = 25 The brackets is just 4-xy after substituting in the square of eqn 1. The rest is like solution 1 where we solve for xy and substitute into eqn 1.
i wrote y=2-x, obtained a quartic and applied newton raphson for it. starting from 0, after 3 or 4 iterations i got something like -0.41421 which looked oddly similar to sqrt(2) after the decimal point.
Ok until 1.38 time. Then putting (x^2 +y^2) = (x +y) ^2 - 2xy and simplifying I find this : (x+y) ^5=(x^5+y^5) - 5x^2y^2(x+y) +5xy(x+y)^3 and so xy =-1 v xy = 5. Then find x and y by a simple system : x^2 + y^2 = 4 - 2xy.. =6 and x + y = 2.... x =1+ 2^0,5 V y = 1- 2^0,5
You introduced a new variable in the second method, but it seems like it would be easier to just define y= 2-x and put that in the second equation and just solve for x.
It's doesn't seem that much easier to me. You have to factor a cubic polynomial, you can take the factor (x-y) out to make it quartic, while this solution is much quicker.
I did it the first way. The second method is very clever, and it's why I watch videos like this. My brain doesn't often accept the method that makes the problem harder, but in some cases that's the easiest thing to do.
Why did you reject complex solutions? They are quite fitting the system. (1+2i)^5 = 41+38i and (1-2i)^5 = 41-38i. So their sum is equal to 82 and sum of the 1+2i and 1-2i is equal to 2.
@@sdspivey They certainly do exist BUT they are not real numbers as in they aren't in the set of real numbers, just as 1/2 is not an integer complex numbers do not belong to R but they do belong to C.
I only know one way to solve this problem. Substitute x=1+z and y=1-z into the given equation and rearrange to (z+1)^5-(z-1)^5-82=0. Noting that odd powers of z cancel out and using Pascal's Triangle: 1 5 10 10 5 1, we get 2[5z^4+10z^2+1]-82=0 or z^4+2x^2-8=0, which has roots z^2=(-2±6)/2=2 or -4 or z=±√2 or ±2i. It follows that x=1+z=1±√2 or 1±2i and y=1-z=1∓√2 or 1∓2i.
I decided to represent x^5+y^5 as product of x+y and x^4+y^4 and subtracting products appropriately and this allows us to substitute xy=z and then solve the quadratic, and plug into the first eqn to solve for x and y.
Nice problem. I did it the first way, solving for xy. I also noticed that xy has to be negative, because if x and y are two positive numbers that add up to 2, then x^5 + y^5 cannot be larger than 32. So we can reject the xy = 5 solution right away and assume xy = -1.
I was totally expecting the first technique. By contrast, that second technique blew my mind. It is so simple, yet so elegant. I can’t believe it wasn’t part of my equation solving arsenal. Well, it is now!
Check IMO2019 shortlist's A3
I did it another way. Factorised the second eqn
(x+y)(x^4-..........+y ^4) = 82
First brackets is 2. So
x^4 - x^3.y + x^2 y^2 - x.y^3 + y^4 = 41
Subtract from this (x+y)^4, expanded, which is just 16, and we get after factorising out -5xy
-5xy(x^2 + xy + y^2) = 25
The brackets is just 4-xy after substituting in the square of eqn 1. The rest is like solution 1 where we solve for xy and substitute into eqn 1.
Hello there mcwulf, consider look to my channel too for similar math olympiad problems. Thanks and regards.
The second method is something I use often so it was nice to see it here!
The second one was as so unusual!
Very interesting substitution
That second method was very clever. Conjugates. Of course.
Wonderful approach
Hello there Badri! consider look to my channel too for similar math olympiad problems. Thanks and regards.
i wrote y=2-x, obtained a quartic and applied newton raphson for it. starting from 0, after 3 or 4 iterations i got something like -0.41421 which looked oddly similar to sqrt(2) after the decimal point.
The quartic can be factored into the form (y^2+ay+b)*(y^2+cy+d).
@@ranshen1486 nice, good to know
Ok until 1.38 time. Then putting (x^2 +y^2) = (x +y) ^2 - 2xy and simplifying I find this : (x+y) ^5=(x^5+y^5) - 5x^2y^2(x+y) +5xy(x+y)^3 and so xy =-1 v xy = 5. Then find x and y by a simple system : x^2 + y^2 = 4 - 2xy.. =6 and x + y = 2.... x =1+ 2^0,5 V y = 1- 2^0,5
2 nd method was very nice
x=1+2i,y=1-2i(e viceversa)... x=1+sqrt2,y=1-sqrt2 è viceversa
Obrigado por esse conteúdo !
You introduced a new variable in the second method, but it seems like it would be easier to just define y= 2-x and put that in the second equation and just solve for x.
It's doesn't seem that much easier to me. You have to factor a cubic polynomial, you can take the factor (x-y) out to make it quartic, while this solution is much quicker.
it just looks so nice when the a^5 cancels
I really like the 2nd approach.
Hello there! consider look to my channel too for similar math olympiad problems. Thanks and regards.
I did it the first way. The second method is very clever, and it's why I watch videos like this. My brain doesn't often accept the method that makes the problem harder, but in some cases that's the easiest thing to do.
Why did you reject complex solutions? They are quite fitting the system. (1+2i)^5 = 41+38i and (1-2i)^5 = 41-38i. So their sum is equal to 82 and sum of the 1+2i and 1-2i is equal to 2.
It's because the question asked for real number solutions. He says it in the first 5 seconds of the video.
@@ropenutter6321 Complex numbers ARE real. Just as real as negative numbers.
@@sdspivey They certainly do exist BUT they are not real numbers as in they aren't in the set of real numbers, just as 1/2 is not an integer complex numbers do not belong to R but they do belong to C.
@@ropenutter6321 oh, yeah, I missed that. That explains everything.
4:00 where do you get z^2 - 2z + 5 from?
This is going in my list of tools. Thanks
I only know one way to solve this problem. Substitute x=1+z and y=1-z into the given equation and rearrange to (z+1)^5-(z-1)^5-82=0. Noting that odd powers of z cancel out and using Pascal's Triangle: 1 5 10 10 5 1, we get 2[5z^4+10z^2+1]-82=0 or z^4+2x^2-8=0, which has roots z^2=(-2±6)/2=2 or -4 or z=±√2 or ±2i. It follows that x=1+z=1±√2 or 1±2i and y=1-z=1∓√2 or 1∓2i.
Genial. Merci
❤
I decided to represent x^5+y^5 as product of x+y and x^4+y^4 and subtracting products appropriately and this allows us to substitute xy=z and then solve the quadratic, and plug into the first eqn to solve for x and y.
Nice problem. I did it the first way, solving for xy. I also noticed that xy has to be negative, because if x and y are two positive numbers that add up to 2, then x^5 + y^5 cannot be larger than 32. So we can reject the xy = 5 solution right away and assume xy = -1.
Why can't x^5+y^5 be larger than 32?
Damn the second solution was much smarter
У меня получилось, что x^5+y^5=0. Сейчас посмотрю видео.
That is very clever
hi
You didn’t even say why you wrote 10 like that. You need not to skip step to get your point across. Be sure to make more clear.
Both the methods are praiseworthy .Thank you ,genius professor !!!