You made a mistake at 4:34 when you said x*y*z*(x+y+z) = x^2 * y^2 * z + x^2 * y * z^2 + x * y^2 * z^2 , but somehow the mistake did not prevent you from finding some correct solutions to the equations. Hmmm, I do not have time to look more closely.
The mistake stayed in all the equations equal to 12 until around the 11:15 mark. Then he says "I'll spare you the details" and he writes down the equation for 12 correctly (from his notes) all in terms of s.
Im sure there must be an easier way but i don't have time to find it. But you could go with my standard hint for such problems try x=1. which gives y^n+z^n =2^n for n=2,3,5 so {y,z}={0,2} is clearly a possibility and should get you some marks in an exam. but perhaps not in a competition.
Here's a solution that changes the variables from x,y,z to s,m,p where s = x + y + z m = x*y + x*z + y*z p = x*y*z Note that these substitutions correspond to Vieta's formula for a cubic, so if we know s, m, p then we can get x, y, z as the roots of the cubic equation t^3 - s*t^2 + m*t - p = 0 Then the given equations are: 1: s^2 - 2*m = 5 2: s^3 - 3*m*s + 3*p = 9 3: s^5 + 5*(m*s - p)*(m - s^2) = 33 The first two are not difficult, but the third one is a little tricky. I'll leave the steps as an exercise for the reader! Anyway, those equations are straightforward to solve. The first equation yields: m = (s^2 - 5) / 2 which can be substituted into the second equation and solved for p to get: p = (s^3 - 15*s + 18) / 6 and those can be substituted into the third equation to get (after some factoring): (s - 3)^2 * (s^3 + 6*s^2 + 2*s + 3) = 0 The cubic factor results in solutions for x,y,z that include imaginary numbers, so I will not continue with that factor. The other factor gives us s = 3 and the formulas above give p = 0 m = 2 Using the cubic equation mentioned earlier t^3 - 3*t^2 + 2*t = 0 = t*(t - 1)*(t - 2) which has roots 0, 1, 2 and so the values for x,y,z are the set {x,y,z} = {0,1,2}
Partitioning a natural numbet into Three square numbers hold the key. Here (2)^2 < 5 < 3^2 and 5 > 1^2 + 1^2 + 1^ 2 These Inequalities implies Highest partition is > 1^2 and each of the partition is < 3^2 Hereby Highest partition = 2^2 Left over task is to partition 5- 2^2 into two square numbers Hereby 5 = 2^2 + 1^2 + 0^2 is only feasible answer to partitioning Given equations being symmetric under any permuation of x, y, z one can begin with x^2 = 2^2 y^2 = 1^2 z^2 = 0^2 Substituting these values in the cubic and quintic equations one gets x* x^2 + y* y^2 + z * z^2 = 9 or 4x + y = 9 x* x^4 + y* y^4+ z * z^4 = 33 or 16x + y = 33 x = 2, y = 1 being a unique solution to these and of course z^2 = 0 i.e z= 0 Hereby all possible permutations of (2,1,0) is a feasible solution
Consider m=x*y+x*z+y*z then 2*m=s^2-5. Consider polynomial on t. t^3-s*t^2+m*s-p. For all s root of the quintic of video on s. Other solutions are roots the polynomial on t. t^3-s*t^2+m*t-p=t^3-s*t^2+(s^2-5)t/2+s/6*(s*(s^2-5)-10*s+8). By X J Will see. The quintic in s is (s-3)^2*(s^3+6*s^2+2*s+3). But s^3+6*s^2+2*s+3=(s+2)^3-10*(s+2)+15. Irreducible by Einseintein by 5. Thus the solution are complicated (a priori roots of nonic). (This nonic is solvable by radicals).
There's a serious error at 9.44min. You have factored out -xyz and written wrongly inside parenthesis (x+y+z), which actually should be (xy+yz+zx). How do you proceed further with this step wrong?
You made an important mistake at min 4:34. Though that dsnt change the result, becouse only of a matter of luck, becouse due to the solution set: xy + xz - yz = x + y + x.... As i said its just luck with this solt.set
With a guess that x=0, we get an overdetermined system so why not Linear Algebra? y^2+z^2=5, y^3+z^3=9, y^5+z^5=33 leads to two 2X2 matrix equations: [(x^2 , y^2),( 0 , 1)].[(1 , y),(1 , z)] = [(5 , 9),(1,z)] ... ➀ and [(x^2 , y^2),( 0 , 1)].[(1 , y^3),(1 , z^3)] = [(5 , 33),(1,z^3)]... ➁ noticing that the leading matrix is the same in both cases? [(x^2 , y^2),( 0 , 1)] = [(5 , 9),(1,z)].[(1 , y),(1 , z)]^-1 substitute into second matrix eqn ➁ get... [(5 , 9),(1,z)].[(1 , y),(1 , z)]^-1.[(1 , y^3),(1 , z^3)] = [(5 , 33),(1,z^3)]...➂ equating determinants of LHS to RHS of ➂ get, (5z-9)(1)(z^3-y^3)-(5z^3-33)=0 ... ➃ which is a quartic in "z" but cubic in "y"? z =2, y=1 solves by inspection, finally *{x,y,z} = {0,1,2}* is at least one solution set, the only one since cubic in "y".🤔
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You made a mistake at 4:34 when you said x*y*z*(x+y+z) = x^2 * y^2 * z + x^2 * y * z^2 + x * y^2 * z^2 , but somehow the mistake did not prevent you from finding some correct solutions to the equations. Hmmm, I do not have time to look more closely.
because he didn't use that middle one.
The mistake stayed in all the equations equal to 12 until around the 11:15 mark. Then he says "I'll spare you the details" and he writes down the equation for 12 correctly (from his notes) all in terms of s.
I’m glad you said something. That was driving me crazy.
Im sure there must be an easier way but i don't have time to find it.
But you could go with my standard hint for such problems try x=1.
which gives y^n+z^n =2^n
for n=2,3,5
so {y,z}={0,2} is clearly a possibility and should get you some marks in an exam. but perhaps not in a competition.
Here's a solution that changes the variables from x,y,z to s,m,p where
s = x + y + z
m = x*y + x*z + y*z
p = x*y*z
Note that these substitutions correspond to Vieta's formula for a cubic, so if we know s, m, p then
we can get x, y, z as the roots of the cubic equation
t^3 - s*t^2 + m*t - p = 0
Then the given equations are:
1: s^2 - 2*m = 5
2: s^3 - 3*m*s + 3*p = 9
3: s^5 + 5*(m*s - p)*(m - s^2) = 33
The first two are not difficult, but the third one is a little tricky. I'll leave the steps as an exercise for the reader!
Anyway, those equations are straightforward to solve. The first equation yields:
m = (s^2 - 5) / 2
which can be substituted into the second equation and solved for p to get:
p = (s^3 - 15*s + 18) / 6
and those can be substituted into the third equation to get (after some factoring):
(s - 3)^2 * (s^3 + 6*s^2 + 2*s + 3) = 0
The cubic factor results in solutions for x,y,z that include imaginary numbers, so I will not continue with that factor. The other factor gives us
s = 3 and the formulas above give
p = 0
m = 2
Using the cubic equation mentioned earlier
t^3 - 3*t^2 + 2*t = 0 = t*(t - 1)*(t - 2)
which has roots 0, 1, 2 and so the values for x,y,z are the set
{x,y,z} = {0,1,2}
4:27 "-xyz(x+y+z)" is an error. it should have been "-xyz(xy + xz + yz)"
Yes,but he did find xy+xz+yz in terms of S, so i guess the final eqution for S is right.
@@yoav613 yeah because he didn't use that middle one.
Right!
The error at 4:27 makes the entire solution invalid.
At first sight {0; 1; 2}
Wow!
x^2 + y^2 + z^2 = 5
x^3 + y^3 + z^3 = 9
x^5 + y^5 + z^5 = 33
Partitioning 5 into two or more square numbers being the issue
and 1^2 + 1^2 + 1^2 < 5
1^2 + 2^2 = 5
So 0^2 + 1^2 + 2^2 = 5
Again 0^3 + 1^3 + 2^3 = 9
And 0^5 + 1^5 + 2^5 = 33
Therefore set of feasible solutions of (x,y,z) equals to
{( 0, 1, 2), (0, 2, 1)
(1, 0, 2), (1, 2, 0),
(2, 1, 0), (2, 0,, 1) }
Good approach !
What about negatives?
But this doesn’t show that there are no non integer solutions
Partitioning a natural numbet into Three square numbers hold the key.
Here
(2)^2 < 5 < 3^2
and 5 > 1^2 + 1^2 + 1^ 2
These Inequalities implies
Highest partition is > 1^2
and each of the partition is < 3^2
Hereby
Highest partition = 2^2
Left over task is to partition
5- 2^2 into two square numbers
Hereby
5 = 2^2 + 1^2 + 0^2 is only feasible answer to partitioning
Given equations being symmetric under any permuation of x, y, z one can begin with
x^2 = 2^2
y^2 = 1^2
z^2 = 0^2
Substituting these values in the cubic and quintic equations one gets
x* x^2 + y* y^2 + z * z^2 = 9
or 4x + y = 9
x* x^4 + y* y^4+ z * z^4 = 33
or 16x + y = 33
x = 2, y = 1 being a unique solution to these and of course z^2 = 0 i.e z= 0
Hereby all possible permutations of (2,1,0) is a feasible solution
So do we get more solutions for other values of s?
Consider m=x*y+x*z+y*z then 2*m=s^2-5. Consider polynomial on t. t^3-s*t^2+m*s-p. For all s root of the quintic of video on s. Other solutions are roots the polynomial on t. t^3-s*t^2+m*t-p=t^3-s*t^2+(s^2-5)t/2+s/6*(s*(s^2-5)-10*s+8). By X J Will see. The quintic in s is (s-3)^2*(s^3+6*s^2+2*s+3). But s^3+6*s^2+2*s+3=(s+2)^3-10*(s+2)+15. Irreducible by Einseintein by 5. Thus the solution are complicated (a priori roots of nonic). (This
nonic is solvable by radicals).
There's a serious error at 9.44min.
You have factored out -xyz and written wrongly inside parenthesis (x+y+z), which actually should be (xy+yz+zx). How do you proceed further with this step wrong?
Thanks for your service please bring some questions like this
You must say real solutions. Because the system have more solution. (Real solutions is a condition strong.
Well done for your hard work! It is not easy to make avideo for this problem💯
Thank you 🙌🧡
You made an important mistake at min 4:34. Though that dsnt change the result, becouse only of a matter of luck, becouse due to the solution set: xy + xz - yz = x + y + x.... As i said its just luck with this solt.set
Can you do something about the symmetry tho,
2^2+1=5
2^3+1=9
2^5+1=32
good observation
Thanks a Yottabyte for making these videos ❤🧡😸
Glad you like them! 🤗
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Təşəkkürlər. 🧡
Thanks for this vidéo
What about the other solutions?!
great job
Thank you!
Why doesn't consider complex solution
I think you made a mistake at 11.33
When u realize that youtube teaches better than your teacher lol......😆
Use Newton's Identities
With a guess that x=0,
we get an overdetermined system so why not Linear Algebra?
y^2+z^2=5,
y^3+z^3=9,
y^5+z^5=33
leads to two 2X2 matrix equations:
[(x^2 , y^2),( 0 , 1)].[(1 , y),(1 , z)] = [(5 , 9),(1,z)] ... ➀
and
[(x^2 , y^2),( 0 , 1)].[(1 , y^3),(1 , z^3)] = [(5 , 33),(1,z^3)]... ➁
noticing that the leading matrix is the same in both cases?
[(x^2 , y^2),( 0 , 1)] = [(5 , 9),(1,z)].[(1 , y),(1 , z)]^-1
substitute into second matrix eqn ➁ get...
[(5 , 9),(1,z)].[(1 , y),(1 , z)]^-1.[(1 , y^3),(1 , z^3)] = [(5 , 33),(1,z^3)]...➂
equating determinants of LHS to RHS of ➂ get,
(5z-9)(1)(z^3-y^3)-(5z^3-33)=0 ... ➃
which is a quartic in "z" but cubic in "y"?
z =2, y=1 solves by inspection,
finally *{x,y,z} = {0,1,2}* is at least one solution set, the only one since cubic in "y".🤔
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