## Poll

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1 member has voted

Wizard Joined: Oct 14, 2009
• Posts: 21598
March 30th, 2020 at 11:47:33 AM permalink
This puzzle is definitely beer worthy. If you're looking for something a little less challenging, I recommend the easy math puzzles thread. I will admit I was stuck on this one and needed some help on a method I didn't know to solve it. Now, I have what I believe is the answer, but it would be nice to have it confirmed. That said, here is the problem:

Given:
x + y + z = 1
x^2 + y^2 + z^2 = 4
x^3 + y^3 + z^3 = 9

What is x^4 + y^4 + z^4 ?

Usual rules:

1. Please don't just plop a URL to a solution elsewhere until a winner here has been declared.
2. All those who have won a beer previously are asked to not post answers or solutions for 24 after this posting. Past winners who must chime in early, may PM me.
3. Beer to the first satisfactory answer and solution, subject to rule 2.

I will provide hints if everybody seems to be stuck.
Last edited by: Wizard on Mar 30, 2020
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard Joined: Oct 14, 2009
• Posts: 21598
March 30th, 2020 at 2:39:33 PM permalink
I got a PM from someone who was under the impression I didn't know the answer to my own question. As always, there is a chance I could be wrong, but I do have what I think is a correct answer. I am not crystal clear why the method I followed works, which I hope can be a topic of discussion once we at least come to a majority agreement what the answer is.
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard Joined: Oct 14, 2009
• Posts: 21598
March 30th, 2020 at 5:50:00 PM permalink
I received a correct answer and solution by PM from an existing Beer Club member. The clock is ticking for non-members to get in the club before the 24-hour clock expires.
Last edited by: Wizard on Mar 30, 2020
It's not whether you win or lose; it's whether or not you had a good bet.
ssho88 Joined: Oct 16, 2011
• Posts: 427
March 30th, 2020 at 7:00:05 PM permalink

x^4 + y^4 + z^4 = 97/6 = 16.16666 ?

I will show the calculations if correct.

Wizard Joined: Oct 14, 2009
• Posts: 21598
March 30th, 2020 at 7:38:40 PM permalink
Quote: ssho88

x^4 + y^4 + z^4 = 97/6 = 16.16666 ?

I will show the calculations if correct.

Yes, I agree. May I see the calculations?
It's not whether you win or lose; it's whether or not you had a good bet.
ssho88 Joined: Oct 16, 2011
• Posts: 427
March 30th, 2020 at 7:42:18 PM permalink
Quote: Wizard

Quote: ssho88

x^4 + y^4 + z^4 = 97/6 = 16.16666 ?

I will show the calculations if correct.

Yes, I agree. May I see the calculations?

x + y + z = 1-----------------------Eq1
x^2 + y^2 + z^2 = 4-----------------Eq2
x^3 + y^3 + z^3 = 9-----------------Eq3

Eq1 * Eq2
(x + y + z)*(x^2 + y^2 + z^2) = 1 * 4 = 4

(x^3 + y^3 + z^3) + yx^2 + xy^2 + xz^2 + zx^2 + zy^2 + yz^2 = 4

Substitute Eq3 into above equation,
9 + yx^2 + xy^2 + xz^2 + zx^2 + zy^2 + yz^2 = 4

yx^2 + xy^2 + xz^2 + zx^2 + zy^2 + yz^2 = -5 -----------------Eq4

Eq1 * Eq1
(x + y + z)*(x + y + z) = 1

(x^2 + y^2 + z^2) + 2(xy + xz + yz) = 1

Substitute Eq2 into above equation,
4 + 2(xy + xz + yz) = 1

(xy + xz + yz) = -3/2 -----------------Eq5

Eq1 * Eq5
(x + y + z)*(xy + xz + yz) = 1 * -3/2 = -3/2

(yx^2 + xy^2 + xz^2 + zx^2 + zy^2 + yz^2) + 3(xyz) = -3/2

Substitute Eq4 into above equation,

So, -5 + 3(xyz) = -3/2

(xyz) = 7/6 -----------------Eq6

Eq2 * Eq5
(x^2 + y^2 + z^2)*(xy + xz + yz) = 4*-3/2 = -6

(yx^3 + zx^3 + xy^3 + zy^3 +xz^3 + yz^3) + xyz^2 + xzy^2 + yzx^2 = -6

(yx^3 + zx^3 + xy^3 + zy^3 +xz^3 + yz^3) + xyz(x + y + z) = -6

Substitute Eq6 and Eq1 into above equation,
(yx^3 + zx^3 + xy^3 + zy^3 +xz^3 + yz^3) + 7/6(1) = -6

(yx^3 + zx^3 + xy^3 + zy^3 +xz^3 + yz^3) = -6 - 7/6 = -43/6 -----------------Eq7

Eq1 * Eq3
(x + y + z)*(x^3 + y^3 + z^3) = 9

x^4 + y^4 + z^4 + (yx^3 + zx^3 + xy^3+ zy^3 + xz^3+ yz^3) = 9

Substitute Eq7 into above equation,
x^4 + y^4 + z^4 + (-43/6) = 9

x^4 + y^4 + z^4 = 9 + 43/6 = 97/6 = 16.166666

Last edited by: ssho88 on Mar 30, 2020
ThatDonGuy Joined: Jun 22, 2011
• Posts: 4103
April 4th, 2020 at 7:04:15 AM permalink
Wizard, how did you solve it?

Here's my solution - it's similar to ssho88's:

(x + y + z)^2 = x^2 + y^2 + z^2 + 2 (xy + xz + yz)
1 = 4 + 2 (xy + xz + yz)
xy + xz + yz = -3/2

(x + y + z)^3 = x^3 + y^3 + z^3 + 3 (x^2 y + x^2 z + y^2 x + y^2 z + z^2 x + z^2 y) + 6 xyz
1 = 9 + 3 (x^2 x + x^2 y + x^2 z + y^2 x + y^2 y + y^2 z + z^2 x + z^2 y + z^2 z - x^3 - y^3 - z^3) + 6 xyz
-8 = 3 ((x^2 + y^2 + z^2)(x + y + z) - (x^3 + y^3 + z^3)) + 6 xyz
-8 = 3 (4 * 1 - 9) + 6 xyz
xyz = (15 - 8) / 6 = 7/6

Let S represent x^4 + y^4 + z^4:

(x^2 + y^2 + z^2)^2 = x^4 + y^4 + z^4 + 2 (x^2 y^2 + x^2 z^2 + y^2 z^2)
16 = S + 2 (x^2 y^2 + x^2 z^2 + y^2 z^2)
2 (x^2 y^2 + x^2 z^2 + y^2 z^2) = 16 - S
6 (x^2 y^2 + x^2 z^2 + y^2 z^2) = 48 - 3S

(x + y + z)^4 = x^4 + y^4 + z^4 + 4 (x^3 y + x^3 z + y^3 x + y^3 z + z^3 x + z^3 y) + 6 (x^2 y^2 + x^2 z^2 + y^2 z^2)
+ 12 (x^2 yz + y^2 xz + z^2 xy)
1 = S + 4 (x^3 x + x^3 y + x^3 z + y^3 x + y^3 y + y^3 z + z^3 x + z^3 y + z^3 z - x^4 - y^4 - z^4) + 48 - 3S + 12 xyz (x + y + z)
1 = S + 4 ((x^3 + y^3 + z^3)(x + y + z) - (x^4 + y^4 + z^4)) + 48 - 3S + 12 * 7/6
1 = S + 4 (9 - S) + 48 - 3S + 14
1 = 98 - 6S

x^4 + y^4 + z^4 = S = 97/6

Wizard Joined: Oct 14, 2009
• Posts: 21598
April 4th, 2020 at 5:30:42 PM permalink
Quote: ThatDonGuy

Wizard, how did you solve it?

Yes. I'm proud to say I solved it for the general case without using the Newton Girard identities. It was quite a mess and took a while.

My solution looks similar to yours, just eyeballing it.

I will post something soon, as I plan to make an Ask the Wizard question out of it.
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard Joined: Oct 14, 2009
• Posts: 21598
April 6th, 2020 at 9:18:30 PM permalink
I was feeling a little bored so decided to solve the problem for the general case. Let me state is as follows:

Let:
a = x + y + z
b = x2 + y2 + z2
c = x3 + y3 + z3

Solve for x4 + y4 + z4 in terms of a, b, and c.

Let d = x4 + y4 + z4

Let's start with a and b to find values for expressions involving both a and b.

a2 = (x + y + z)2 =
x2 + y2 + z2 + 2*(xy + xz + yz) =
b + 2(xy + xz + yz)
(1) xy + xz + yz = (a2 - b)/2

a * b = (x + y + z)*( x2 + y2 + z2)
= x3 + y3 + z3 + xy2 + xz2 + yx2 + yz2 + zx2 + zy2
= c + xy2 + xz2 + yx2 + yz2 + zx2 + zy2
(2) xy2 + xz2 + yx2 + yz2 + zx2 + zy2 = ab - c

a3 = (x + y + z)3 =
x3 + y3 + z3 + 3*( xy2 + xz2 + yx2 + yz2 + zx2 + zy2) + 6xyz =
c + 3*(ab - c) + 6xyz
6xyz = a3 - c - 3*(ab-c)
6xyz = a3 + 2c - 3*ab
(3) xyz = (a3 + 2c - 3ab)/6

b2 = (x2 + y2 + z2)2 = x4 + y4 + z4 + 2*( x2y2 + x2z2 + y2z2)
(4) x2y2 + x2z2 + y2z2 = (b2 - d)/2

a2 * b = (x2 + y2 + z2 + 2*(xy + xz + yz)) * (x2 + y2 + z2) =
x4 + y4 + z4 + 2*(x2y2 + x2z2 + y2z2) + 2(x3y + x3z + y3x + y3z + z3x + z3y) + 2(x2yz + xy2z + xyz2) =
d + 2*(x2y2 + x2z2 + y2z2) + 2(x3y + x3z + y3x + y3z + z3x + z3y) + 2xyz*(x+y+z) =
d + 2*(x2y2 + x2z2 + y2z2) + 2(x3y + x3z + y3x + y3z + z3x + z3y) + a(a3 + 2c - 3ab)/3 =
d + (b2 - d) + 2(x3y + x3z + y3x + y3z + z3x + z3y) + a(a3 - 3ab + 2c)/3 =
b2 + 2(x3y + x3z + y3x + y3z + z3x + z3y) + a(a3 - 3ab+ 2c)/3

2(x3y + x3z + y3x + y3z + z3x + z3y) = a2 * b - b2 - a(a3 - 3ab+ 2c)/3
x3y + x3z + y3x + y3z + z3x + z3y = a2 * b/2 - b2/2 - a4/6 + a2b/2 - ac/3
(5) x3y + x3z + y3x + y3z + z3x + z3y = a2 * b - b2/2 - a4/6 - ac/3

a4 = (x + y + z)4 =
x4 + y4 + z4 + 4*(x3y + x3z + y3x + y3z + z3x + z3y) + 6*(x2y2 +x2z2 + y2z2) + 12*(x2yz + xy2z + xyz2) =
d + 4*( a2b - b2/2 - a4/6 - ac/3) + 6*((b2 - d)/2) + 12*xyz*(x+y+z) =
d + 4a2b - 2b2 - (2/3)*a4 - (4/3)ac + 3*b2 - 3d + 12*((a3 + 2c - 3ab)/6)*a =
d + 4a2b - 2b2 - (2/3)*a4 - (4/3)ac + 3b2 - 3d + 2a4 + 4ac - 6a2b =
4/3*a4 + (8/3)ac - 2a2b + b2

2d = 1/3*a4 + (8/3)ac - 2a2b
d = a4/6 + (4/3)ac - a2b + b2/2

Going back to the original problem were a = 1, b = 4, and c = 9:

d = 14/6 + (4/3)1*9 - 12*4 + 42/2 = 97/6 = 16 1/6.
Last edited by: Wizard on Apr 7, 2020
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard 