Thread Rating:

## Poll

No votes (0%) | |||

No votes (0%) | |||

1 vote (100%) | |||

No votes (0%) | |||

No votes (0%) | |||

1 vote (100%) | |||

1 vote (100%) | |||

1 vote (100%) | |||

1 vote (100%) | |||

1 vote (100%) |

**1 member has voted**

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:

I will provide hints if everybody seems to be stuck.

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:

- Please don't just plop a URL to a solution elsewhere until a winner here has been declared.
- 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.
- Beer to the first satisfactory answer and solution, subject to rule 2.
- Please put answers and solutions in spoiler tags.

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.

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.

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.

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.

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.

March 30th, 2020 at 7:42:18 PM
permalink

Quote:WizardQuote: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

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

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

April 4th, 2020 at 5:30:42 PM
permalink

Quote:ThatDonGuyWizard, 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.

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 = x

c = x

Solve for x

Let d = x

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

a

x

b + 2(xy + xz + yz)

(1) xy + xz + yz = (a

a * b = (x + y + z)*( x

= x

= c + xy

(2) xy

a

x

c + 3*(ab - c) + 6xyz

6xyz = a

6xyz = a

(3) xyz = (a

b

(4) x

a

x

d + 2*(x

d + 2*(x

d + (b

b

2(x

x

(5) x

a

x

d + 4*( a

d + 4a

d + 4a

4/3*a

2d = 1/3*a

d = a

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

d = 1

Let:

a = x + y + z

b = x

^{2}+ y^{2}+ z^{2}c = x

^{3}+ y^{3}+ z^{3}Solve for x

^{4}+ y^{4}+ z^{4}in terms of a, b, and c.Let d = x

^{4}+ y

^{4}+ z

^{4}

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

a

^{2}= (x + y + z)

^{2}=

x

^{2}+ y

^{2}+ z

^{2}+ 2*(xy + xz + yz) =

b + 2(xy + xz + yz)

(1) xy + xz + yz = (a

^{2}- b)/2

a * b = (x + y + z)*( x

^{2}+ y

^{2}+ z

^{2})

= x

^{3}+ y

^{3}+ z

^{3}+ xy

^{2}+ xz

^{2}+ yx

^{2}+ yz

^{2}+ zx

^{2}+ zy

^{2}

= c + xy

^{2}+ xz

^{2}+ yx

^{2}+ yz

^{2}+ zx

^{2}+ zy

^{2}

(2) xy

^{2}+ xz

^{2}+ yx

^{2}+ yz

^{2}+ zx

^{2}+ zy

^{2}= ab - c

a

^{3}= (x + y + z)

^{3}=

x

^{3}+ y

^{3}+ z

^{3}+ 3*( xy

^{2}+ xz

^{2}+ yx

^{2}+ yz

^{2}+ zx

^{2}+ zy

^{2}) + 6xyz =

c + 3*(ab - c) + 6xyz

6xyz = a

^{3}- c - 3*(ab-c)

6xyz = a

^{3}+ 2c - 3*ab

(3) xyz = (a

^{3}+ 2c - 3ab)/6

b

^{2}= (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})

(4) x

^{2}y

^{2}+ x

^{2}z

^{2}+ y

^{2}z

^{2}= (b

^{2}- d)/2

a

^{2}* b = (x

^{2}+ y

^{2}+ z

^{2}+ 2*(xy + xz + yz)) * (x

^{2}+ y

^{2}+ z

^{2}) =

x

^{4}+ y

^{4}+ z

^{4}+ 2*(x

^{2}y

^{2}+ x

^{2}z

^{2}+ y

^{2}z

^{2}) + 2(x

^{3}y + x

^{3}z + y

^{3}x + y

^{3}z + z

^{3}x + z

^{3}y) + 2(x

^{2}yz + xy

^{2}z + xyz

^{2}) =

d + 2*(x

^{2}y

^{2}+ x

^{2}z

^{2}+ y

^{2}z

^{2}) + 2(x

^{3}y + x

^{3}z + y

^{3}x + y

^{3}z + z

^{3}x + z

^{3}y) + 2xyz*(x+y+z) =

d + 2*(x

^{2}y

^{2}+ x

^{2}z

^{2}+ y

^{2}z

^{2}) + 2(x

^{3}y + x

^{3}z + y

^{3}x + y

^{3}z + z

^{3}x + z

^{3}y) + a(a

^{3}+ 2c - 3ab)/3 =

d + (b

^{2}- d) + 2(x

^{3}y + x

^{3}z + y

^{3}x + y

^{3}z + z

^{3}x + z

^{3}y) + a(a

^{3}- 3ab + 2c)/3 =

b

^{2}+ 2(x

^{3}y + x

^{3}z + y

^{3}x + y

^{3}z + z

^{3}x + z

^{3}y) + a(a

^{3}- 3ab+ 2c)/3

2(x

^{3}y + x

^{3}z + y

^{3}x + y

^{3}z + z

^{3}x + z

^{3}y) = a

^{2}* b - b

^{2}- a(a

^{3}- 3ab+ 2c)/3

x

^{3}y + x

^{3}z + y

^{3}x + y

^{3}z + z

^{3}x + z

^{3}y = a

^{2}* b/2 - b

^{2}/2 - a

^{4}/6 + a

^{2}b/2 - ac/3

(5) x

^{3}y + x

^{3}z + y

^{3}x + y

^{3}z + z

^{3}x + z

^{3}y = a

^{2}* b - b

^{2}/2 - a

^{4}/6 - ac/3

a

^{4}= (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 + xy

^{2}z + xyz

^{2}) =

d + 4*( a

^{2}b - b

^{2}/2 - a

^{4}/6 - ac/3) + 6*((b

^{2}- d)/2) + 12*xyz*(x+y+z) =

d + 4a

^{2}b - 2b

^{2}- (2/3)*a

^{4}- (4/3)ac + 3*b

^{2}- 3d + 12*((a

^{3}+ 2c - 3ab)/6)*a =

d + 4a

^{2}b - 2b

^{2}- (2/3)*a

^{4}- (4/3)ac + 3b

^{2}- 3d + 2a

^{4}+ 4ac - 6a

^{2}b =

4/3*a

^{4}+ (8/3)ac - 2a

^{2}b + b

^{2}

2d = 1/3*a

^{4}+ (8/3)ac - 2a

^{2}b

d = a

^{4}/6 + (4/3)ac - a

^{2}b + b

^{2}/2

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

d = 1

^{4}/6 + (4/3)1*9 - 1

^{2}*4 + 4

^{2}/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.

April 7th, 2020 at 8:27:32 PM
permalink

Here is my PDF solution.

It's not whether you win or lose; it's whether or not you had a good bet.