MHB Find the value of x^5 + y^5 + z^5.

[anemone]

Knowing that the system

$x+y+z=3$

$x^3+y^3+z^3=15$

$x^4+y^4+z^4=35$

has a real solution x, y, and z for which $x^2+y^2+z^2<10$, find the value of $x^5+y^5+z^5$ for that solution.

Hi all,

I found this problem very interesting and wanted to post it here as one of the challenging problem and I have solved it and am looking forward to seeing your method and solution too.

I will post my solution in a few days, so that everyone interested can have a chance to demonstrate how they would solve it.
anemone

 

[Opalg]

Re: Find the value of x^5+y^5+z^5.

Knowing that the system

$x+y+z=3$

$x^3+y^3+z^3=15$

$x^4+y^4+z^4=35$

has a real solution x, y, and z for which $x^2+y^2+z^2<10$, find the value of $x^5+y^5+z^5$ for that solution.

Hi all,

I found this problem very interesting and wanted to post it here as one of the challenging problem and I have solved it and am looking forward to seeing your method and solution too.

I will post my solution in a few days, so that everyone interested can have a chance to demonstrate how they would solve it.
anemone

I used Newton's identities to solve this problem. Let $\lambda^3 -a\lambda^2+b\lambda-c=0$ be the equation whose solutions are $x,\ y$ and $z$, and let $p_k = x^k+y^k+z^k$, for $k=1,2,3,\ldots$. Newton's identities say that $$\begin{aligned}a &= p_1 \\ 2b &= ap_1-p_2 \\ 3c &= bp_1-ap_2 + p_3 \\ 0 &= cp_1-bp_2+ap_3 - p_4 \\ 0 &= 0p_1 - cp_2 + bp_3-ap_4 + p_5 .\end{aligned}$$ Inserting the values that we are given, the first four identities become $$\begin{aligned}a &= 3 \\ 2b &= 9-p_2 \\ 3c &= 3b-3p_2 + 15 \\ 0 &= 3c-bp_2+45 - 35 .\end{aligned}$$ Substitute the values for $b$ and $c$ from the second and third of these equations into the final one, we get $$0 = \tfrac32(9-p_2) - 3p_2 + 15 - \tfrac12(9-p_2)p_2 + 10 = \tfrac12(p_2^2 - 18p_2 + 77) = \tfrac12(p_2-7)(p_2-11).$$ That gives possible values of $7$ or $11$ for $p_2$. But we are told that $p_2<10$, so we must have $p_2=7.$

Substituting that value for $p_2$ it is easy to check that $b=1$, $c=-1$, and then the last of the Newton identities above gives $\boxed{p_5 = 83}.$

[The cubic equation in $\lambda$ then becomes $\lambda^3 -3\lambda^2+\lambda+1=0$, which has solutions $\lambda = 1$ and $\lambda = 1\pm\sqrt2.$ So those are the values of $x$, $y$ and $z$.]

 

[anemone]

Re: Find the value of x^5+y^5+z^5.

I used Newton's identities to solve this problem. Let $\lambda^3 -a\lambda^2+b\lambda-c=0$ be the equation whose solutions are $x,\ y$ and $z$, and let $p_k = x^k+y^k+z^k$, for $k=1,2,3,\ldots$. Newton's identities say that $$\begin{aligned}a &= p_1 \\ 2b &= ap_1-p_2 \\ 3c &= bp_1-ap_2 + p_3 \\ 0 &= cp_1-bp_2+ap_3 - p_4 \\ 0 &= 0p_1 - cp_2 + bp_3-ap_4 + p_5 .\end{aligned}$$ Inserting the values that we are given, the first four identities become $$\begin{aligned}a &= 3 \\ 2b &= 9-p_2 \\ 3c &= 3b-3p_2 + 15 \\ 0 &= 3c-bp_2+45 - 35 .\end{aligned}$$ Substitute the values for $b$ and $c$ from the second and third of these equations into the final one, we get $$0 = \tfrac32(9-p_2) - 3p_2 + 15 - \tfrac12(9-p_2)p_2 + 10 = \tfrac12(p_2^2 - 18p_2 + 77) = \tfrac12(p_2-7)(p_2-11).$$ That gives possible values of $7$ or $11$ for $p_2$. But we are told that $p_2<10$, so we must have $p_2=7.$

Substituting that value for $p_2$ it is easy to check that $b=1$, $c=-1$, and then the last of the Newton identities above gives $\boxed{p_5 = 83}.$

[The cubic equation in $\lambda$ then becomes $\lambda^3 -3\lambda^2+\lambda+1=0$, which has solutions $\lambda = 1$ and $\lambda = 1\pm\sqrt2.$ So those are the values of $x$, $y$ and $z$.]

Awesome!

I used an elementary method to solve it because I don't know anything about the Newton Identities, and you can expect my solution to be lengthy and tedious. But, my answer is different from yours and I am certain that it must be my answer that is wrong.

Could you please tell me where did I do wrong?

Here is my solution:

We're told that

$x+y+z=3$

$\therefore (x+y+z)^2=3^2=(x+y+z)(x+y+z)=x^2+y^2+z^2+2(xy+yz+xz)$

We get

$9=x^2+y^2+z^2+2(xy+yz+xz)$---(1)

Now, we cube the expression $x+y+z$, we get:

$(x+y+z)^3=3^3=x^3+y^3+z^3+3x^2y+3x^2z+3xy^2+3xz^2+3y^2z+3yz^2+6xyz$

$12=3y(x^2+z^2)+3z(x^2+y^2)+3x(y^2+z^2)+6xyz$

Rearrange it and do some algebraic trick, the equation above becomes:

$12=3y(x^2+z^2+y^2-y^2)+3z(x^2+y^2+z^2-z^2)+3x(y^2+z^2+x^2-x^2)+6xyz$

$12=3y(x^2+z^2+y^2-y^2)+3z(x^2+y^2+z^2-z^2)+3x(y^2+z^2+x^2-x^2)+6xyz$

$12=3y(x^2+z^2+y^2)-3y^3+3z(x^2+y^2+z^2)-3z^3+3x(y^2+z^2+x^2)-3x^2+6xyz$

$12=3(x^2+y^2+z^2)(x+y+z)-3(x^3+y^3+z^3)+6xyz$

$12=3(x^2+y^2+z^2)(3)-3(15)+6xyz$

$57=9(x^2+y^2+z^2)+6xyz$

$19=3(x^2+y^2+z^2)+2xyz$---(2)

Similarly, we raise the equation $x+y+z$ to the fourth power yields

$(x+y+z)^4=3^4=x^4+y^4+z^4+4x^3y+4xy^3+4xy^3+4xz^3+4yz^3+6x^2y^2+6x^2z^2+6y^2x^2$

$81=35+4x^3y+4xy^3+4xy^3+4xz^3+4yz^3+6x^2y^2+6x^2z^2+6y^2x^2$

$46=4y(x^3+z^3)+4z(x^3+y^3)+4x(y^3+z^3)+6(x^2y^2+x^2z^2+y^2z^2)$

$46=4y(x^3+z^3+y^3-y^3)+4z(x^3+y^3+z^3-z^3)+4x(y^3+z^3+x^3-x^3)+6(x^2y^2+x^2z^2+y^2z^2)$

$46=4(x^3+z^3+y^3)(x+y+z)-4(x^4+y^4+z^4)+6(x^2y^2+x^2z^2+y^2z^2)$

$46=4(15)(3)-4(35)+6(x^2y^2+x^2z^2+y^2z^2)$

$6=6(x^2y^2+x^2z^2+y^2z^2)$

$1=x^2y^2+x^2z^2+y^2z^2$---(3)

From (1) and (2), we get:

$19=3(9-2(xy+yz+xz))+2xyz$

$19=27-6(xy+yz+xz)+2xyz$

$6(xy+yz+xz)=8+2xyz$

$3(xy+yz+xz)=4+xyz$---(4)

Next, we square the expression $xy+yz+xz$ and obtain:

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

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

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

$\therefore (xy+yz+xz)^2=1+6xyz$ since $x^2y^2+x^2z^2+y^2z^2=1$

From (4) and the equation above, we could further simplify them to get

$9(xy+yz+xz)^2=(4+xyz)^2$

$9(1+6xyz)=16+8xyz+(xyz)^2$

$9+54xyz=16+8xyz+(xyz)^2$

$(xyz)^2-46xyz+7=0$

$\displaystyle xyz=23 \pm 3\sqrt{58}$ (This doesn't seem right...)

Since $x^2+y^2+z^2<10$, $\displaystyle xyz=23 - 3\sqrt{58}$.

Finally, we raise the equation $x+y+z$ to the fifth power, this gives

$(x+y+z)^5=3^5=x^5+y^5+z^5+5x^4y+5x^4z+5xy^4+5xz^4+5y^4z+5yz^4$

$+10x^3y^2+10x^3z^2+10x^2y^3+10x^2z^3+10y^3z^2+10y^2z^3+20xyz(x^2+y^2+z^2)+30xyz(xy+yz+zx)$

$3^5=-14(x^5+y^5+z^5)+5(x^4+y^4+z^4)(x+y+z)+10(x^3+y^3+z^3)(x^2+y^2+z^2)$

$+20xyz(x^2+y^2+z^2)+30xyz(xy+xz+yz)$

$3^5=-14(x^5+y^5+z^5)+5(35)(3)+10(15)(x^2+y^2+z^2)+20xyz(x^2+y^2+z^2)+30xyz(xy+xz+yz)$

$3^5=-14(x^5+y^5+z^5)+5(35)(3)+150(x^2+y^2+z^2)+20xyz(x^2+y^2+z^2)+30xyz(xy+xz+yz)$

$3^5=-14(x^5+y^5+z^5)+5(35)(3)+(x^2+y^2+z^2)(150+20xyz)+30xyz(xy+xz+yz)$

$3^5=-14(x^5+y^5+z^5)+5(35)(3)+\left(\frac{19-2xyz}{3}\right)(150+20xyz)+30xyz\left(\frac{4xyz}{3}\right)$

By substituting the value of $\displaystyle xyz=23 - 3\sqrt{58}$ into the equation above, we get

$x^5+y^5+z^5=\frac{-2214+780\sqrt{58}}{42}$

I am aware that this is just way too tiring to even look at,(Tmi) never mind, Opalg, I will use the Newton Identities instead, thanks for participating in this particular problem.:)

Thanks.

 

[Nono713]

Re: Find the value of x^5+y^5+z^5.

The symbols... they confuse me...

Congratulations for sticking with your method to the bitter end though Anemone. I shamefully admit I probably would've thrown the whole thing out of the window a third of the way through and looked for an easier way (Smoking)​

By the way, you can use \tag{some number} in LaTeX to number your equations :)

 

[MarkFL]

Re: Find the value of x^5+y^5+z^5.

...

Congratulations for sticking with your method to the bitter end though Anemone. I shamefully admit I probably would've thrown the whole thing out of the window a third of the way through and looked for an easier way (Smoking)​

...

Yes, I too must applaud your tenacity...and willingness to share your work even though you believe it to be incorrect. Most would have run for the hills, tail tucked...myself included! (Rofl)

 

[anemone]

Re: Find the value of x^5+y^5+z^5.

The symbols... they confuse me...

Congratulations for sticking with your method to the bitter end though Anemone. I shamefully admit I probably would've thrown the whole thing out of the window a third of the way through and looked for an easier way (Smoking)​

That's so nice of you to say so, Bacterius..

By the way, you can use \tag{some number} in LaTeX to number your equations :)

I see...I don't realize that. I'll keep this in mind! Thanks!

Yes, I too must applaud your tenacity...and willingness to share your work even though you believe it to be incorrect. Most would have run for the hills, tail tucked...myself included! (Rofl)

Hey Mark, it was you who encouraged me to post the silliest solution of mine here, hehehe...

 

[MarkFL]

Re: Find the value of x^5+y^5+z^5.

I may have given you a nudge, but you were the one that made the final decision. (Yes)

 

[Opalg]

Re: Find the value of x^5+y^5+z^5.

Could you please tell me where did I do wrong?

Here is my solution:

We're told that

$x+y+z=3$

$\therefore (x+y+z)^2=3^2=(x+y+z)(x+y+z)=x^2+y^2+z^2+2(xy+yz+xz)$

We get

$9=x^2+y^2+z^2+2(xy+yz+xz)$---(1)

Now, we cube the expression $x+y+z$, we get:

$(x+y+z)^3=3^3=x^3+y^3+z^3+3x^2y+3x^2z+3xy^2+3xz^2+3y^2z+3yz^2+6xyz$

$12=3y(x^2+z^2)+3z(x^2+y^2)+3x(y^2+z^2)+6xyz$

Rearrange it and do some algebraic trick, the equation above becomes:

$12=3y(x^2+z^2+y^2-y^2)+3z(x^2+y^2+z^2-z^2)+3x(y^2+z^2+x^2-x^2)+6xyz$

$12=3y(x^2+z^2+y^2-y^2)+3z(x^2+y^2+z^2-z^2)+3x(y^2+z^2+x^2-x^2)+6xyz$

$12=3y(x^2+z^2+y^2)-3y^3+3z(x^2+y^2+z^2)-3z^3+3x(y^2+z^2+x^2)-3x^2+6xyz$

$12=3(x^2+y^2+z^2)(x+y+z)-3(x^3+y^3+z^3)+6xyz$

$12=3(x^2+y^2+z^2)(3)-3(15)+6xyz$

$57=9(x^2+y^2+z^2)+6xyz$

$19=3(x^2+y^2+z^2)+2xyz$---(2)
Everything is correct up to here!

Similarly, we raise the equation $x+y+z$ to the fourth power yields

$(x+y+z)^4=3^4=x^4+y^4+z^4+4x^3y+4xy^3+4xy^3+4xz^3+4yz^3+6x^2y^2+6x^2z^2+6y^2x^2$
That is where it goes wrong. The expansion of $\color{red}{(x+y+z)^4}$ should be
$\color{red}{ x^4+y^4+z^4 + 4x^3y+4xy^3+4x^3z+4xz^3+4y^3z+4yz^3 +6x^2y^2+6x^2z^2+6y^2z^2 + 12x^2yz+12xy^2z+12xyz^2}.$

As you can see, things get very complicated when you raise a trinomial to a large power! That is why the Newton identities are so useful. They save you from having to make very long calculations by splitting them into manageable steps.

 

[Albert1]

Re: Find the value of x^5+y^5+z^5.

In fact if students have no idea about Newton's identities they can still solve this problem, in a similar way like Opalg did (very innovative (Yes))

 

[anemone]

Re: Find the value of x^5+y^5+z^5.

As you can see, things get very complicated when you raise a trinomial to a large power! That is why the Newton identities are so useful. They save you from having to make very long calculations by splitting them into manageable steps.

Thanks for pointing this out for me, Opalg!:) I couldn't believe that I made such a silly mistake while multiplying out the trinomials.(Headbang)

Now, I agree wholeheartedly that Newton identities are very useful and it's never too late to learn a new concept in maths, I guess.