- Thread starter
- #1

#### mente oscura

##### Well-known member

- Nov 29, 2013

- 172

Prove:

[tex]\dfrac{a^5+b^5+c^5}{5}=\dfrac{a^3+b^3+c^3}{3} \ \dfrac{a^2+b^2+c^2}{2}[/tex]

Regards.

- Thread starter mente oscura
- Start date

- Thread starter
- #1

- Nov 29, 2013

- 172

Prove:

[tex]\dfrac{a^5+b^5+c^5}{5}=\dfrac{a^3+b^3+c^3}{3} \ \dfrac{a^2+b^2+c^2}{2}[/tex]

Regards.

- Admin
- #2

I believe we require the restriction that $a+b+c=0$. Note that for $a=b=c=1$ the equation does not hold.

Here is my solution, which I have slightly modified by changing the variables, from my solution to essentially the same problem posted by

http://mathhelpboards.com/challenge-questions-puzzles-28/prove-s_5-5=-s_3-3s_2-2-a-6692.html

If we view $S_n$ as a recursive algorithm, we see that it must come from the characteristic equation:

\(\displaystyle (r-a)(r-b)(r-c)=0\)

\(\displaystyle r^3-(a+b+c)r^2+(ab+ac+bc)r-abc=0\)

Since $a+b+c=S_1=0$, we obtain the following recursion:

\(\displaystyle S_{n+3}=-(ab+ac+bc)S_{n+1}+abcS_{n}\)

Now, observing we may write:

\(\displaystyle (a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)\)

\(\displaystyle 0=S_2+2(ab+ac+bc)\)

\(\displaystyle -(ab+ac+bc)=\frac{S_2}{2}\)

Also, we find:

\(\displaystyle (a+b+c)^3=-2\left(a^3+b^3+c^3 \right)+3\left(a^2+b^2+c^2 \right)(a+b+c)+6abc\)

\(\displaystyle 0=-2S_3+6abc\)

\(\displaystyle abc=\frac{S^3}{3}\)

And so our recursion may be written:

\(\displaystyle S_{n+3}=\frac{S_2}{2}S_{n+1}+\frac{S_3}{3}S_{n}\)

Letting $n=2$, we then find:

\(\displaystyle S_{5}=\frac{S_2}{2}S_{3}+\frac{S_3}{3}S_{2}\)

\(\displaystyle S_{5}=\frac{5}{6}S_2S_{3}\)

\(\displaystyle \frac{S_5}{5}=\frac{S_3}{3}\cdot\frac{S_2}{2}\)

Shown as desired.

- Mar 10, 2012

- 835

Thanks to I Like Serena to point out a mistake in my original post.

Prove:

[tex]\dfrac{a^5+b^5+c^5}{5}=\dfrac{a^3+b^3+c^3}{3} \ \dfrac{a^2+b^2+c^2}{2}[/tex]

Regards.

As Mark mentioned, we require $a+b+c=0$ to prove the above.

Let $a,b$ and $c$ be roots of $x^3+\beta x-\gamma=0$.

I will use $\sum a$ to denote $a+b+c$, $\sum a^2$ to denote $a^2+b^2+c^2$, and so on.

Further, let's use $\sum ab$ to denote $ab+bc+ca$.

Then $\sum a^2=(\sum a)^2-2(\sum ab)=-2\beta$.

Thus $$\sum a^2=-2\beta$$

Since $x^3+\beta x-\gamma=0$ is satisfied by $a,b,c$, we get $\sum a^3+\beta(\sum a)-3\gamma=0$, giving $$\sum a^3=3\gamma$$.

Also $x^2(x^3+\beta x-\gamma)=0$ is satisfied by $a,b,c$.

So $x^5+\beta x^3-\gamma x^2=0$ is satisfied by $a,b,c$.

Thus $\sum a^5+\beta(\sum a^3)-\gamma(\sum a^2)=0$.

This leads to $$\sum a^5=-3\beta\gamma-2\beta\gamma=-5\beta\gamma$$.

From here the desired equality easily follows.

As Mark mentioned, we require $a+b+c=0$ to prove the above.

Let $a,b$ and $c$ be roots of $x^3+bx-c=0$.

I will use $\sum a$ to denote $a+b+c$, $\sum a^2$ to denote $a^2+b^2+c^2$, and so on.

Further, let's use $\sum ab$ to denote $ab+bc+ca$.

Then $\sum a^2=(\sum a)^2-2(\sum ab)=-2b$.

Thus $$\sum a^2=-2b$$

Since $x^3+bx-c=0$ is satisfied by $a,b,c$, we get $\sum a^3+b(\sum a)-3c=0$, giving $$\sum a^3=3c$$.

Also $x^2(x^3+bx-c)=0$ is satisfied by $a,b,c$.

So $x^5+bx^3-cx^2=0$ is satisfied by $a,b,c$.

Thus $\sum a^5+b(\sum a^3)-c(\sum a^2)=0$.

This leads to $$\sum a^5=-3bc-2bc=-5bc$$.

From here the desired equality easily follows.

Last edited:

- Admin
- #4

- Mar 5, 2012

- 9,416

Nice and elegant.As Mark mentioned, we require $a+b+c=0$ to prove the above.

Let $a,b$ and $c$ be roots of $x^3+bx-c=0$.

I will use $\sum a$ to denote $a+b+c$, $\sum a^2$ to denote $a^2+b^2+c^2$, and so on.

Further, let's use $\sum ab$ to denote $ab+bc+ca$.

Then $\sum a^2=(\sum a)^2-2(\sum ab)=-2b$.

Thus $$\sum a^2=-2b$$

Since $x^3+bx-c=0$ is satisfied by $a,b,c$, we get $\sum a^3+b(\sum a)-3c=0$, giving $$\sum a^3=3c$$.

Also $x^2(x^3+bx-c)=0$ is satisfied by $a,b,c$.

So $x^5+bx^3-cx^2=0$ is satisfied by $a,b,c$.

Thus $\sum a^5+b(\sum a^3)-c(\sum a^2)=0$.

This leads to $$\sum a^5=-3bc-2bc=-5bc$$.

From here the desired equality easily follows.

However, shouldn't your equation be $x^3+(\sum ab)x-abc=0$?

With this modification, the same argument works out.

- Mar 10, 2012

- 835

Ah! What I meant was let $x^3+\beta x-\gamma =0$ be the cubic whose roots are $a,b$ and $c$. There was a glaring clash in notation which I didn't see.Nice and elegant.

However, shouldn't your equation be $x^3+(\sum ab)x-abc=0$?

With this modification, the same argument works out.

Last edited:

- Mar 22, 2013

- 573

Clever use of powersum identities, caffeinemachine!

- Admin
- #7

- Mar 5, 2012

- 9,416

Inspired by

\begin{array}{}

p_k &=& \sum a^k \\

e_k &=& \text{sum of all distinct products of k distinct variables}

\end{array}

after which we have the identities:

\begin{array}{}

p_1 &=& e_1 \\

p_2 &=& e_1p_1 - 2e_2 \\

p_3 &=& e_1p_2 - e_2p_1 + 3e_3 \\

p_5 &=& e_1p_4 - e_2p_3 + e_3p_2 - e_4p_1 + 5e_5

\end{array}

Following

\begin{array}{}

Σ a &=& Σ a

&=& 0 \\

\textstyle Σ a^2 &=& Σ a Σ a - 2Σ ab

&=& -2 Σ ab \\

Σ a^3 &=& Σ a Σ a^2 - Σ ab Σ a + 3abc

&=& 3abc \\

Σ a^5 &=& Σ a Σ a^4 - Σ ab Σ a^3 + abc Σ a^2 - 0 + 0 \\

&=& -Σ ab \cdot 3abc + abc \cdot -2 Σ ab \\

&=& -5abc Σ ab

\end{array}

The equality to prove follows immediately.

- Mar 10, 2012

- 835

Thanks.Clever use of powersum identities, caffeinemachine!

- Thread starter
- #9

- Nov 29, 2013

- 172

Hello.

I'am omitted the restriction that [tex]a, \ b, \ c \in{Z} \ and \ a+b+c=0[/tex]. I'm sorry.

This problem, to be propose in another forum of mathematics in the Spanish language, and I resolved it in the following way: A little to the brute force.

[tex](a+b+c)^3=a^3+b^3+c^3+3(a+b)(a+c)(b+c)[/tex]

[tex](a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc[/tex]

[tex]\dfrac{a^5+b^5+c^5}{5}=-(a+b)(a+c)(b+c)(a^2+b^2+c^2+ab+ac+bc)[/tex]

[tex]\dfrac{a^5+b^5+c^5}{5}=-[-\dfrac{a^3+b^3+c^3}{3} \ (a^2+b^2+c^2+ab+ac+bc)][/tex]

[tex]\dfrac{a^5+b^5+c^5}{5}=-[-\dfrac{a^3+b^3+c^3}{3}(a^2+b^2+c^2-\dfrac{a^2+b^2+c^2}{2})][/tex]

[tex]\dfrac{a^5+b^5+c^5}{5}=\dfrac{a^3+b^3+c^3}{3} \ \dfrac{a^2+b^2+c^2}{2}[/tex]

I expose hidden, in case someone wants to keep trying the prove.

Regards.