# prove (a^5 + b^5 + c^5)/5 = (a^3 + b^3 + c^3)/3 * (a^2 + b^2 + c^2)/2

#### mente oscura

##### Well-known member
Hello.

Prove:

$$\dfrac{a^5+b^5+c^5}{5}=\dfrac{a^3+b^3+c^3}{3} \ \dfrac{a^2+b^2+c^2}{2}$$

Regards.

#### MarkFL

Staff member
Re: prove (a^5+b^5+c^5)/5=(a^3+b^3+c^3)/3 * (a^2+b^2+c^2)/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 anemone here:

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

Let $S_n=a^n+b^n+c^n$, where $S_1=0$.

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.

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Re: prove (a^5+b^5+c^5)/5=(a^3+b^3+c^3)/3 * (a^2+b^2+c^2)/2

Hello.

Prove:

$$\dfrac{a^5+b^5+c^5}{5}=\dfrac{a^3+b^3+c^3}{3} \ \dfrac{a^2+b^2+c^2}{2}$$

Regards.
Thanks to I Like Serena to point out a mistake in my original post.
CORRECTED POST:
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.

ORIGINAL POST:
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:

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: prove (a^5+b^5+c^5)/5=(a^3+b^3+c^3)/3 * (a^2+b^2+c^2)/2

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.
Nice and elegant.
However, shouldn't your equation be $x^3+(\sum ab)x-abc=0$?
With this modification, the same argument works out.

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Re: prove (a^5+b^5+c^5)/5=(a^3+b^3+c^3)/3 * (a^2+b^2+c^2)/2

Nice and elegant.
However, shouldn't your equation be $x^3+(\sum ab)x-abc=0$?
With this modification, the same argument works out.
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.

Last edited:

#### mathbalarka

##### Well-known member
MHB Math Helper
Re: prove (a^5+b^5+c^5)/5=(a^3+b^3+c^3)/3 * (a^2+b^2+c^2)/2

Clever use of powersum identities, caffeinemachine!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: prove (a^5+b^5+c^5)/5=(a^3+b^3+c^3)/3 * (a^2+b^2+c^2)/2

Inspired by caffeinemachine, I decided to explore this a bit more.
I've looked up Newton's Identities, that define:
\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 caffeinemachine's notation, this becomes (with $Σ a = 0$):
\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.

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Re: prove (a^5+b^5+c^5)/5=(a^3+b^3+c^3)/3 * (a^2+b^2+c^2)/2

Clever use of powersum identities, caffeinemachine!
Thanks.

#### mente oscura

##### Well-known member
Re: prove (a^5+b^5+c^5)/5=(a^3+b^3+c^3)/3 * (a^2+b^2+c^2)/2

Hello.

I'am omitted the restriction that $$a, \ b, \ c \in{Z} \ and \ a+b+c=0$$. 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.

$$(a+b+c)^5=a^5+b^5+c^5+5(a+b)(a+c)(b+c)(a^2+b^2+c^2+ab+ac+bc)$$

$$(a+b+c)^3=a^3+b^3+c^3+3(a+b)(a+c)(b+c)$$

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

$$\dfrac{a^5+b^5+c^5}{5}=-(a+b)(a+c)(b+c)(a^2+b^2+c^2+ab+ac+bc)$$

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

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

$$\dfrac{a^5+b^5+c^5}{5}=\dfrac{a^3+b^3+c^3}{3} \ \dfrac{a^2+b^2+c^2}{2}$$

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

Regards.