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Found this lovely identity the other day, and thought it was fun enough to share as a problem:

If $a+b+c=0$ then show $$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2.$$

There are, of course, brute force techniques for showing this, but I'm hoping for something elegant.

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    $\begingroup$ Why would somebody downvote this. I am upvoting it.. $\endgroup$ – Mark Fischler Feb 18 '16 at 6:03
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    $\begingroup$ See math.stackexchange.com/questions/851985/… $\endgroup$ – lab bhattacharjee Feb 18 '16 at 6:04
  • $\begingroup$ Nobody downvoted. @MarkFischler (Or if the did, they quickly retracted.) $\endgroup$ – Thomas Andrews Feb 18 '16 at 6:05
  • $\begingroup$ Nice, @labbhattacharjee ! $\endgroup$ – Thomas Andrews Feb 18 '16 at 6:06
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    $\begingroup$ @MarkFischler I accidentally downvoted it while meaning to press the favorite button. Sorry. $\endgroup$ – S.C.B. Feb 18 '16 at 6:09
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Let $T_{m}$ be $a^m+b^m+c^m$.

Let $k=-ab-bc-ca$, and $l=abc$.

Note that this implies $a,b,c$ are solutions to $x^3=kx+l$.

Using Newton's Identity, note the fact that $T_{m+3}=kT_{m+1}+lT_{m}$(which can be proved by summing $x^3+kx+l$)

It is not to difficult to see that $T_{2}=2k$, $T_3=3l$, from $a+b+c=0$.

From here, note that $T_{4}=2k^2$ using the identity above.

In the same method, note that $T_{5}=5kl$.

From here, note $T_{7}=5k^2l+2k^2l=7k^2l$ from $T_{m+3}=kT_{m+1}+lT_{m}$ . Therefore, the equation simplifies to showing that $k^2l \times l=(kl)^2$, which is true.

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  • $\begingroup$ Not clear why this answer has so few upvotes. It's definitely the easiest and least brute-force answer. Maybe if you were explicit about why $T_{m+3}=kT_{m+1}+lT_m$ - maybe spell out what $k,l$ are? (They look like "constants, since you use $k,l,m$, when what they really are is polynomials in $a,b,c$.) $\endgroup$ – Thomas Andrews Feb 19 '16 at 14:29
  • $\begingroup$ @Thomas Andrews Noted. Hope my edit clarifies matters. $\endgroup$ – S.C.B. Feb 19 '16 at 14:34
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    $\begingroup$ Perhaps people who are not comfortably familiar with Newton's identities look at this solution as pulling a rabbit out of a hat, and would not upvote it for that reason (although I am, because it throws light on the case I started looking at, with 4 variables). $\endgroup$ – Mark Fischler Feb 20 '16 at 22:10
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Useful identities:

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

$(y-z)^{5}+(z-x)^{5}+(x-y)^{5}= 5(y-z)(z-x)(x-y)(x^{2}+y^{2}+z^{2}-yz-zx-xy)$

$(y-z)^{7}+(z-x)^{7}+(x-y)^{7}= 7(y-z)(z-x)(x-y)(x^{2}+y^{2}+z^{2}-yz-zx-xy)^{2}$

By letting $a=y-z,b=z-x,c=x-y$.

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Start off with the fact that $$a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) = - 2(ab+bc+ca) $$

Since $a+b+c=0$ we know that ${a,b,c}$ are the three roots of some cubic missing the $x^2$ term: $$ x^3+kx+m = 0 $$ with $ab+bc+ca = k$. And by the way, that says that $$a^2+b^2+c^2=-2k$$

Now start chaining upward, expressing $a^n+b^n+c^n$ in terms of $m$ and $k$. For $n=3$ we add the cubic expression with $x=a$ to that with $x=b$ and $x=c$ to get $$ a^3+ b^3 + c^3 +k(a+b+c)+(m+m+m) = 0 \\ a^3+ b^3 + c^3 +3m = 0 \\ a^3+ b^3 + c^3 = -3m $$ Now for $n=4$ we multiply each of the equations by $a$, $b$ and $c$ respectively before adding them: $$ a^4+ b^4 + c^4 +k(a^2+b^2+c^2)+m(a+b+c) = 0 \\ a^4+ b^4 + c^4 -2k^2 = 0 \\ a^4+ b^4 + c^4 = 2k^2 $$ Now for $n=5$ we multiply each of the equations by $a^2$, $b^2$ and $c^2$ respectively before adding them: $$ a^5+ b^5 + c^5 +k(a^3+b^3+c^3)+m(a^2+b^2+c^2) = 0 \\ a^5+ b^5 + c^5 -3mk -2mk = 0 \\ a^5+ b^5 + c^5 = 5mk $$ For this problem we can afford to skip $n=6$. $$ a^7+ b^7 + c^7 +k(a^5+b^5+c^5)+m(a^4+b^4+c^4) = 0 \\ a^7+ b^7 + c^7 +5mk^2 +2mk^2 = 0 \\ a^7+ b^7 + c^7 = -7mk^2 $$ Then your identity reads $$ \left( -\frac{3m}{3} \right) \left( -\frac{7mk^2}{7} \right) = \left( -\frac{5mk}{5} \right)^2 \\ \left( -m \right) \left( -mk^2 \right) = (mk)^2 $$

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Here's the symmetric polynomial approach.

$a^n+b^n+c^n$ is a symmetric homogeneous polynomial of degree $n$. So it can be expresses as a linear combination of polynomials $s_1^is_2^js_3^k$ where $i+2j+3k=n$, and $s_1=a+b+c,$ $s_2=ab+bc+ac,$ $s_3=abc$ are the elementary symmetric polynomials.

Now, $s_1=a+b+c=0$ implies that $a^n+b^n+c^n$ can be written as a linear combination of $s_2^js_3^k$ with $2j+3k=n$.

It the case where $n=2,3,4,5,7$, there is only one pair $j,k$ such that $2j+3k=n$, so $a^n+b^n+c^n$ becomes a monomial in $s_2,s_3$.

When $n=2$, this means $a^2+b^2+c^2=k_2s_2$ for some constant $k_2$. Setting $(a,b,c)=(2,-1,-1)$, we see $k_2=-2$.

When $n=3$, this means $a^3+b^3+c^3=k_3s_3$ for some constant $k_3$. Setting $a=2,b=c=-1$, we get $k_3=3$.

Similarly $a^5+b^5+c^5=k_5s_2s_3$, and again we use $(a,b,c)=(2,-1,-1)$ to get that $k_5=-5$.

Likewise, $a^7+b^7+c^7=7s_2^2s_3$.

Similar formula under the same condition:

$$\frac{a^2+b^2+c^2}{2}\frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5}\\ \frac{a^2+b^2+c^2}{2}\frac{a^5+b^5+c^5}{5} = \frac{a^7+b^7+c^7}{7} $$

More generally, if $a,b,c\in\mathbb Z$ with $a+b+c=0$ and $n$ is relatively prime to $6$ then $a^n+b^n+c^n$ is divisible by $s_2s_3=(ab+ac+bc)abc$.


There's actually an expression for $a^n+b^n+c^n$ with explicit coefficients:

$$a^n+b^n+c^n = \sum_{i+2j+3k=n} (-1)^j\frac{n}{i+j+k}\binom{i+j+k}{i,j,k} s_1^is_2^js_3^k\tag{1}$$

I confess I found that formula when looking at Fermat's Last many many years ago. It occurred to me at the time that Fermat can be expresses, for odd $n$, as

Given odd positive integer $n>1$. Then for integers $a,b,c$, $a^n+b^n+c^n=0$ if and only if $a+b+c=0$ and $abc=0$.

which seemed to imply it is a question about symmetric polynomials.

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  • $\begingroup$ Nice, but what does $\binom{i+j+k}{i,j,k}$ mean? $\endgroup$ – vesszabo Feb 28 '16 at 14:14
  • $\begingroup$ It's a multinomial. @vesszabo $\endgroup$ – Thomas Andrews Feb 28 '16 at 18:39
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$$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=-2(ab+bc+ca)$$

The key here is the identity (for all $n\in\mathbb Z_{\ge 3}$):

$$a^n+b^n+c^n=(a+b+c)\left(a^{n-1}+b^{n-1}+c^{n-1}\right)-$$

$$-(ab+bc+ca)\left(a^{n-2}+b^{n-2}+c^{n-2}\right)+abc\left(a^{n-3}+b^{n-3}+c^{n-3}\right)$$

Therefore: $$a^3+b^3+c^3=3abc\\ a^4+b^4+c^4=2(ab+bc+ca)^2\\ a^5+b^5+c^5=-5abc(ab+bc+ca)\\a^7+b^7+c^7=7abc(ab+bc+ca)^2$$

Therefore:

$$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2\\\frac{a^2+b^2+c^2}{2}\frac{a^3+b^3+c^3}{3}=\frac{a^5+b^5+c^5}{5}\\ \frac{a^2+b^2+c^2}{2}\frac{a^5+b^5+c^5}{5}=\frac{a^7+b^7+c^7}{7}\\$$

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If we remember the well-known formulas

$(x+y)^3 - (x^3 + y^3) = 3xy(x+y)$

$(x+y)^5 - (x^5 + y^5) = 5xy(x+y)(x^2 + xy + y^2)$

$(x+y)^7 - (x^7 + y^7) = 7xy(x+y)(x^2 + xy + y^2)^2$

then your identity becomes the observation that there are equal powers of the $xy(x+y)$ and $(x^2+xy+y^2)$ factors on both sides. (Under the change of notation $a,b,c \to -(x+y),x,y$)

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Let $a,b,c$ be roots of $$x^3+mx-n=0$$
From Vieta formula we get: $$ 0 = (a+b+c)^2 = a^2+b^2+c^2+2m$$ so $$\boxed{a^2+b^2+c^2 =-2m}$$ Further $$x^3 = n-mx\implies a^3+b^3+c^3 = 3n-m(a+b+c) =3n $$ so $$\boxed{a^3+b^3+c^3 =3n}$$

Also $$x^5 = nx^2-mx^3 = nx^2-mn+m^2x$$ so $$a^5+b^5+c^5 = n(a^2+b^2+c^2)-3mn = -5mn$$ so $$\boxed{a^5+b^5+c^5 =-5mn}$$

Finally $$x^7 = x(n-mx)^2 = n^2x-2mnx^2+m^2x^3 $$ $$=n^2x-2mnx^2+m^2n-m^3x$$ so $$\boxed{a^7+b^7+c^7 = 7m^2n}$$

and we are done.

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