0

For $k = 0,1,2,3,4,5,6,7,8$, we have the equality,

$$(-5)^k + (-119)^k + (-101)^k + (-215)^k + (-197)^k + 43^k + 157^k + 31^k + 217^k + 169^k\\ =\\ (-47)^k + (-161)^k + (-35)^k + (-221)^k + (-173)^k + 1^k + 115^k + 97^k + 211^k + 193^k$$

I was very surprised to see the following more general identity (I have google it, and not found it)： post

\begin{align*} &(m^2-mn-n^2)^k+(15m^2-25mn-21n^2)^k+(13m^2-71mn+7n^2)^k\\ &+(27m^2-95mn-13n^2)^k+(-11m^2-27mn-33n^2)^k+(17m^2-129mn+71n^2)^k\\ &+(3m^2-105mn+91n^2)^k+(29m^2-101mn+51n^2)^k+(-m^2-49mn+79n^2)^k\\ &+(-13m^2-27mn+59n^2)^k\\ &=(-m^2+3mn-13n^2)^k+(13m^2-21mn-33n^2)^k+(-13m^2-25mn+7n^2)^k\\ &+(17m^2-77mn-21n^2)^k+(29m^2-99mn-n^2)^k+(15m^2-125mn+59n^2)^k\\ &+(m^2-101mn+79n^2)^k+(3m^2-55mn+51n^2)^k+(-11m^2-31mn+71n^2)^k\\ &+(27m^2-99mn+91n^2)^k \end{align*}

where the example is just the particular case $m,n = 1,2$.

Maybe this identity is new result? Because we only know some background:

http://mathworld.wolfram.com/DiophantineEquation5thPowers.html

http://mathworld.wolfram.com/DiophantineEquation6thPowers.html

http://mathworld.wolfram.com/DiophantineEquation7thPowers.html

http://mathworld.wolfram.com/DiophantineEquation8thPowers.html

Can someone explain this identity's secret?

PS: this post author is Zipei Nie, see Zipei Nie

Tito Piezas III
  • 60,745
  • 3
    I don't see how anything meaningful can be gained by using this equality. As $k$ can take only a few ($9$) different values, it looks like the author has just tweaked the coefficients so that the LHS and RHS happen to agree for those values of $k$. – JiK Mar 23 '15 at 14:30
  • This might (?) be related to the well known construction of disjoint sets of small integers such that their respective power sums agree up to a certain power. See this question for the gory details. But here the sizes of the two sets are much smaller (OTOH they don't cover an unbroken interval of values), so may be it isn't at all related? – Jyrki Lahtonen Mar 23 '15 at 14:39
  • @JyrkiLahtonen,maybe it. –  Mar 23 '15 at 14:42
  • 3
    This looks spectacularly unmotivated, and it's not surprising that such an identity exists, given that there are $20$ polynomials in it and only $9$ only values of $k$. Also, the coefficients of $m^2$ and $n^2$ on either side of the equation agree (though are permuted), so there's presumably a lot of cancellation. – anomaly Mar 29 '15 at 08:04
  • @JiK: It's hardly meaningless. Kindly see my answer. – Tito Piezas III Mar 29 '15 at 09:34
  • @JyrkiLahtonen: It's more on using certain theorems on the Tarry-Escott problem as I elaborated in my answer. – Tito Piezas III Mar 29 '15 at 13:56

1 Answers1

3

The identity used two theorems from equal sums of like powers and is hardly meaningless nor unmotivated. The $(k,10,10)$ equation above can be labelled as,

$$x_1^k+x_2^k+\dots+x_{10}^k = y_1^k+y_2^k+\dots+y_{10}^k$$

Notice that we have,

$$S_1 = x_1+y_6 = x_2+y_7 = x_3+y_8 = x_4+y_9 = x_5+y_{10}$$

$$S_2 = x_6+y_1 = x_7+y_2 = x_8+y_3 = x_9+y_4 = x_{10}+y_5$$

$$S_1=S_2=2(8m^2-63mn+29n^2)$$

In the study of the Prouhet-Tarry-Escott Problem, this is what is known as a symmetric solution and it is a clue how Zipie Nie found it. Assume the notation,

$$[a_1,a_2,\dots,a_m]^k = a_1^k + a_2^k +\dots +a_m^k$$

Theorem 1: (Tarry-Escott) If,

$$[a_1,a_2,\dots,a_m]^k = [b_1,b_2,\dots,b_m]^k,\;\; \text{for}\, k = 1,3,\dots, 2n-1$$

then for any constant $c$,

$$[c+a_1,\dots, c+a_m, c-b_1,\dots, c-b_m]^k = [c-a_1,\dots, c-a_m, c+b_1,\dots, c+b_m]^k\\ \text{for}\, k=1,2,3,\dots, 2n$$

This explains why sums of appropriate terms from either side of the equality is a constant. We can then find the "seed" identity of $a_i,b_i$ that Nie used as the $(k,5,5)$,

$$(-7 m^2 + 62 m n - 30 n^2)^k + (7 m^2 + 38 m n - 50 n^2)^k + (5 m^2 - 8 m n - 22 n^2)^k + (19 m^2 - 32 m n - 42 n^2)^k + (-19 m^2 + 36 m n - 62 n^2)^k\\=\\ (-9 m^2 + 66 m n - 42 n^2)^k + (5 m^2 + 42 m n - 62 n^2)^k + (-21 m^2 + 38 m n - 22 n^2)^k + (9 m^2 - 14 m n - 50 n^2)^k + (21 m^2 - 36 m n - 30 n^2)^k$$

for $k=1,3,5,7.$

In turn, this 7th degree multi-grade, for a suitable permutation of the $a_i,b_i$, belongs to a family by Gloden-Sinha,

Theorem 2: (Gloden-Sinha) If,

$$u_1^k+u_2^k+u_3^k = v_1^k+v_2^k+v_3^k,\;\;\text{for}\,k = 2,4\\ \text{where}\;\;u_1+u_2-u_3 = 2(v_1+v_2-v_3) \neq 0\tag1$$

or alternatively,

$$(a+4e)^k + (b+4e)^k + (a+b)^k = (c+2e)^k + (d+2e)^k + (c+d)^k,\;\; \text{for}\,k = 2,4$$

where $e\neq 0$ then,

$$(-a-b-e)^k + (-c-e)^k + (b+e)^k + (a+e)^k + (c+d+e)^k = \\(3e)^k + (a+b+3e)^k + (-a-3e)^k + (-b-3e)^k + (d+e)^k,\;\; \text{for}\, k = 1,3,5,7$$

So in a nutshell, what Nie essentially did (and is no trivial feat) was to find a solution to $(1)$ in terms of quadratic forms, of which several are known already.

P.S. For similar "mysteries", see also my MathOverflow answer regarding Ramanujan's 6-10-8 Identity which uses even higher powers.

Community
  • 1
Tito Piezas III
  • 60,745