Generalization of the identity $2(m^2+n^2)=(m+n)^2+(m-n)^2$ for cubic case

Question

I am very excited to see the following beautiful identity:

$$2(m^2+n^2)=(m+n)^2+(m-n)^2.$$

I wonder if I can generalize it for cubic case like the expression as follows:

$$a(n_{1}^3+n_{2}^3+n_{3}^3)=(\pm n_1\pm n_2 \pm n_3)^3+(\pm n_1\pm n_2 \pm n_3)^3+\cdots+(\pm n_1\pm n_2 \pm n_3)^3$$

where $a$ is some constant and $\pm$ means you are free to choose any sign.

I tried it by calculating $(n_1+n_2+n_3)^3+(n_1-n_2+n_3)^3+(n_1+n_2-n_3)^3$ but could not able to get an expression like $a(n_{1}^3+n_{2}^3+n_{3}^3)$. Any hint or an explicit equation will be a great help.

Answer 1

It's quite simple to see that no such identity holds, which breaks my heart. I'll try to find some pieces of it and put it back together at the end of this answer.

Ok, ONE trivial identity holds , which is pretty obvious : if you sum up every kind of expression of the form $(\pm n_1 \pm n_2 \pm n_3)^3$ by taking all $\pm$ combinations, you'll get zero, simply because $(-a)^3 = -a^3$ for all $a$ and every expression above can be paired to its negative counterpart in the summation.

In fact , because of the pairing, we can restrict the possible $\pm$ combinations to $4$ instead of $8$, because half of them are a negative of the other.

But apart from that, we can't have an identity of the form : $$ a(n_1^3+n_2^3+n_3^3) = a_0(n_1+ n_2+ n_3)^3 + a_1(-n_1+n_2+n_3)^3 + a_2(n_1-n_2+n_3)^3 + a_3(n_1+n_2-n_3)^3 $$ for any $a \neq 0$. Let's prove it by contradiction : we proceed to assume that this equality holds, and then demonstrate another equality from here which is clearly false.

First, match the coefficients of the $n_i^3$ together. From the multinomial expansions of the sums of cubes, we know that the coefficient of $n_i^3$ on the right hand side is $(a_0+a_1+a_2+a_3) - 2a_i$ (which is basically saying that you subtract the $a_i$ corresponding to the $n_i$, but the rest get added). But all the $n_i^3$ have coefficient $a$ on the LHS. This reasoning, along with simple algebra, leads to the conclusion that $a_1=a_2=a_3$ and $a = a_0+a_1$.

Thus, we are left with the identity : $$ (a_0+a_1)(n_1^3+n_2^3+n_3^3) = a_0(n_1+n_2+n_3)^3 + a_1[(-n_1+n_2+n_3)^3 + (n_1-n_2+n_3)^3 + (n_1+n_2-n_3)^3] $$

Transpose to get : $$ (a_0+a_1)(n_1^3+n_2^3+n_3^3) - a_0(n_1+n_2+n_3)^3 = a_1[(-n_1+n_2+n_3)^3 + (n_1-n_2+n_3)^3 + (n_1+n_2-n_3)^3] $$

Let's compare the coefficients of $n_1n_2n_3$ on both sides now. On the LHS, it is seen to be $-6a_0$. On the RHS, it is found to be $-18a_1$. (All this can be seen from the multinomial expansion). It follows that $a_0 = 3a_1$ and therefore $a = 4a_1$.

All this clearly proves that if $a_1 = 0$ then the identity has $0$ on both sides, hence makes no sense. If $a_1 \neq 0$ then we can cancel out $a_1$ from both sides of that equation, and we get : $$ 4(n_1^3+n_2^3+n_3^3) - 3(n_1+n_2+n_3)^3 = (-n_1+n_2+n_3)^3 + (n_1-n_2+n_3)^3 + (n_1+n_2-n_3)^3 $$

So if any identity of the kind we want holds true, it MUST be this identity. Proving that this identity isn't true is fairly clear : we just find one triple $n_1,n_2,n_3$ for which it's false. I took the most obvious triple that came to my mind (note that I couldn't take all of $n_1,n_2,n_3$ equal!) So I took $n_1=n_2 = 2$ and $n_3 = 1$ and I got the LHS as $-307$, while the RHS came out as as something much bigger.

Thus we complete the contradiction : and no such identity exists.

---

For comparison, note that the identical identity with squares does work out , but things over there are slightly different. First of all ,we have $(-a)^2 = a^2$ so while opposing terms (in $(\pm x \pm y)^2$) don't cancel out, they do add up which means any identity is going to look like : $$ a(x+y)^2 + b(x-y)^2 = c(x^2+y^2) $$ Comparing the coefficients of $x^2$ or $y^2$ tells you that $a+b = c$ ,and comparing the coefficients of $xy$ tells you that $a=b$. So once again, dividing by $a$ tells you that the only identity that can be true is $(x+y)^2+(x-y)^2 = 2(x^2+y^2)$, which turns out to be true, instead of false.

---

There are, however, many identities regarding sum of cubes : these don't have arbitrary bases, though. Here is a minefield of algebraic identities, not just about cubes but about general polynomials of varying degrees in many variables. This is enough to get you through a life of boredom and broken hearts.

Another interesting question that comes out from the question you proposed , is linked with sums of squares : the very interesting identity $2(m^2+n^2) = (m+n)^2+(m-n)^2$ actually brings the following interpretation :

If $K$ is a sum of two integer squares, then $2K$ is also a sum of two integer squares.

Indeed, this interpretation hasn't been afforded to cubes yet, as far as I know. I believe that representation as a sum of three cubes hasn't been characterised, while representation as a sum of two cubes and four or more cubes are certainly better known. Following a perusal of these results, I had come to the conclusion that no identities such as the one used in the squared case were available there, and so the kind of identity proposed wasn't going to work out.

Answer 2

With some $k \ge 1$ terms, for $1 \le i \le k$, have

$$t_i = a_i n_1 + b_i n_2 + c_i n_3 \tag{1}\label{eq1A}$$

where $a_i = \pm 1$, $b_i = \pm 1$ and $c_i = \pm 1$. Cubing, expanding and simplifying using that $a_i^2 = b_i^2 = c_i^2 = 1$ gives (note using the multinomial theorem allows skipping to the third line below)

$$\begin{equation}\begin{aligned} t_i^3 & = (a_i n_1 + (b_i n_2 + c_i n_3))^3 \\ & = a_i^3n_1^3 + 3a_i^2 n_1^2(b_i n_2 + c_i n_3) + \\ & \; \; \; \; 3a_i n_1(b_i n_2 + c_i n_3)^2 + (b_i n_2 + c_i n_3)^3 \\ & = a_i^3n_1^3 + 3a_i^2 b_i n_1^2 n_2 + 3a_i^2c_in_1^2n_3 + 3a_ib_i^2n_1n_2^2 + 6a_i b_i c_i n_1 n_2 n_3 + \\ & \; \; \; \; 3a_ic_i^2n_1n_3^2 + b_i^3n_2^3 + 3b_i^2c_i n_2^2n_3 + 3b_ic_i^2 n_2n_3^2 + c_i^3n_3^3 \\ & = a_i n_1^3 + 3b_i n_1^2 n_2 + 3c_in_1^2n_3 + 3a_in_1n_2^2 + 6a_i b_i c_i n_1 n_2 n_3 + \\ & \; \; \; \; 3a_i n_1 n_3^2 + b_i n_2^3 + 3c_i n_2^2 n_3 + 3b_i n_2 n_3^2 + c_i n_3^3 \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Next, summing all of the cubed terms gives

$$\begin{equation}\begin{aligned} \sum_{i=1}^{k}t_i^3 & = \left(\sum_{i=1}^{k}a_i\right) n_1^3 + 3\left(\sum_{i=1}^{k}b_i\right) n_1^2 n_2 + 3\left(\sum_{i=1}^{k}c_i\right)n_1^2n_3 + \\ & \; \; \; \; 3\left(\sum_{i=1}^{k}a_i\right)n_1n_2^2 + 6\left(\sum_{i=1}^{k}a_i b_i c_i\right)n_1 n_2 n_3 + 3\left(\sum_{i=1}^{k}a_i\right) n_1 n_3^2 + \\ & \; \; \; \; \left(\sum_{i=1}^{k}b_i\right) n_2^3 + 3\left(\sum_{i=1}^{k}c_i\right) n_2^2 n_3 + 3\left(\sum_{i=1}^{k}b_i\right) n_2 n_3^2 + \left(\sum_{i=1}^{k}c_i\right) n_3^3 \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Since we want the coefficient of the $n_1 n_2^2$ term (i.e., the first one on the second line) to be $0$, this requires that $\sum_{i=1}^{k}a_i = 0$. However, this means the coefficient of the $n_1^3$ term will also be $0$. Similarly, we also require that $\sum_{i=1}^{k}b_i = \sum_{i=1}^{k}c_i = 0$, so the coefficients of $n_2^3$ and $n_3^3$ would also be $0$.

Thus, the only possible value of your $a$ is $0$.

Answer 3

The OP seeks to find an identity of form,

$$n_1^3+n_2^3+n_3^3=(\pm n_1\pm n_2 \pm n_3)^3+(\pm n_1\pm n_2 \pm n_3)^3+\cdot \cdot \cdot+(\pm n_1\pm n_2 \pm n_3)^3$$

As the other answers have pointed out, it is quite impossible to do so. However, if we relax certain conditions, then the nearest we can approach this is by combining Ryley's Theorem and Boutin's Identity.

Ryley's Theorem states every rational number can be expressed as a sum of three rational cubes. One version is the identity,

$$\big(n - 1\big)^3 + \left(\frac{-n(n^2 + 3) + 1}{(n + 1)^2 - n}\right)^3 + \left(\frac{3n(n+1)}{(n + 1)^2 - n}\right)^3=9n$$

while the cubic case of Boutin's Identity yields,

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

We simply equate $9n=3abc$. Thus, for any non-zero $n$, we can find infinitely many rational $(a,b,c)$. And multiplying by a scaling factor, it is possible to make both the LHS (three cubes) and RHS (four cubes) as integers.

Example. Let $9n=3abc=180$, and assume $(a,b,c,n)=(3,4,5,20)$. So,

$$\big(20-1\big)^3+\left(\frac{-20(20^2 + 3) + 1}{(20 + 1)^2 - 20}\right)^3 + \left(\frac{60(20+1)}{(20 + 1)^2 - 20}\right)^3 =\\ \left(\frac{3+4+5}2\right)^3+\left(\frac{-3+4-5}2\right)^3+ \left(\frac{-3-4+5}2\right)^3+\left(\frac{3-4-5}2\right)^3$$

Or equivalently,

$$19^3 + \left(-\frac{7939}{421}\right)^3 + \left(\frac{1260}{421}\right)^3 = 6^3+(-2)^3+(-1)^3+(-3)^3 =180$$

Answer 4

I am going to take this problem in a different direction.

We have $2(m^2+n^2)=(m-n)^2+(m+n)^2$. We could also recover this from the Gaussian integer product

$(1+i)(m+ni)=(m-n)+(m+n)i$

and taking the Euclidean norm.

Let us try this with Eisenstein integers. The norm of $m-n\omega$ is $m^2+mn+n^2$. Therefore, we apply the Eisenstein integer product

$(1-\omega)(m-n\omega)=(m-n)-(m+2n)\omega$

and from the norm

$2(m^2+mn+n^2)=(m-n)^2+(m-n)(m+2n)+(m+2n)^2.$

The reader can verify this condition.

So we can choose among many possible generalizations by just plugging in a cubic integer domain. We select $\mathbb{Z}[\sqrt[3]2]$ with norm

$||m+n\sqrt[3]2+p\sqrt[3]4||=m^3+2n^3+4p^3-6mnp.$

Use $1+\sqrt[3]2$ as our multiplier (note that this has positive norm greater than $1$ whereas $1+\sqrt[3]2+\sqrt[3]4$ would not). So

$(1+\sqrt[3]2)(m+n\sqrt[3]2+p\sqrt[3]4)=(m+2p)+(m+n)\sqrt[3]2+(n+p)\sqrt[3]4$

and so

$3(m^3+2n^3+4p^3-6mnp)=(m+2p)^3+2(m+n)^3+4(n+p)^3-6(m+2p)(m+n)(n+p).$

For example, $m=n=p=1$ would give

$3(1+2+4-6)=27+(2×8)+(4×8)-6(3×2×2)$

and calculation reveals that this holds.