A (k,d)-multigrade is a diophantine equation of the form $\displaystyle \sum_1^d m_i^j=\sum_1^d n_i^j$, that is valid for all powers $\displaystyle j=1,2,...k$. We can call k, the highest valid exponent, the "level" and d, the dimension of each vector $\displaystyle \vec{m},\vec{n}$ the "order." Consider altering the multigrade by another vector $\displaystyle \vec{a}$ as such: $\displaystyle \sum_1^d a_im_i^j=\sum_1^d a_in_i^j$, provided $\displaystyle m_i<m_{i+1},n_i<n_{i+1}$ for all i. In another thread, Wonderboy1953 conjectured that for the vector $\displaystyle \vec{a}=\{1,2,3,...,d\}$, no multigrade exists such that its altered version holds on any level greater than 2. That is, given a (k,d)-multigrade $\displaystyle \vec{m},\vec{n}$, $\displaystyle \sum_1^d i(m_i^3-n_i^3)\neq 0$. This is equivalent to the following formulation using matrices...

Find two vectors $\displaystyle \vec{m},\vec{n}$,

Define $\displaystyle M=\begin{bmatrix} 1 & 1 & ... & 1 \\ m_1+n_1 & m_2+n_2 & ... & m_d+n_d \\ m_1^2+m_1n_1+n_1^2 & m_2^2+m_2n_2+n_2^2 & ... & m_d^2+m_dn_d+n_d^2 \end{bmatrix}$

Define $\displaystyle A=\begin{bmatrix}m_1-n_1 & m_2-n_2 & ... & m_d-n_d\end{bmatrix}$ and $\displaystyle B=\begin{bmatrix}a_1(m_1-n_1) & a_2(m_2-n_2) & ... & a_d(m_d-n_d)\end{bmatrix}$

Such that $\displaystyle MA^T=MB^T=\vec{0}$. In general, A and B are independent of each other, so the nullspace of matrix M has at least dimension 2, requiring $\displaystyle d\geq5$.

In search of a counterexample, I have verified that when $\displaystyle \vec{a}=\{1,2,3,4,5\}$, no d=5 solutions exist for vectors $\displaystyle \vec{m},\vec{n}$ with components less than 50. Is anyone interested in pursuing a general proof?

2. Originally Posted by Media_Man
A (k,d)-multigrade is a diophantine equation of the form $\displaystyle \sum_1^d m_i^j=\sum_1^d n_i^j$, that is valid for all powers $\displaystyle j=1,2,...k$. We can call k, the highest valid exponent, the "level" and d, the dimension of each vector $\displaystyle \vec{m},\vec{n}$ the "order." Consider altering the multigrade by another vector $\displaystyle \vec{a}$ as such: $\displaystyle \sum_1^d a_im_i^j=\sum_1^d a_in_i^j$, provided $\displaystyle m_i<m_{i+1},n_i<n_{i+1}$ for all i. In another thread, Wonderboy1953 conjectured that for the vector $\displaystyle \vec{a}=\{1,2,3,...,d\}$, no multigrade exists such that its altered version holds on any level greater than 2. That is, given a (k,d)-multigrade $\displaystyle \vec{m},\vec{n}$, $\displaystyle \sum_1^d i(m_i^3-n_i^3)\neq 0$. This is equivalent to the following formulation using matrices...

Find two vectors $\displaystyle \vec{m},\vec{n}$,

Define $\displaystyle M=\begin{bmatrix} 1 & 1 & ... & 1 \\ m_1+n_1 & m_2+n_2 & ... & m_d+n_d \\ m_1^2+m_1n_1+n_1^2 & m_2^2+m_2n_2+n_2^2 & ... & m_d^2+m_dn_d+n_d^2 \end{bmatrix}$

Define $\displaystyle A=\begin{bmatrix}m_1-n_1 & m_2-n_2 & ... & m_d-n_d\end{bmatrix}$ and $\displaystyle B=\begin{bmatrix}a_1(m_1-n_1) & a_2(m_2-n_2) & ... & a_d(m_d-n_d)\end{bmatrix}$

Such that $\displaystyle MA^T=MB^T=\vec{0}$. In general, A and B are independent of each other, so the nullspace of matrix M has at least dimension 2, requiring $\displaystyle d\geq5$.

In search of a counterexample, I have verified that when $\displaystyle \vec{a}=\{1,2,3,4,5\}$, no d=5 solutions exist for vectors $\displaystyle \vec{m},\vec{n}$ with components less than 40. Is anyone interested in pursuing a general proof?

Damn, now I understand what was this about! It looks interesting and now I can see why wonderboy was asking about some relation with FLT, which I can see now.
I've no idea about this stuff, but I might be getting a little wet in it in the future.

Tonio

3. ## Media_Man

My library won't let me email you (something about an Outlook Express problem). Sorry.

4. ## d=5 solved

Wonderboy, etc:

Previously I used matrices to prove your conjecture is true for all multigrades up to order 4. I have just managed to prove the conjecture also holds for all order 5 multigrades as well. The proof is quite long and involves proving that the roots of a cubic polynomial are always zero. As you know, the general formula for cubic roots is ridiculously complicated, making a direct proof impossible, but I was able to find a shortcut that allowed me to reduce it to a quadratic.

I will post the proof shortly. In the meantime, anyone interested in this problem should know that if a counterexample exists, it must have at least six variables on either side of the equation (with no two variables on opposite sides equal, causing trivial cancellation). This substantially increases the space of candidate solutions (to $\displaystyle \mathbb{R}^{12}$), rendering computer aid virtually useless in searching for counterexamples.

As a side note, none of my proofs so far require solutions in integers. Therefore, a more general form may be true for all real numbers. I shall formally write up my two proofs, before tackling the d=6 case.

-- Media_(is_the_)Man

5. ## Comment

Thank you Media_Man for following up on this. I've anticipated that disproving by counterexample through the computer may not be fruitful. A mathematical proof for, or against, is what's needed here.