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**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?