The equation ap^3+bq^3+cr^3+ds^3 = 0

Theorem 2:  “Likewise, given one non-trivial solution to the cubic,

a_1y_1^3+a_2y_2^3+a_3y_3^3+\dots+a_ny_n^3 = 0

then an infinite more can be found.”

Proof:  As before, we will start with a particular example and derive the rest by induction. The following is identically true,

ax_1^3+bx_2^3+cx_3^3+dx_4^3 = (ay_1^3+by_2^3+cy_3^3+dy_4^3)\,(ay_1^3-by_2^3)^3


\begin{aligned} x_1 &=y_1(ay_1^3+2by_2^3)\\ x_2 &= -y_2(2ay_1^3+by_2^3)\\ x_3 &= y_3(ay_1^3-by_2^3)\\ x_4 &= y_4(ay_1^3-by_2^3) \end{aligned}

Thus, if one has initial {y_1, y_2, y_3, y_4} such that the RHS is zero, this leads to a second, the {x_1, x_2, x_3, x_4}. By iteration, these can be used to generate a third, and so on.  The reason why {x_3, x_4}  have a common factor will be clear in a moment.

This is by A. Desboves, but it is not hard to generalize it to n cubes. The basis is the identity,

\text{(1)}\;\; a(ax^4+2bxy^3)^3 + b(-2ax^3y-by^4)^3 = (ax^3+by^3)(ax^3-by^3)^3

If the first factor of the RHS, ax^3+by^3, can be expressed as a sum of n cubes,

ax^3+by^3 = c_1z_1^3+c_2z_2^3+\dots +c_nz_n^3

it is a simple matter of distributing the second factor of the RHS among the n cubes so that (1) assumes the form,

au_1^3+bu_2^3 = c_1v_1^3+c_2v_2^3+\dots +c_nv_n^3

which explains the common factor of {x_3, x_4} in the 4-cube identity.  Hence from {x, y, z_i}, we get new solutions {u_i, v_i} leading to further solutions, ad infinitum.

There doesn’t seem to be any example of a diagonal quartic surface,

a_1y_1^4+a_2y_2^4+a_3y_3^4+\dots+a_ny_n^4 = 0

which has been proven to have only one non-trivial and primitive integer solution.  Based upon the quadratic and cubic cases, it is tempting to speculate that if it has one, then there may be in fact an infinity.


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