The equation ap^2+bq^2+cr^2+ds^2 = 0

In the previous post, it was discussed that an initial non-trivial integer solution to the diagonal quartic surface,

ax_1^4+bx_2^4+cx_3^4+dx_4^4 = 0

apparently does not help in determining if the equation has an infinite number of primitive integer solutions.  However, it is different for its quadratic and cubic counterparts.

Theorem 1:  “In general, given one non-trivial solution to the quadratic,

a_1y_1^2+a_2y_2^2+\dots+a_ny_n^2 = 0

then an infinite more can be found.”

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

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

then an infinite more can be found.”

Proof of Theorem 1  (Theorem 2 will be discussed in the next post):

We’ll start with n = 3,4 and prove the rest by induction. There is the identity,

ax_1^2+bx_2^2+cx_3^2 = (ay_1^2+by_2^2+cy_3^2)\,(az_1^2+bz_2^2+cz_3^2)^2

where,

(x_1, x_2, x_3) = (uy_1-vz_1, uy_2-vz_2, uy_3-vz_3)

and,

(u,v) = (az_1^2+bz_2^2+cz_3^2,\, 2(ay_1z_1+by_2z_2+cy_3z_3)\,)

Thus, if one has an initial solution (y_1, y_2, y_3), then the RHS of the identity becomes zero, and one gets a parametrization in the (x_1, x_2, x_3) for three free variables (z_1, z_2, z_3). An example is given in this page, form 2b.  For n = 4, it is just a generalization,

ax_1^2+bx_2^2+cx_3^2+dx_4^2 = (ay_1^2+by_2^2+cy_3^2+dy_4^2)(az_1^2+bz_2^2+cz_3^2+dz_4^2)^2

where,

(x_1, x_2, x_3, x_4) = (uy_1-vz_1, uy_2-vz_2, uy_3-vz_3, uy_4-vz_4)

and,

(u,v) = (az_1^2+bz_2^2+cz_3^2+dz_4^2,\, 2(ay_1z_1+by_2z_2+cy_3z_3+dy_4z_4)\,)

The pattern is easily seen for n = 5,6, ad infinitum.  If only it was that easy for the quartic case.

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