Entry 15

Ramanujan gave the following \(4\)th power identities
$$(2x^2+12xy-6y^2)^4+(2x^2-12xy-6y^2)^4+(4x^2-12y^2)^4+(4x^2+12y^2)^4+(3x^2+9y^2)^4=(5x^2+15y^2)^4$$ $$(6x^2-44xy-18y^2)^4+(8x^2+40xy-24y^2)^4+(14x^2-4xy-42y^2)^4+(9x^2+27y^2)^4+(4x^2+12y^2)^4=(15x^2+45y^2)^4$$ Expressed as the sextuple $$z_1^4+z_2^4+z_3^4+z_4^4+z_5^4=z_6^4$$ note that the two obey \(z_1+z_2=z_3\). It can be shown that, just like its \(3\)rd power counterpart discussed in Entry 13, there are infinitely many such formulas. Use the identity (by yours truly) $$(ax^2+2u_1xy-3ay^2)^k+(bx^2-2u_2xy-3by^2)^k+\big((a+b)x^2-2u_3xy-3(a+b)y^2\big)^k = \big(a^k+b^k+(a+b)^k\big)(x^2+3y^2)^k$$ for \(k = 2,4\) and where \(u_1 =a+2b,\,u_2=2a+b,\,u_3=a-b\). Thus all one needs is to find an initial sextuple of form \(z_1+z_2=z_3\) like $$6^4+8^4+(6+8)^4+9^4+4^4=15^4$$ $$2^4+44^4+(2+44)^4+39^4+52^4=65^4$$ and distributing the \(\text{RHS}\) of the identity will yield a quadratic parameterization.