Ramanujan gave the following 4th power identities
(2x2+12xy−6y2)4+(2x2−12xy−6y2)4+(4x2−12y2)4+(4x2+12y2)4+(3x2+9y2)4=(5x2+15y2)4 (6x2−44xy−18y2)4+(8x2+40xy−24y2)4+(14x2−4xy−42y2)4+(9x2+27y2)4+(4x2+12y2)4=(15x2+45y2)4 Expressed as the sextuple z41+z42+z43+z44+z45=z46 note that the two obey z1+z2=z3. It can be shown that, just like its 3rd power counterpart discussed in Entry 13, there are infinitely many such formulas. Use the identity (by yours truly) (ax2+2u1xy−3ay2)k+(bx2−2u2xy−3by2)k+((a+b)x2−2u3xy−3(a+b)y2)k=(ak+bk+(a+b)k)(x2+3y2)k for k=2,4 and where u1=a+2b,u2=2a+b,u3=a−b. Thus all one needs is to find an initial sextuple of form z1+z2=z3 like 64+84+(6+8)4+94+44=154 24+444+(2+44)4+394+524=654 and distributing the RHS of the identity will yield a quadratic parameterization.
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