Noam Elkies[1] mentioned that diophantine equation a^6+2b^6+125c^6=2d^6 has infinitely many integer solutions using his identity. (t^2+t-1)^3+(t^2-t-1)^3=2t^6-2. Substitute t^2+t-1=au^2 and t^2-t-1=bv^2 to his identity, then we obtain a^3x^6+b^3y^6 = 2w^6 - 2z^6. We show that a^3x^6+b^3y^6+ 2z^6 = 2w^6 has infinitely many integer solutions for some (a,b).1.Introductiont^2+t-1=au^2..............................................(1) t^2-t-1=bv^2..............................................(2) Example of (a,b)=(11,5). From equation (1) we obtain a parametric solution as follows. (t,u)=( (4+33k^2-22k)/(-1+11k^2), -(1+11k^2-7k)/(-1+11k^2) ). k is arbitrary. Substitute t=(4+33k^2-22k)/(-1+11k^2) to equation (2) and let k=U, then we obtain quartic equation (3). V^2 = 3025U^4-6050U^3+3795U^2-990U+95.....................(3) This quartic equation is birationally equivalent to an elliptic curve below. Y^2 = X^3+39325X+5656750..................................(4) U = (Y+55X-15125)/(110X-54450) V = 1/220*(X^3-1485X^2-24200Y-30779375-39325X)/((X-495)^2) X = 110V+6050U^2-6050U+1265 Y = 12100UV+665500U^3-998250U^2+417450U-6050V-54450. Elliptic curve (4) has rank 2 and generators P1(X,Y)=(15,2500) and P2(X,Y)=(-15935/196,3804375/2744). Hence we can obtain infinitely many rational solutions for simultaneous equation {t^2+t-1=au^2 , t^2-t-1=bv^2}. Consequently, diophantine equation 1331x^6+125y^6+2z^6 = 2w^6 has infinitely many integer solutions. Example : Solution of using by P1(X,Y)=(15,2500). Solution of equation (3) is (U,V)=(59/264, -11525/6336). So k=59/264 and we obtain t=-927/571, then u=sqrt((t^2+t-1)/a) and v=sqrt((t^2-t-1)/b). (u,v,t)=(19/571,461/571,-927/571). Finally, we obtain (x,y,z,w)=(19,461,571,927). Similaly, we can obtain infinitely many integer solutions by computing multiplication of P1 and P2. Q.E.D.2.MethodWe searched the solution of simultaneous equation {t^2+t-1=au^2 , t^2-t-1=bv^2}. If a=a0n^2, then t^2+t-1=au^2 becomes to t^2+t-1=a0(nu)^2. Solution of t^2+t-1=au^2 is included in solution of t^2+t-1=a0(nu)^2. Hence we consider a,b are square free number. (a,b)=(5,1), (11,5), (19,11), (29,19), (31,1), (31,11), (41,29). a<50 and a>b. Thus diophantine equation a^3x^6+b^3y^6+ 2z^6 = 2w^6 has infinitely many integer solutions for above (a,b).3.Results(a,b)=(19,11): 6859x^6+1331y^6+2z^6 = 2w^6Rank 2. (x,y,z,w)=(39079,44801,21509,161276).(a,b)=(29,19): 24389x^6+6859y^6+2z^6 = 2w^6Rank 1. (x,y,z,w)=(43009,124129,247031,483965).(a,b)=(31,1): 29791x^6+y^6+2z^6 = 2w^6Rank 1. (x,y,z,w)=(71711,205729,163827,357365).(a,b)=(31,11): 29791x^6+1331y^6+2z^6 = 2w^6Rank 2. (x,y,z,w)=(118739,587701,963238,1745285).(a,b)=(41,29): 68921x^6+24389y^6+2z^6 = 2w^6Rank 2. (x,y,z,w)=(930829,2080789,4486751,10033614).[1]: Noam Elkies, The ABC's of Number Theory, Harvard Math. Rev. 1, 64-76, 2007.4.Reference

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