We show that X^4 + Y^4 + Z^8 = 2W^4 has the parametric solutions.1.IntroductionDiophantine equation X^4 + Y^4 + Z^8 = 2W^4 has the parametric solutions. a,b are arbitrary. Proof. X^4 + Y^4 + Z^8 = 2W^4..........................................................(1) We consider the Proth's identity: A^4 + B^4 + (A+B)^4 = 2(A^2 + AB + B^2)^2.....(2) Let A^2 + AB + B^2 = u^2........................................................(3) We can obtain a parametric solution of equation (3) easily. (A,B,u)=(a^2-b^2, b^2+2ab, b^2+ab+a^2). By using this solution, we can obtain three parametric solutions of equation (1) as follows.2.TheoremCase 1: A is square numberX^8 + Y^4 + Z^4 - 2W^4=0........................................................(4) We can obtain a parametric solution of a^2-b^2=v^2. (a,b,v)=(p^2+q^2, 2pq, p^2-q^2). Hencw, we obtain a parametric solution of equation (4) below. X = p^2-q^2 Y = 4pq(-pq+p^2+q^2) Z = (p^2+q^2)(q^2-4pq+p^2) W = q^4-2pq^3+6p^2q^2-2p^3q+p^4 p,q are arbitrary.Case 2: B is square numberX^4 + Y^8 + Z^4 - 2W^4=0............................(5) We can obtain a parametric solution of b^2+2ab=v^2. (a,b,v)=(1/2(2p+1)/p, p, p+1). Hencw, we obtain a parametric solution of equation (5) below. X = (2p^2-2p-1)(2p^2+2p+1) Y = 2(p+1)p Z = (2p+1)(4p^2+2p+1) W = 1+4p+6p^2+4p^3+4p^4 p is arbitrary.Case 3: A+B is square numberX^4 + Y^4 + Z^8 - 2W^4=0...........................(6) We can obtain a parametric solution of a^2+2ab=v^2. (a,b,v)=(1, 2p(-1+p), -1+2p). Hencw, we obtain a parametric solution of equation (6) below. X = (2p^2-2p+1)(2p^2-2p-1) Y = 4p(-1+p)(p^2-p+1) Z = -1+2p W = 1-2p+6p^2-8p^3+4p^4 p is arbitrary. Q.E.D.

HOME