Last time, we showed n^4=a^4+b^4+c^4-d^4-e^4 has a parametric solution. This time, we show n^5=a^5+b^5+c^5-d^5-e^5-f^5-g^5 has a parametric solution for any n. This means that any fifth power can be represented by sum and difference of seven fifth powers. We used Lander's[1] method.1.IntroductionA diophantine equation n^5=a^5+b^5+c^5-d^5-e^5-f^5-g^5 has a parametric solution as follows. a = 12n^2+271n+1530 b = 8n^2+179n+990 c = 4n^2+109n+720 d = 4n^2+79n+390 e = 12n^2+269n+1500 f = 19n+210 g = 8n^2+191n+1140 n is arbitrary. Proof. A1^5+A2^5+A3^5+A4^5=B1^5+B2^5+B3^5+B4^5.............................(1) Let A1=u1+v1+w1, A2=u1-v1-w1, A3=u2+v2-w2, A4=u2-v2+w2, B1=u1+v1-w1, B2=u1-v1+w1, B3=u2+v2+w2, B4=u2-v2-w2. Then equation (1) is reduced to simultaneous equation (2). { u1v1w1=u2v2w2, u1^2+v1^2+w1^2 = u2^2+v2^2+w2^2 }..................(2) Let (u1,v1,w1)=(x,uy,v) and (u2,v2,w2)=(y,vx,u), the we obtain (1-v^2)x^2+(u^2-1)y^2+v^2-u^2=0.....................................(3) Let (u,v)=(3,2), we can obtain a parametric solution of equation (1) as follows. (A1,A2,A3,A4)=( -12+2k, 18+16k^2-34k, 12-16k^2-26k, -18+38k ) (B1,B2,B3,B4)=( -32k^2+2k, 6+48k^2-34k, -6+32k^2-26k, -48k^2+38k ) Substitute k = 6+1/2n to above equation, then we obtain a parametric solution of n^5=a^5+b^5+c^5-d^5-e^5-f^5-g^5. a = 12n^2+271n+1530 b = 8n^2+179n+990 c = 4n^2+109n+720 d = 4n^2+79n+390 e = 12n^2+269n+1500 f = 19n+210 g = 8n^2+191n+1140 Q.E.D.@2.Theorem[1]: L.J.Lander,Geometric aspects of diophantine equations involving equal sums of powers, Amer. Math. Monthly 75 (1968)3.Reference

HOME