It's showed that there is a parametric solution of n^3+a^3=b^3+c^3 on MathStackExchange website[1]. We show n^4+a^4+b^4=c^4+d^4+e^4 has a parametric solution for any n.1.IntroductionA diophantine equation n^4+a^4+b^4=c^4+d^4+e^4 has a parametric solution as follows. a = (n^2+nm+m^2)x+n+m b = (n^2+nm+m^2)x+m c = (m^2+2nm)x+n+m d = (n+2m)nx+n e = (m^2-n^2)x+m m,n,x are arbitrary. Proof. (a3x+c2)^4+(a3x+c1+c2)^4+c1^4-((a1+a2)x+(c1+c2))^4-(a1x+c1)^4-(a2x+c2)^4 =2x(c1a2-c1a3+2a1c1+a1c2+xa1^2+2a2c2+a2a1x+xa2^2-2a3c2-xa3^2) (a1^2x^2+a1xc2+2a1xc1+2c2^2+2c1c2+2c1^2+a2xc1+2a2xc2+a1x^2a2+a2^2x^2+a3xc1+2a3xc2+a3^2x^2) Then we obtain a1 = (2c2+c1)c1, a2 = c2^2-c1^2, a3 = c1^2+c1c2+c2^2. Let c1=n and c2=m, then we obtain a parametric solution as follows. a = (n^2+nm+m^2)x+n+m b = (n^2+nm+m^2)x+m c = (m^2+2nm)x+n+m d = (n+2m)nx+n e = (m^2-n^2)x+m Q.E.D.@2.Theoremx=1 n^4 + (n^2+2n+2)^4 + (n^2+n+2)^4 = (2+3n)^4 + (n^2+3n)^4 + (2-n^2)^4. n is arbitrary. (m,n)=(2,1) 1 = (8x+3)^4 + (5x+1)^4 + (3x+2)^4 - (7x+3)^4 - (7x+2)^4. x is arbitrary.3.Examples[1]: StackExchange: Find a,b,c such that n^3+a^3=b^3+c^34.Reference

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