1.Introduction

Ajai Choudhry,etal.[1] showed that equation (x1^4 + x2^4 )(y1^4 + y2^4 ) = z1^4 + z2^4 has infinitely many parametric solutions.

We show that (x1^3 + x2^3 )(y1^3 + y2^3 ) = z1^3 + z2^3 has a parametric solution.
     
     
2.Theorem
     
Diophantine equation (x1^3 + x2^3 )(y1^3 + y2^3 ) = z1^3 + z2^3 has a parametric solution.

x1 = 4a^6+(-6c^3+8b^3-6c^2)a^3+4b^6+(-6c^3-6c^2)b^3+3c^4+6c^5-c^6
x2 = -4a^6+(-6c^3-8b^3+6c^2)a^3-4b^6+(-6c^3+6c^2)b^3+c^6+6c^5-3c^4
y1 = a
y2 = b
z1 = -c(8a^6+(16b^3-12c^2)a^3-12b^3c^2+3c^4+c^6+8b^6)
z2 = -c(4a^6+8a^3b^3+4b^6-c^6-3c^4)
a,b,c are arbitrary.
 
Proof.

(x1^3 + x2^3 )(y1^3 + y2^3 ) = z1^3 + z2^3........................................(1)

Substitute x1=t+1, x2=t-1, y1=a, y2=b, z1=pt+c, z2=t-c to equation (1), we obtain

(2a^3+2b^3-p^3-1)t^3+(-3cp^2+3c)t^2+(6a^3+6b^3-3c^2p-3c^2)t=0.....................(2)

Let p = -(-2a^3-2b^3+c^2)/(c^2), then we obtain

t = -6c^3(-b^3-a^3+c^2)/(-4b^6-8a^3b^3+6b^3c^2+c^6+6a^3c^2-4a^6-3c^4).

Hence we obtain a parametric solution as follows.

x1 = 4a^6+(-6c^3+8b^3-6c^2)a^3+4b^6+(-6c^3-6c^2)b^3+3c^4+6c^5-c^6
x2 = -4a^6+(-6c^3-8b^3+6c^2)a^3-4b^6+(-6c^3+6c^2)b^3+c^6+6c^5-3c^4
y1 = a
y2 = b
z1 = -c(8a^6+(16b^3-12c^2)a^3-12b^3c^2+3c^4+c^6+8b^6)
z2 = -c(4a^6+8a^3b^3+4b^6-c^6-3c^4)

Q.E.D.


3.Reference

[1]. Ajai Choudhry,Iliya Bluskov and Alexander James,  A quartic diophantine equation inspired by Brahmaguptafs identity
https://arxiv.org/abs/2010.08133





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