ax^3+by^3+cz^3 = dxyz


We show diophantine equation ax^3+by^3+cz^3 = dxyz has infinitely many integer solutions.


ax^3+by^3+cz^3 = dxyz......................................................................................................................(1)

Let x=pt+1, y=t+1, z=t-1, d=c-a-b, equation (1) reduces to

(b-cp+ap+ap^3+bp+c)t^2+(-4c+4b+a+3ap^2)t+3c-bp+2ap+3b+cp=0.

Since this quadratic in t must have rational solutions, the discriminant must be square number.

v^2 = (a^2+4ab-4ac)p^4+(-12ac-12ab)p^3+(-2a^2-20ac+20ab+4c^2-8cb+4b^2)p^2+(-20ac-20ab-8b^2+8c^2)p+a^2-8ac+8ab+4b^2-56cb+4c^2................(2)

This quartic equation is birationally equivalent to an elliptic curve below.

Y^2-4(7ac+7ab+b^2-c^2)YX/(-b+2a+c)-288a(c^5+b^5-3bc^4+2b^2c^3+4a^3b^2-3b^4c+2b^3c^2+ba^4-4a^3c^2+a^4c-4ac^4+4ab^4+19ab^3c-19abc^3+21a^2b^2c+21a^2c^2b+6a^2b^3+6a^2c^3)Y/((-b+2a+c)^3)
= X^3-8(a^4+11a^3c-11a^3b-35a^2bc-a^2c^2-a^2b^2-7ab^3-5ab^2c+7ac^3+5abc^2+2bc^3-4b^2c^2+2b^3c)X^2/((-b+2a+c)^2)
-64a(-24a^5bc+a^7+80a^4b^3+25ac^6-66a^2c^5-80a^4c^3+95a^3b^4+25ab^6+95a^3c^4+66a^2b^5-10a^6c+39a^5c^2+39a^5b^2+796ab^3c^3+66abc^5-321a^3c^2b^2-1470a^2b^2c^3-138a^4c^2b+372a^2b^4c+430a^3b^3c
+1470a^2b^3c^2+138a^4b^2c+430a^3c^3b-372a^2c^4b-489ab^4c^2-489ab^2c^4-4c^7+10a^6b-84c^5b^2+28c^6b-28b^6c+84b^5c^2-140b^4c^3+66b^5ca+140c^4b^3+4b^7)X/((-b+2a+c)^4)
+512a(-7839a^4b^5c^2+63885a^4b^3c^4-63885a^4b^4c^3+7839a^4b^2c^5-6258a^6b^4c-1527a^7c^3b+3573a^7b^2c^2+46263a^5b^3c^3-1527a^7b^3c+4941a^6b^3c^2+6258a^6bc^4+179c^9a^2+567c^8a^2b-72a^9c^2+1002a^4c^7
-1332a^6b^5-179b^9a^2-531a^3b^8+1332a^6c^5-531a^3c^8-1002b^7a^4-1356a^5c^6+226b^9ac-894a^7c^4-396b^8ac^2-486a^8c^2b+486b^2a^8c-14418a^3b^6c^2-3333a^3c^7b+6486a^4c^6b-567b^8a^2c+5940b^4a^2c^5-4590c^7a^2b^2
-8343a^5c^5b-396c^8ab^2+3624b^6ac^4-3810b^6a^2c^3-744b^7ac^3+288b^8c^3-72b^9c^2+8b^10c-672b^7c^4+226c^9ba+3624c^6ab^4+3810c^6a^2b^3-28b^10a-288c^8b^3+672c^7b^4+1008c^5b^6-1008c^6b^5+72c^9b^2-8c^10b-28c^10a
-5364c^5ab^5+a^10c+a^11-ba^10-8343a^5b^5c-26919a^5b^4c^2+21807a^3b^5c^3+21807a^3b^3c^5-894a^7b^4-7050a^3b^4c^4+366a^8c^3-4941a^6b^2c^3-366b^3a^8-26919a^5b^2c^4+161a^9bc-6486a^4b^6c-3333a^3b^7c-14418a^3c^6b^2
+4590b^7a^2c^2-5940b^5a^2c^4-72a^9b^2-744c^7ab^3-1356a^5b^6)/((-b+2a+c)^6)


Using a rational solution Q(p,v)=( -3(c+b)/(-b+2a+c), 4(-c^3+3c^2a-3a^2c+a^3+3c^2b+21abc+3a^2b-3b^2c+3b^2a+b^3)/((-b+2a+c)^2) ), then we obtain an integer solution of equation (1) below.

(x,y,z)=(-c-b, a+c, a-b).

The corresponding point is P(X,Y)=( 8(a^4+11a^3c-11a^3b-35a^2bc-a^2c^2-a^2b^2-7ab^3-5ab^2c+7ac^3+5abc^2+2bc^3-4b^2c^2+2b^3c)/((-b+2a+c)^2), 64a^2b+64a^2c+64c^2a-64b^2a-64b^2c-64c^2b ).

Using 2P(X,Y) we obtain other integer solution of equation (1) below.

(x,y,z)=( -(b+c)(bc^3+3abc^2+b^3c-a^3c+6a^2bc-3ab^2c+a^3b),
           (a+c)(ac^3+3abc^2+a^3c+6ab^2c-b^3c-3a^2bc+ab^3),
           (b-a)(a^3b+3a^2bc+ab^3+ac^3+6abc^2+3ab^2c+bc^3) ).

This point P is of infinite order, and the multiples kP, k = 2, 3, ...give infinitely many points.

Hence we can obtain infinitely many integer solutions for equation (1).


Example: 2x^3 + 4y^3 + 9z^3 = 3xyz

(a,b,c,d)=(2,4,9,3)

P: (x,y,z)=(-13,11,-2)

2P: (x,y,z)=(-35074, 23771, 9718)