1.Introduction


By Choudhry[1], it seems that x(x+y)(x+2y)(x+3y) = u(u+v)(u+2v)(u+3v) has follwing 6 parametric solutions.
       x                    y              u                    v
 [ p(3p^4+q^4)        -p(p^4-q^4)         4p^4q            -q(p^4-q^4) ]
 [ 2p(3p^4-q^4)       -p(2p^4-5q^4)      13p^4q           -q(7p^4+2q^4)]
 [ 2p(3p^4+4q^4)      -p(2p^4+7q^4)     -13p^4q            q(5p^4-2q^4)]
 [ 2p(3p^4-4q^4)      -p(2p^4-5q^4)       7p^4q           -q(5p^4-2q^4)]
 [ 6p(p^4+q^4)        -p(2p^4+7q^4)     -15p^4q            q(7p^4+2q^4)]
 [   p+3q                  -q              p                    q      ]
    
We give two degree 3 parametric solutions of x(x+y)(x+2y)(x+3y) = u(u+v)(u+2v)(u+3v).


2.Theorem
        
     

     x(x+y)(x+2y)(x+3y) = u(u+v)(u+2v)(u+3v) has two parametric solutions.
     
     Case 1:
       x = 2b^2(a-b)
       y = -a^2b+b^3-b^2a-3a^3
       u = -2b^2(b+a)
       v = -a^2b+b^3+b^2a+3a^3
     
     Case 2:
       x = -64ba^2+42b^2a-9b^3
       y = -64a^3+120ba^2-58b^2a+9b^3
       u = -256a^3+296ba^2-120b^2a+18b^3
       v = 128a^3-136ba^2+46b^2a-6b^3
 

Proof.

x(x+y)(x+2y)(x+3y) = u(u+v)(u+2v)(u+3v)......................................(1)

x = pt+m, y = t+n, u = pt-m,  v = t-n .......................................(2)

Substitute (2) to (1), and simplifying (1), we obtain

(8mp^3+36mp^2+12p^3n+44mp+44p^2n+12m+36pn)t^3
+(36mn^2+12pn^3+8m^3p+12m^3+36m^2pn+44m^2n+44mpn^2)t=0.......................(3)

To obtain a solution of (3), we have to find the rational solution of (4).

w^2=(-784p^2-64p^4-384p^3-144-624p)m^4+(-384p^4-3984p^2-2144p^3-528-2800p)nm^3
   +(-3696p-784p^4-3984p^3-432-6464p^2)n^2m^2
   +(-2800p^3-1440p-3696p^2-624p^4)n^3m
   +(-144p^4-528p^3-432p^2)n^4...............................................(4)
   
Let {m,n}={-p,2p+3}, right hand side of (4) becomes to 192p^2(4p+9)(4p-5)(p+1)^2.

Hence, let p=2b^2/(a^-b^2), then m=-p=-2b^2/(a^-b^2), n=(3a^2+b^2)/(a^-b^2), and t=b/a.
   
Substitute m, n, and t to (2), and obtain a parametric solution of case 1.            


Let {m,n}={-3p,2p-1}, right hand side of (4) becomes to 64p^3(2p+3)^2(p+2)^3.

Hence, let p=(64+9s^2-42s)/(-16+3s^2), then m=-3p=-3(64+9s^2-42s)/(-16+3s^2), n=3(48+5s^2-28s)/(-16+3s^2), and t=3(s-4)/(3s-4).
   
Substitute m, n, t, and s=b/a to (2), and obtain a parametric solution of case 2.            

   
Q.E.D.　
 
　　                  
       
3.Examples


Case 1:
            a,b     x  x+y   x+2y  x+3y    u  u+v  u+2v u+3v
           [2, 1]  [2, -27,  -56,  -85]  [-6,  17,  40,  63]
           [3, 1], [1, -22,  -45,  -68]  [-2,  17,  36,  55]
           [3, 2]  [8, -95, -198, -301]  [-40, 43, 126, 209]



Case 2:
            a,b      x    x+y   x+2y  x+3y     u     u+v  u+2v u+3v
           [2, 1]  [-181, -320, -459, -598]  [-1086, -520, 46, 612]
           [3, 1]  [-153, -424, -695, -966]  [-1530, -742, 46, 834]
           [3, 2]  [ -30,  -38,  -46,  -54]  [ -120,  -57,  6,  69]




4.Reference

[1].Ajai Choudhry: Symmetric diophantine equations, Rocky Mountain Journal of Mathematics. 34(2004)