The equation y2=x3+k has been studied including Diophantus and Fermat by the various people.

This time,I solve this equation in the Diophantus way,especially parameter solutions.



            1.  y2=x3+16(9a2-4a+4)(a-1)2 has four solutions.

                This equation has very interesting property.

                (x1,y1)=(4a, 8-12a+12a2 )
                (x2,y2)=(4a-4, 12a(a-1) )
                (x3,y3)=(8-8a, 12(a-1)(a-2) )
                (x4,y4)=(4a2-4a, 4(a-1)(2a2-a+2) )

                Example.

                    a     y        x        k


                    2    322 =   83+     512    
                    2    242 =   43+     512
                    2     02 =  -83+     512
                    2    322 =   83+     512

                    3    802 =  123+    4672
                    3    722 =   83+    4672
                    3    242 = -163+    4672
                    3   1362 =  243+    4672

                    4   1522 =  163+   19008
                    4   1442 =  123+   19008
                    4    722 = -243+   19008
                    4   3602 =  483+   19008

                    5   2482 =  203+   53504
                    5   2402 =  163+   53504
                    5   1442 = -323+   53504
                    5   7522 =  803+   53504

            2a.  ( a2(a2-1)2 )2 = ( a2(a2-1) )3 - ( a(a2-1) )2a2(a2-1)

                We obtain a  parameter solution of y2=x3-n2x.
                Substitute x=na,y=n to above equation and obtain

                y=a2(a2-1)2
                x=a2(a2-1)
                n=a(a2-1) is always congruent number.

                Example.

                    a         y         x             n

                    2        362 =     123 -          62*   12
                    3       5762 =     723 -         242*   72
                    4      36002 =    2403 -         602*  240
                    5     144002 =    6003 -        1202*  600
                    6     441002 =   12603 -        2102* 1260
                    7    1128962 =   23523 -        3362* 2352
                    8    2540162 =   40323 -        5042* 4032
                    9    5184002 =   64803 -        7202* 6480
                   10    9801002 =   99003 -        9902* 9900
                 
           


            2b.  ( (a2-b2)(a2+b2)2 )2=( (a2+b2)2 )3-( 2ab(a2+b2 ))2( a2+b2 )2

                We obtain a  parameter solution of y2=x3-n2x.
                Substitute x=t,y=mt to above equation and obtain

                y=(a2-b2)(a2+b2)2
                x=( a2+b2 )2
                n=2ab(a2+b2) is always congruent number.

                Example.

                    a   b      y          x           n

                    2   1      752=     253 -         202*   25
                    3   1     8002=    1003 -         602*  100
                    3   2     8452=    1693 -        1562*  169
                    4   1    43352=    2893 -        1362*  289
                    4   2    48002=    4003 -        3202*  400
                    4   3    43752=    6253 -        6002*  625
                    5   1   162242=    6763 -        2602*  676
                    5   2   176612=    8413 -        5802*  841
                    5   3   184962=   11563 -       10202* 1156
                    5   4   151292=   16813 -       16402* 1681

                 


            3.  (1-6b+6b2)2 = (1-4b)3 + 36b4-8b3


                y2=x3+k.....................................(1)
                Substitute x=t+2b,y=t+a to (1) and obtain
                a=6b2, t=1-6b................................(2)
                
                So,substitute (2) to x,y,then we obtain x=1-4b,y=1-6b+6b2 and k=36b4-8b3

                Example.

                    b    y      x      k
                  
                    1    12=  -33 + 28
                    2   132=  -73 + 512
                    3   372= -113 + 2700
                    4   732= -153 + 8704
                    5  1212= -193 + 21500





            4.  (-36a3b2+27a6+8b4)2 = (a(-8b2+9a3))3 + 64b6(b2-a3)

                We can obtain a parameter solution in the same way as the above equation.

                Example.

                    a   b       y       x       k

                    1   2      112=   -233 +  12288
                    1   3     3512=   -633 +  373248
                    1   4    14992=  -1193 +  3932160
                    1   5    41272=  -1913 +  24000000
                    2   1    14482=   1283 -  448
                    2   2     7042=    803 -  16384
                    2   3    -2162=     03 +  46656
                    2   4    -8322=  -1123 +  2097152
                    2   5    -4722=  -2563 +  17000000
                    3   1   187192=   7053 -  1664
                    3   2   159232=   6333 -  94208
                    3   3   115832=   5133 -  839808
                    3   4    61792=   3453 -  2883584
                    3   5     3832=   1293 -  2000000
                    4   1  1082962=  22723 -  4032
                    4   2  1015042=  21763 -  245760
                    4   3   905042=  20163 -  2566080
                    4   4   757762=  17923 -  12582912
                    4   5   579922=  15043 -  39000000
                    5   1  4173832=  55853 -  7936
                    5   2  4040032=  54653 -  495616
                    5   3  3820232=  52653 -  5412096
                    5   4  3519232=  49853 -  28573696
                    5   5  3143752=  46253 -  100000000


            5.  (4a3+4b3)2 = (4ab)3+(4a3-4b3)2

                We obtain a  parameter solution of y2=x3+k2.

                Example.

                    a   b      y      x     k


                    2   1    362=    83 + 282  
                    3   1   1122=   123 + 1042
                    3   2   1402=   243 + 762
                    4   1   2602=   163 + 2522
                    4   2   2882=   323 + 2242
                    4   3   3642=   483 + 1482
                    5   1   5042=   203 + 4962
                    5   2   5322=   403 + 4682
                    5   3   6082=   603 + 3922
                    5   4   7562=   803 + 2442






            6.  ( a3(1+8b3)(-1+b3)2(1-b3)3 )2=( a2(4b3-1)(-1+b3)3 )3+( -3a2b(-1+b3)3 )3

                We obtain a  parameter solution of y2=x3+k3.


                Example.

                    a   b              y             x               k


                    1   2           10924552=     106333 -       20583
                    1   3        25782585922=   18806323 -     1581843
                    2   2           87396402=     425323 -       82323
                    2   3       206260687362=   75225283 -     6327363
                    3   2          294962852=     956973 -      185223
                    3   3       696129819842=  169256883 -    14236563