Notes on the Diophantine Equation A 4  aB 4  C 4  aD 4 By Paul A. Roediger, UTRS, Inc. (paul.a.roediger.ctr@mail.mil) Abstract: A new formulation of the subject equation is presented. Several parametric and semi‐parametric solutions are derived. Originally presented in 1972 [4], two of the then‐new parametric solutions for a  1 were later published in a comprehensive survey of the a  1 case [5]. Following Euler, let A  p  q , C  p  q , D  r  s and B  r  s . The subject equation becomes (1)
Let (2)
Equation (1) becomes (3)




pq p 2  q 2  ars r 2  s 2 p  ry , s  qx and r  qt Observe that (3) can be written as (4)
(5)
a x3  y
 t2 y3  ax
x  y
 t 2 , where y  a x
xy  1
 2
ax  y 2
Solutions of (4) are given by (6)
x 
t2  
,y

at 2  1

Substituting (6) into (5) produces a 2  3 t 4  3a 2  1 t 2  a 3   2 (7)
Solutions of (1) therefore take the form: 
(8)

p  t  a  t 2  1 , q   , r   t , s  t 2   Case 1: a  1 With a  1 equation (7) has the obvious solution   1 ,   t 2  1 , which leads to the trivial solution 



A  D  t  1 t 2  1 , B  C  t  1 t 2  1 . Following up this case, take   1  z , obtaining   
      3
Let   t  1 z  t  1 in equation (9). The constant and linear terms in z cancel, leaving 2
t  1z  3t  t  1z  94 t  1 z , which has the solution z   34tt 11 . t  1 t  18t  and t  6t  1
,  
Accordingly,   1  z 
4t  1
8t  1
p  2t  t  10t  t  4  ,
q   t  1 t  18t  1
(10) r  t  t  1 t  18t  1 , s  2  4t  t  10t  1
(9)
2
2
4
3
2
t 4  1 z 3  3 t 4  t 2  1 z 2  t 2  1 z  t 2  1
2
2
4
2
2
2
2
2
2
2
4
4
2
2
4
4
2 1
4
6
2
4
4
2
2
2
4
6
4
2
2
Table 1: Euler’s 1st Solution of A 4  B 4  C 4  D 4 obtained from (10) t * C A B
D 3 158
‐59
133
134 2 1203
‐76
653
1176 5 3351
‐2338
3494
1623 5/3 17332
529
6673
17236 * The free parameter here is t   B  D 




2
 A  C 3 t 2 1 2 3 2
z  t  1 z  t 2  1 . Equation (9) is also satisfied by taking  
2
2
8 t 1
In this case, the constant, linear and quadratic terms in z cancel, leaving 4
2
2
8t 2  1  t 4  18t 2  1
9t 2  1 z 4 9t 2  1 z 3
4
3
. t  1z  2 2  8 , so that z  
4
9t 2  1
64t  1
Accordingly,   1  z 
t


t 8  92t 6  326t 4  92t 2  1


9 t 2 1

4
 

,
 and  1  t 12  214t 6  2481t 8  2804t 6  2481t 4  214t 2  1
2


27 t  1
2
6
p  3t  t 2  1  t 8  100t 6  190t 4  44t 2  9 
2
q   t12  214t 6  2481t 8  2804t 6  2481t 4  214t 2  1
(11) r  t  t12  214t 6  2481t 8  2804t 6  2481t 4  214t 2  1
s  3  t 2  1  9t 8  44t 6  190t 4  100t 2  1
2
Table 2: Euler’s 2nd Solution of A 4  B 4  C 4  D 4 obtained from (11) t * C A B
D 3 10381
10203
2903
12231 2 1584749
2061373
‐555707
2219449 5 2533177
1123601
1834883
2367869 * The free parameter here is t   B  D 
 A  C Lander [2] provided methods to derive new parametric solutions of A 4  B 4  C 4  D 4 when one starts out with a known one. For example, starting with (10), Euler’s 1st solution, he derived (11), Euler’s 2nd solution, and another of degree nineteen. Brudno [3] obtained two more solutions of degree nineteen and thirty one. Case 2: a  1 With a  1, solving (7) for t 2 eventually yields other parametric solutions as follows: 3 2  1  
, where 2 3
(12) 2
t 
(13) 2   2  1 4  2  1  4  3 2 

2

 2 1
z , (13) becomes With  

(14) 
2
2

1
2
 4 2  4 z 2  1 Since the right side of (14) is a perfect square when z  1 , let z  1  k . A rational k is obtained by setting


2
 1
 2n  k   2   1 , namely k 
2   1n  2 . n2 1
 2  1   n 2   2   1 n  1

 n 2   2   1 n  1
,  
, Accordingly, z 
n2   1
  n 2   1
 
(15) Let t
t 
2

2


 1 2   1  n 2  4  n  2   1
, and n2 1
n 2  3   2n 2  2n  1  2   n 2  1    n  1
2
 2  n 2   1
 v
n2  n  1
. The  3 term in (15) drops out, as does the  2 term if v 
. 
n2

The remaining linear equation has solution  
v 2   n  1
n 2 v 2  2v   n 2  1
t
n

(16) 5
2

n 6  2n5  2n 4  2n3  3n 2  2n  1
n 2  2n3  2n 2  1
n  n5  4n 4  6n3  6n 2  4n  1
n 6  2n5  2n 4  2n3  3n 2  2n  1
, so that ,  2n 4  2n3  2n 2  n  1 n8  4n7  12n 6  20n5  21n 4  16n3  10n 2  4n  1
n3  2n3  2n 2  1 n6  2n5  2n 4  2n3  3n 2  2n  1
, and p  n 4  n  1 n 2  2n  2  n 4  3n3  3n 2  3n  1
q   n 2  n  1 n3  n 2  1 n6  2n5  2n 4  2n3  3n 2  2n  1
r  n n  1 n 2  n  1 n3  n 2  1 n 4  3n3  3n 2  3n  1
s  n  n10  4n9  8n8  10n7  7 n6  2n5  n 4  2n3  3n 2  2n  1
In [5], Zajta surveys the known transformation methods leading to solutions of A 4  B 4  C 4  D 4 . There, several new methods are documented, including so‐called Simple Dual (SD) and Composite Dual (CD) parametric transformations. By applying an unspecified CD transformation to a variant of (10), the simplest of known parametric solutions, he presents the version of (16), P1  u, v  ([5], pp. 651), one gets for n  
uv
. v
n * 1 ‐2 ‐1/2 ‐3/2 1/2 2 ‐3 ‐1/3 Table 3: Some solutions of A 4  B 4  C 4  D 4 obtained from (16) C A B
D 7
157
‐227
239 ‐257
292
193
‐256 502
298
‐497
‐271 ‐6842
9018
‐4903
‐8409 6742
5098
‐9043
8531 ‐10757
18292
‐45883
46136 ‐28997
33237
59777
‐60369 89841
27879
‐90829
‐43307 * The free parameter is n 
 y  x  y  1
x   y  y  1
2
, where x 
DB
AC
,y
AC
DB
Following Lander’s method [2], two more solutions of comparatively small degree may be derived from (16). They are of degree thirteen and fifteen as follows: p  n3  n 2  1 n 2  n  1 n8  4n7  9n6  14n5  14n 4  10n3  6n 2  2n  1
q  n 4  n 2  2n  2  n3  n  1 n 4  2n3  2n 2  n  1
 n12  6n11  19n10  40n9  64n8  80n7  82n6  
r  n 

5
4
3
2
 68n  46n  26n  12n  4n  1

(17) s  n  n3  n 2  1 n3  n  1 n 2  n  1  n 4  2n3  2n 2  n  1
 n14  8n13  32n12  90n11  195n10  320n9  391n8  
p   n  1 

7
6
5
4
3
2
 358n  254n  146n  71n  30n  12n  4n  1 
q  n n  1  n 4  3n3  3n 2  3n  1 n 4  2n3  n  1 n5  5n 4  8n3  5n 2  n  1
(18) r  n  n  1  n 4  3n3  3n 2  3n  1 n6  4n5  9n 4  6n3  3n 2  2n  1
4
s   n 4  2n3  n  1 n5  5n 4  8n3  5n 2  n  1 n6  2n5  2n 4  2n3  3n 2  2n  1
Table 4: More small solutions of A 4  B 4  C 4  D 4 Equation n C A
B
D
1 292
193
257
256 ‐2 ‐2797
248
2131 ‐2524 *(17) ‐1/2
2345
‐2986
3190
1577 1/2 60763 38078
62206 29531 ‐2 ‐239
7
‐227
157 1 4288
4303
3364
4849 (18) ‐1/2
2707
6730
3070 ‐6701 ‐3/2 ‐73703 154522 ‐151394 ‐92839 * Zajta’s P2  u , v  , ([5], pp. 651), is (17) with n  v / u Case 3:   1 (Ad hoc solutions) Parametric solutions of equation (1) abound in the literature. The most remarkable ones have constant a and are typically hard to come by. Other solutions feature non‐constant a which are functionally simple and can provide new and small numerical solutions. An example of this type was noted in [1] by Hayashi: (19) p  u u 2  3 , q  2 u 2  1 , r  u u 2  1 , s  2 , a  u 2  3 
By choosing u 





7
in (19) one obtains 542 4  103 4  514 4  359 4 , a numerical solution first found by Euler. 4
Other interesting solutions of this type may be found by solving (7) for a when   1 and   at 2   . Doing so, and reevaluating (8) yields: (20) p  t at 2  1 , q  at 2   , r  t at 2    , s  t 2  1 , a 
 2  t2
2  3 t 2  1
The   1 restriction is not as limiting as first thought: examination reveals that ac 4 , c 2 , tc, c satisfies (7) 

whenever a,  , t ,   does. Thus, setting   1 is tantamount to restricting  to be a perfect square. Judicious play with  and t in (20) lead to simple a ' s , as found in (19). Some choices, indexed by i , are tabulated below. Note that the first two lead to parametric solutions for a  1 / 4 and a  1 / 9 . Table 5: Some  , t combinations with resulting a 
i i ti ai i 1 1
2
u 1
4
7 2 3u 2  4
u2
u 3 1
u2
1
u


4  3u 2  4 u 3u 2  4
u
5 2
2
2
u u  2 u  2
6 1  4u 2
4
u u
4
9u 4
2

2
1
2
u 2
9 9u 2  16
10
1  6u 2
i ti ai 2
u2
u 4  2u 2  2
2 1 u2
1
u 1

1
u
3u 2  2
2u u 2  1

1 u4
5
u2  9
u2  7
3u 2  5
u u 2  7 
u2  7
u2  3

8 

3
2

11
2 4u 2  1  9u 2
2 u2  2
4u 2  1
8 2  u2
‐ ‐ 


u u2  2
u4

 2  t2
2  3 t 2  1

2 ‐ 2
u2 
9
4
4u 2  1
22  u 2 
‐ Table 6: Resulting p, q, r , s and a for the various  and t of Table 5 i pi qi ri si ai 1 u u 2  4 u2  2 u u 2  2 4u 2  1 1
4
2 u u  16 u  20 u u  20 9u 3 u 2  1 u 1 u u 2  2 4 u 9u 2  1 9u 2  4 u 9u 2  4 1  6u 2 5 u u
3u
2
2
2
 1  4u 4  
 u
6  2

7

16
u


7 8 2
2
2
 2  4u 4  


19u 2  4  

u 2u 2  1 u4  u2 1 3u
9u
 3u
2
  1
 5 
2 
2u u 2  1 2
u u 2  7  
9  u 6  u 4    u 6  3u 4  
 2

 9u  25 
 2

 3u  25 
10  4u 4  
r  2
 9
u
4



11 2 u2 1 2 *12 2u 2  1 

2
u
3u u
2
u
2

2
4
9u 4
1
2
u 2
9u 2  16
1  6u 2
2
 4 2
u2  2
u4
2
 4u 4  
 8 u 2  1 2  u 2 4u  1 u 
2

8 2  u2
19u  4 
1
2u 2  1 u u 4 1 2
u 1
3u 2  2 
1 u4
2
10
u
4
u

1
1 u 2
5





 3u

2

 5 
 u 6  3u 4  
 2

 3u  25 





u  u 2  3 
u
u

2
 1 
4
 6u 2  25 
u2  7
u2  3
 4u 4  
r  2
 
9
u
6


4 u 2  1 3u 2 6u 22  u 2  4u 2  1
22  u 2 
3u 6u 22  u 2  4u 2  1
82  u 2 
4u 4 
9u 2  6
* Line 12 is a rational version of Line 11. u2 
9
4
Table 6 parameters i u 3 7/4 8 1/3 8 3 5 7/4 10 7/16 7 3 7 7/9 3 1 5 1 9 1 7 2 9 3 3 5 8 2 9 2 8 1/2 5 5 7 7 1 1 1 3 1 1/2 1 3/2 1 1/3 1 5 4 1 12 3/2 6 3/2 7 9 7 1/9 2 3/2 2 8/3 2 15/4 2 5/6 Table 7: Numerical solutions of A 4  aB 4  C 4  aD 4
a 1 2 3 4 5 9 A 631 631 1381 2949 10943964 248 16727 4 11 11 37 86 93 277 277 304 444 16897 9 19 47 101 137 219 3 22 197 58879 64151 625 830 2159 2509 B C D 222
222
878
1034
1733885
223
36384
1
2
2
1
997
134
149
149
127
49
3348
4
46
3
77
14
122
0
17
85
15860
34620
77
329
1367
233
558
558
1342
2854
10758915
44
41513
2
7
7
23
1256
63
241
241
268
426
16703
7
61
33
107
103
11
1
4
49
59201
51031
85
250
1513
1105
503
503
997
1797
5558948
257
23532
3
8
8
27
631
136
191
191
193
211
6064
6
32
31
73
88
168
2
19
137
10064
43152
361
503
1519
1435
References: 1.
2.
3.
4.
5.
T. Hayashi, On the Diophantine equation x  y  z  t , Tohoku Math . J., v. 1, 1911‐1912, pp.143‐145. L. J. Lander, Geometric aspects of Diophantine equations involving equal sums of like powers, Amer. Math. Monthly, v. 75, 1968, pp. 1061‐1073. S. Brudno, Some new results on equal sums of like powers, Math. Comp., v. 23, 1969, pp. 877‐880 P. A. Roediger, Notes on the Diophantine Equation A 4  aB 4  C 4  aD 4 , Presented 4
4
4
4
At the Eighteenth Conference of Army Mathematicians, 25 May, 1972 (Unpublished) A. J. Zajta, Solutions of the Diophantine equation A 4  B 4  C 4  D 4 , Math. Comp., v. 41, 1983, pp. 635‐659