1.Introduction

I found many parameter solutions for A14 + A24 + A34 - A44 - A54 - A64 +/- A74 = n. (n< 100)


2. Result
          

(4050x4+2)4 + (4050x4+4)4 + (2025x4+4)4 - (4725x4+4)4 - (2700x4+1)4 - (1350x4-4)4 - (30x)4 = 15

(567x4+8)4 + (162x4+6)4 + (81x4-1)4 - (513x4+8)4 - (432x4+5)4 - (27x4+5)4 - (12x)4 = 47

(63x4+3)4 + (36x4)4 + (18x4+3)4 - (54x4+3)4 - (54x4+2)4 - (27x4-1)4 - (6x)4 = 64

(54x4+4)4 + (54x4+3)4 + (27x4+4)4 - (63x4+4)4 - (36x4+4)4 - (18x4-1)4 - (6x)4 = 80

(4050x4+4)4 + (4050x4)4 + (2025x4-1)4 - (4725x4+3)4 - (2700x4-3)4 - (1350x4)4 - (30x)4 = 95

(8x+3)4 + (5x+1)4 + (3x+2)4 - (7x+3)4 - (7x+2)4 = 1

(8x+2)4 + (5x+3)4 + (3x-1)4 - (7x+3)4 - (7x+1)4 = 16

(8x+1)4 + (5x-2)4 + (3x+3)4 - (7x+2)4 - (7x-1)4 = 81




3. Method

A14 + A24 + A34 - A44 - A54 - A64 +/- A74 = n......(1)


1. Case 1

   (a1x+c1)4+(a2x+c2)4+(a3x+c3)4-(a4x+c4)4-(a5x+c5)4-(a6x+c6)4

   =(a14+a24+a34-a44-a54-a64)x4
   +(-4c4a43-4c5a53+4c1a13+4c2a23+4c3a33-4c6a63)x3
   +(-6c42a42-6c52a52-6c62a62+6c12a12+6c22a22+6c32a32)x2
   +(4c13a1-4c43a4-4c53a5-4c63a6+4c23a2+4c33a3)x
   +c14+c24+c34-c44-c54-c64..........................(2)


   Decide (a1,a2,a3,a4,a5,a6) and (c1,c2,c3,c4,c5,c6) to make the coefficient of x4,x3 and x2 to 0.
   Set n=c14+c24+c34-c44-c54-c64.
   We can obtain a parameter solution of (1).
   n=15,47,64,80,95   
 
2. Case 2

   ((a1+a2)x+(c1+c2))4+(a1x+c1)4+(a2x+c2)4-(a3x+c2)4-(a3x+c1+c2)4
   =(2a24+2a14+4a13a2+6a12a22+4a1a23-2a34)x4
   +(8c2a23-8c2a33+8c1a13-4a33c1+12c1a12a2+12c1a1a22+4c1a23+4c2a13+12c2a12a2+12c2a1a22)x3
   +(-12c22a32-6a32c12-12a32c1c2+12c22a22+12c12a12+12c12a1a2+6c12a22+12c1c2a12+24c1c2a1a2+12c1c2a22+6c22a12+12c22a1a2)x2
   +(-4a3c13-12a3c12c2-12a3c1c22-8c23a3+8c13a1+4c13a2+12c12c2a1+12c12c2a2+12c1c22a1+12c1c22a2+4c23a1+8c23a2)x
   +c14

   Decide (a1,a2,a3) and (c1,c2) to make the coefficient of x4,x3,x2 and x to 0.
   
   Set a1= (c1+2c2)c1,a2= -c12+c22,a3= c22+c1c2+c12, and n=c14.
   We can obtain many parameter solutions of special case of (1).

   ((2c1c2+c22)x+c1+c2)4+((c1+2c2)c1x+c1)4+((-c12+c22)x+c2)4-((c22+c1c2+c12)x+c2)4-((c22+c1c2+c12)x+c1+c2)4 = c14

   For example, the case for n=1,16,81

   (c1,c2)=(1,2): (8x+3)4+(5x+1)4+(3x+2)4-(7x+2)4-(7x+3)4 =1
   
   (c1,c2)=(2,1): (5x+3)4+(8x+2)4+(-3x+1)4-(7x+1)4-(7x+3)4 = 16
 
   (c1,c2)=(-3,2): (8x+1)4+(3x+3)4+(5x-2)4-(7x+2)4-(7x-1)4 = 81

   



4. Example

       (8x+3)4 + (5x+1)4 + (3x+2)4 - (7x+2)4 - (7x+3)4 = 1

       1<=x<=20

       x       
 
       1    114 +   64 +   54 -   94 -  104 = 1
       2    194 +  114 +   84 -  164 -  174 = 1
       3    274 +  164 +  114 -  234 -  244 = 1
       4    354 +  214 +  144 -  304 -  314 = 1
       5    434 +  264 +  174 -  374 -  384 = 1
       6    514 +  314 +  204 -  444 -  454 = 1
       7    594 +  364 +  234 -  514 -  524 = 1
       8    674 +  414 +  264 -  584 -  594 = 1
       9    754 +  464 +  294 -  654 -  664 = 1
      10    834 +  514 +  324 -  724 -  734 = 1
      11    914 +  564 +  354 -  794 -  804 = 1
      12    994 +  614 +  384 -  864 -  874 = 1
      13   1074 +  664 +  414 -  934 -  944 = 1
      14   1154 +  714 +  444 - 1004 - 1014 = 1
      15   1234 +  764 +  474 - 1074 - 1084 = 1
      16   1314 +  814 +  504 - 1144 - 1154 = 1
      17   1394 +  864 +  534 - 1214 - 1224 = 1
      18   1474 +  914 +  564 - 1284 - 1294 = 1
      19   1554 +  964 +  594 - 1354 - 1364 = 1
      20   1634 + 1014 +  624 - 1424 - 1434 = 1