1.Introduction

I proved that there were always infinitely many parameter solutions of A4 +B4 +C4 +D4 +E4 =F4
in the last time.

For example,one of the parameter solutions is
(2x2+36x-54)4+(2x2-36x-54)4+(4x2-108)4+(4x2+108)4+(3x2+81)4=(5x2+135)4

This time,I proved that there were many other types of parameter solutions of A4 +B4 +C4 +D4 +E4 =F4.



2.Theorem
       
     　There are always  many parameter solutions of A4 +B4 +C4 +D4 +E4 =F4.
      

     
Proof.
     We use the famous identity a4+b4+(a+b)4=2(a2+ab+b2)2.

     2(a2+ab+b2)2=((a2+ab+b2)2+1)2-(a2+ab+b2)4-1.

     So, we get a identity a4+b4+(a+b)4+(a2+ab+b2)4+1=((a2+ab+b2)2+1)2.........(1)

     We find the rational number (a,b) so that (a2+ab+b2)2+1 becomes the square number.

     Let take a2+ab+b2=n/m (m,n are integer)..................................(2)

     (a2+ab+b2)2+1=(n2+m2)/m2.

     Set m=2uv,n=u2-v2 (u,v are integer).

     (a2+ab+b2)2+1=(u2+v2)2/(2uv)2

     So,we have to find the rational solutions of a2+ab+b2=(u2-v2)/(2uv).

     Under the condition sqrt(m2+n2) < 100,
     Pythagorean triples are (m,n,sqrt(m2+n2))=(4,3,5),(8,15,17),(12,5,13),(12,35,37),
     (16,63,65),(20,21,29),(24,7,25),(28,45,53),(40,9,41),(48,55,73),(56,33,65),(60,11,61),
     (80,39,89),(84,13,85).

     Among the above Pythagorean triples,(n/m)=(3/4),(63/16),(13/84) are the rational solutions of
     a2+ab+b2=n/m.

     Then,we can obtain a parameter solution of a2+ab+b2=n/m.

     Substitute (a,b) to (1),and set A=a,B=b,C=a+b,D=n/m,E=1,F=sqrt(m2+n2)/m.

     Consequently,we obtain a parameter solution of A4 +B4 +C4 +D4 +E4 =F4.

     There are many rational solutions of a2+ab+b2=(u2-v2)/(2uv).
     So,there are many parameter solutions of A4 +B4 +C4 +D4 +E4 =F4.

     Under the condition sqrt(m2+n2) < 1000,(n/m)=(133/156),(27/364),(325/228),(37/684),
     (589/300),(301/900),(675/52),(925/372) are solutions of (2).
Q.E.D.　
 
　　                  
       
3.Example

I show a few examples.

1. (n/m)=(3/4)
   
(-4-4K+2K2)4 + (2-4K-4K2)4 + (-2-8K-2K2)4 + (3+3K+3K2)4 + (4+4K+4K2)4 = (5+5K+5K2)4

2. (n/m)=(13/84)

(-34-64k+2k2)4 + (32-4k-34k2)4 + (-2-68k-32k2)4 + (13+13k+13k2)4 + (84+84k+84k2)4 = (85+85k+85k2)4

3. (n/m)=(63/16)

(-36-48k+12k2)4 + (24-24k-36k2)4 + (-12-72k-24k2)4 + (63+63k+63k2)4 + (16+16k+16k2)4 = (65+65k+65k2)4

4. (n/m)=(325/228)

(-290-500k+40k2)4 + (250-80k-290k2)4 + (-40-580k-250k2)4 + (325+325k+325k2)4 + (228+228k+228k2)4 = (397+397k+397k2)4

  I have never seen the example that F is not the multiple of 5. 

  Example: 

  k      A      B        C       D        E        F

  1    2504+    404 +   2904 +   3254 +   2284 =   3974
  2   11304+  10704 +  22004 +  22754 +  15964 =  27794
  3    1104+   2004 +   3104 +   3254 +   2284 =   3974
  4    5504+  15704 +  21204 +  22754 +  15964 =  27794
  5   17904+  74004 +  91904 + 100754 +  70684 = 123074
  6   18504+ 106704 + 125204 + 139754 +  98044 = 170714
  7    6104+  48404 +  54504 +  61754 +  43324 =  75434
  8   17304+ 189504 + 206804 + 237254 + 166444 = 289814
  9   15504+ 239604 + 255104 + 295754 + 207484 = 361274
 10    4304+  98504 + 102804 + 120254 +  84364 = 146894