1.Introduction

I found interesting identity a4+b4+(a+b)4=2(a2+ab+b2)2 in the Dickson's book([1]. Dickson).

Because some application examples of the above identity are introduced,
I think that I will try to put it in order as a theorem here.



Ramanujan's 4th power identity:
(8s2+40st-24t2)4+(6s2-44st-18t2)4+(14s2-4st-42t2)4+(9s2+27t2)4+(4s2+12t2)4 =(15s2+45t2)4　　

New identity this time:
(-14-16k+6k2)4+(8-12k-14k2)4+(-6-28k-8k2)4+(4+4k+4k2)4+(9+9k+9k2)4=(15+15k+15k2)4

We can transform above two identities each other.

[1]. Dickson: HISTORY OF THE THEORY OF NUMBERS VOL.2



2.Theorem
      
     
    
    n:integer

　　If n > 2, there are always infinitely many parameter solutions of A14+A24+A34+...+An-14=An4.

    Condition

    If n > 2,there is always a solution of a14+a24+a34+...+an-14=an4.
    There are x,y,z such that ax+ay=az
    x,y,z < n
     
Proof.

a1,a2,...,an:integer
a,b:rational number
n:integer


Let C be a rational conic with equation f(a,b)=a2+ab+b2.........................(1)
D=-3<0 then C is ellipse.
Set a0=ax,b0=ay
Take P(a0,b0) a rational point on C,then we can get many rational points infinitely.
We parametrise the rational point on C.
Let a=a0+t,b=b0+kt..............................................................(2)
then substitute (2) to a2+ab+b2=a02+a0b0+b02.
Solve for t,and obtain t=g(k)...................................................(3)
g(k) is rational function of t.
Substitute (3) to (2),then

a=a0+g(k)=ha(k)
b=b0+kg(k)=hb(k)................................................................(4)

By condition
a14+a24+a34+...+an-14=an4........................................................(5)
ax4+ay4+az4=a4+b4+(a+b)4

Substitute (4) to (5),then

ha(k)4+hb(k)4+(ha(k)+hb(k))4+...+an-14=an4
Simplifying above equation,we obtain a parameter solution for (5).
       
Moreover, because we can obtain the (a1,a2,...,an) infinitely by one of parameter solutions,
there are infinitely many parameter solutions. 
       
Q.E.D.　
 
　　                  
       
3.Example

I show a few examples.

    
Case n=5


44+64+84+94+144=154..............................................(6)
6+8=14
Let a=6,b=8 then a2+ab+b2=148
We can find the other solution of a4+b4+(a+b)4=64+84+144,
if we find the rational solution of a2+ab+b2=148.
Take a=6+t,b=8+kt then we obtain
    a=6-2(10+11k)/(1+k+k2)
    b=8-2k(10+11k)/(1+k+k2) 
as the rational solution of a2+ab+b2=148.

Substitute a,b to (6),and obtain a parameter solution.

(-14-16k+6k2)4+(8-12k-14k2)4+(-6-28k-8k2)4+(4+4k+4k2)4+(9+9k+9k2)4=(15+15k+15k2)4

            k

            1    84+    64+   144+    44+    94=   154
            2   224+   724+   944+   284+   634=  1054
            3    84+  1544+  1624+   524+  1174=  1954
            4    64+   884+   824+   284+   634=  1054
            5   564+  4024+  3464+  1244+  2794=  4654
            6  1064+  5684+  4624+  1724+  3874=  6454
            7   564+  2544+  1984+   764+  1714=  2854
            8  2424+  9844+  7424+  2924+  6574= 10954
            9  3284+ 12344+  9064+  3644+  8194= 13654
           10  1424+  5044+  3624+  1484+  3334=  5554
           11  5364+ 18184+ 12824+  5324+ 11974= 19954
           12  6584+ 21524+ 14944+  6284+ 14134= 23554
           13  2644+  8384+  5744+  2444+  5494=  9154
           14  9384+ 29044+ 19664+  8444+ 18994= 31654
           15 10964+ 33224+ 22264+  9644+ 21694= 36154
           16  4224+ 12564+  8344+  3644+  8194= 13654
           17 14484+ 42424+ 27944+ 12284+ 27634= 46054
           18 16424+ 47444+ 31024+ 13724+ 30874= 51454
           19  6164+ 17584+ 11424+  5084+ 11434= 19054
           20 20664+ 58324+ 37664+ 16844+ 37894= 63154

Take (a1,a2,a3,a4,a5,a6)=(22,72,94,28,63,105) and obtain other parameter solution below.
(-94-144k+22k2)4+(72-44k-94k2)4+(-22-188k-72k2)4+(28+28k+28k2)4+(63+63k+63k2)4-(105+105k+105k2)4


    
Case n=6
  

24+64+84+24+74+124=134...........................................(7)
2+6=8

Let a=2,b=6 then a2+ab+b2=52
We can find the other solution of a4+b4+(a+b)4=24+64+84,
if we find the rational solution of a2+ab+b2=52.
Take a=2+t,b=6+kt then we obtain
    a=2-2(5+7k)/(1+k+k2)=2(-4-6k+k2)/(1+k+k2)
    b=6-2k(5+7k)/(1+k+k2)=-2(-3+2k+4k2)/(1+k+k2)
as the rational solution of a2+ab+b2=52.

Substitute a,b to (7),and obtain a parameter solution.

(-8-12k+2k2)4+(6-4k-8k2)4+(-2-16k-6k2)4+(2+2k+2k2)4+(7+7k+7k2)4+(12+12k+12k2)4=(13+13k+13k2)4

1<=k<=20
            k

            1    64+    24+    84+    24+    74+   124=   134 
            2   244+   344+   584+   144+   494+   844=   914
            3    24+    64+    84+    24+    74+   124=   134
            4    84+   464+   544+   144+   494+   844=   914
            5   184+  2144+  2324+   624+  2174+  3724=  4034
            6    84+  3064+  3144+   864+  3014+  5164=  5594
            7    24+  1384+  1364+   384+  1334+  2284=  2474
            8   244+  5384+  5144+  1464+  5114+  8764=  9494
            9   464+  6784+  6324+  1824+  6374+ 10924= 11834
           10   244+  2784+  2544+   744+  2594+  4444=  4814
           11  1024+ 10064+  9044+  2664+  9314+ 15964= 17294
           12  1364+ 11944+ 10584+  3144+ 10994+ 18844= 20414
           13   584+  4664+  4084+  1224+  4274+  7324=  7934
           14  2164+ 16184+ 14024+  4224+ 14774+ 25324= 27434
           15  2624+ 18544+ 15924+  4824+ 16874+ 28924= 31334
           16    84+   544+   464+   144+   494+   844=   914
           17  3664+ 23744+ 20084+  6144+ 21494+ 36844= 39914
           18  4244+ 26584+ 22344+  6864+ 24014+ 41164= 44594
           19  1624+  9864+  8244+  2544+  8894+ 15244= 16514
           20  5524+ 32744+ 27224+  8424+ 29474+ 50524= 54734


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