1. Introduction

It is said that Indian mathematician Ramanujan (1887-1920) discovered the following identities.

(8s2+40st-24t2)4+(6s2-44st-18t2)4+(14s2-4st-42t2)4+(9s2+27t2)4+(4s2+12t2)4 =(15s2+45t2)4　　

(2m2-12mn-6n2)4+(2m2+12mn-6n2)4+(4m2-12n2)4+(4m2+12n2)4+(3m2+9n2)4 =(5m2+15n2)4

I proved the following theorem and found infinitly many identities like above Ramanujan's identities.

(2m2-12mn-6n2)4+(2m2+12mn-6n2)4+(4m2-12n2)4+(4m2+12n2)4+(3m2+9n2)4 =(5m2+15n2)4　is always made
under a certain condition. 




2.Theorem
      
      x,y,z,w,s and t are assumed to be an integer. 

　　　
    The parameter solution of x4 +y4 +z4 +w4 +s4 =t4 exists infinitely. 

　　  　x=(a1x2+b1x-c1)
        y=(a1x2-b1x-c1)
        z=(2a1x2-2c1)
        w=(2a1x2+2c1)
        s=(a2x2+c2)
        t=((a1+a2)x2+(c1+c2))
 　

　　　

      Condition

      a1,a2,b1,c1,c2: integer
      a1,c1:even number
      a2=3a1/2
      c2=3c1/2
      b12=12a1c1
      



Proof.
  
As (a1x2+b1x-c1)4+(a1x2-b1x-c1)4+(2a1x2-2c1)4+(2a1x2+2c1)4+(a2x2+c2)4=((a1+a2)x2+(c1+c2))4 
To adjust the coefficient of x to 0,decide a1,a2,b1,c1,c2.

Expanding and simplifying above equation
　　
(-4a1a23-4a13a2+33a14-6a12a22)x8
+(-4a13c2-12a12a2c2-12a1a22c1-12a12a2c1-12*a1a22c2-4a23c1+12a12b12-12a1^3c1)x6
+(2b14-12a1a2c12-12a1a2c22-6a22c12-12a12c1c2-24a1b12c1+198a12c12-12a2^2c1c2-24a1a2c1c2-6a12c22)x4
+(-12a1c1c22-12a2c1c22-12a1c13+12b12c12-12a2c12c2-4a2c13-12a1c12c2-4a1c23)x2
-6c12c22-4c1c23+33c14-4c13c2

First of all,coefficient of x8=0,then -4a1a23-4a13a2+33a14-6a12a22=a1(-2a2+3a1)(2a22+6a1a2+11a12)=0
so,it becomes a2=3a1/2 
c2, similarly becomes c2=3c1/2 from the paragraph of the constant, too. 
As for a1 and c1 should be the even number at the same time. 

Next, if a2=3a1/2 and c2=3c1/2 are substituted, 
the coefficient of x6=-144a13c1+12a12b12=-12a12(12a1c1-b12) 
coefficient of x6=0 then  b12=12a1c1. 

The coefficient of x4 and x2 is also similar to x6. 

Because all coefficients were adjusted to 0 by this, it is understood that the parameter 
solution exists. 
Moreover, because a1 and c1 that filled b12=12a1c1 are able to be chosen infinitely, there 
are infinitely many parameter solutions. 

Q.E.D.

 
                  
       
3.Example


I show a few examples.


(2x2+36x-54)4+(2x2-36x-54)4+(4x2-108)4+(4x2+108)4+(3x2+81)4=(5x2+135)4

(2x2+48x-96)4+(2x2-48x-96)4+(4x2-192)4+(4x2+192)4+(3x2+144)4=(5x2+240)4
　
(2x2+72x-216)4+(2x2-72x-216)4+(4x2-432)4+(4x2+432)4+(3x2+324)4=(5x2+540)4

(2x2+96x-384)4+(2x2-96x-384)4+(4x2-768)4+(4x2+768)4+(3x2+576)4=(5x2+960)4

(2x2+108x-486)4+(2x2-108x-486)4+(4x2-972)4+(4x2+972)4+(3x2+729)4=(5x2+1215)4

(6x2+96x-128)4+(6x2-96x-128)4+(12x2-256)4+(12x2+256)4+(9x2+192)4=(15x2+320)4

(6x2+192x-512)4+(6x2-192x-512)4+(12x2-1024)4+(12x2+1024)4+(9x2+768)4=(15x2+1280)4