1.Introduction

Previously, only one solution was given for the fixed coefficients for ax^4 + by^4 + cz^4 + dw^4 + eu^4= 0,
This time we were able to give infinitely many integer solutions.

We show that ax^4 + by^4 + cz^4 + dw^4 + eu^4= 0 has always infinitely many integer solutions by supposing {a+b+c+d+e= 0, d=-c}.


2.Theorem
        
     

    ax^4 + by^4 + cz^4 + dw^4 + eu^4 = 0 has always infinitely many integer solutions as follows.
    
    x = c((a^5-a^3c^2)p^6+(-6a^2bc^2+6a^4b)qp^5+(14a^3b^2-6ab^2c^2+9a^2bc^2-a^4b)q^2p^4+(12ab^2c^2-8a^2bc^2-4a^3b^2+16a^2b^3)q^3p^3
        +(9ab^4-6a^2b^3-13ab^2c^2+2a^2bc^2)q^4p^2+(-4ab^4+4ab^2c^2-2b^3c^2+2b^5)q^5p+(-b^5+b^3c^2)q^6)
    
    y = c((a^5-a^3c^2)p^6+(-4a^2bc^2-2a^5+4a^4b+2a^3c^2)qp^5+(6a^3b^2-2ab^2c^2+13a^2bc^2-9a^4b)q^2p^4
        +(-12a^2bc^2-16a^3b^2+8ab^2c^2+4a^2b^3)q^3p^3+(6a^2bc^2+ab^4-14a^2b^3-9ab^2c^2)q^4p^2+(6ab^2c^2-6ab^4)q^5p+(-b^5+b^3c^2)q^6)
    
    z = (a^6-a^4c^2)p^6+(-6a^3bc^2+6ba^5)qp^5+(-2ac^3b^2-2a^2bc^3+6a^3bc^2-9a^2b^2c^2+15a^4b^2)q^2p^4
        +(4ac^3b^2+4a^2bc^3-4ab^3c^2+20a^3b^3-4a^3bc^2+12a^2b^2c^2)q^3p^3+(-9a^2b^2c^2-2a^2bc^3+6ab^3c^2+15a^2b^4-2ac^3b^2)q^4p^2+(-6ab^3c^2+6ab^5)q^5p+(b^6-b^4c^2)q^6
    
    w = (-a^6+a^4c^2)p^6+(6a^3bc^2-6ba^5)qp^5+(-2ac^3b^2-2a^2bc^3-6a^3bc^2+9a^2b^2c^2-15a^4b^2)q^2p^4
        +(4ac^3b^2+4a^2bc^3+4ab^3c^2-20a^3b^3+4a^3bc^2-12a^2b^2c^2)q^3p^3+(9a^2b^2c^2-2a^2bc^3-6ab^3c^2-15a^2b^4-2ac^3b^2)q^4p^2+(6ab^3c^2-6ab^5)q^5p+(-b^6+b^4c^2)q^6
    
    u = c((a^5-a^3c^2)p^6+(4a^4b-4a^2bc^2)qp^5+(a^4b+3a^2bc^2+6a^3b^2-2ab^2c^2)q^2p^4+(4a^2b^3+4a^3b^2)q^3p^3
        +(3ab^2c^2-2a^2bc^2+ab^4+6a^2b^3)q^4p^2+(4ab^4-4ab^2c^2)q^5p+(-c^2b^3+b^5)q^6)
    
    condition: {a+b+c+d+e= 0, d=-c}.
    a,b,c,p,q are arbitrary.           

 
Proof.

ax^4 + by^4 + cz^4 + dw^4 + e = 0................................(1)

Put x = pt+1, y = qt+1, z = rt+1, w = st+1.......................(2)

Let a+b+c+d+e= 0.

Substitute (2) to equation (1), and simplifying equation (1), we obtain

(bq^4+ap^4+cr^4+ds^4)t^4
+(4cr^3+4ds^3+4bq^3+4ap^3)t^3
+(6ds^2+6cr^2+6bq^2+6ap^2)t^2
+(4ap+4ds+4bq+4cr)t=0............................................(3)

By the assumption, e=-a-b-c-d.

Equating to zero the coefficient of t, then we obtain s = -(ap+bq+cr)/d.

Equating to zero the coefficient of t^2, let d=-c then we obtain

r = -1/2(a^2p^2+2apbq+b^2q^2-cbq^2-ap^2c)/(c(ap+bq))  ...........(4)

t = - coefficient of t^3 / coefficient of t^4.(omitted since result is tedious)

Substitute t to equation (2), and remove common factors, then obtain a parametric solution.

Thus, we can obtain infinitely many integer solutions by supposing {a+b+c+d+e= 0, d=-c}.

   
Q.E.D.　
 
　　                  
       
3.Examples



Case1: (a,b,c,d,e)=(1,1,1-1,-2), x^4+y^4+z^4-w^4-2u^4=0.

x = -2p^2-2pq+q^2
y = -2q^2+p^2-2pq
z = p^2+q^2+4pq
w = -2p^2-2q^2-2pq
u = p^2+q^2+pq

Case2: (a,b,c,d,e)=(1,2,3,-3,-3), x^4+2y^4+3z^4-3w^4-3u^4=0.

x = 3(p^4+14qp^3+27p^2q^2-10pq^3-5q^4)
y = 3(p^4+8qp^3-6q^2p^2-25q^3p-5q^4)
z = p^4+14qp^3+87q^2p^2+50q^3p+10q^4
w = p^4+14qp^3+6q^2p^2+50q^3p+10q^4
u = 3(p^4+10qp^3+18q^2p^2+20q^3p+5q^4)

p,q are arbitrary.