1.Introduction



I show two parametric solutions of x^4 + y^4 + z^4 = w^2.

First,I show the Diophantus's parametric solution, next show a new solution by using Diophantus's one.



2. Result

 


   
    Case 1:  x^4 + y^4 + z^4 = w^2(Diophantus)

             x = 2mn(m^2 - n^2)   

             y = (m^2 - n^2)(m^2 + n^2)

             z = 2mn(m^2 + n^2)

             w = m^8 + 14m^4n^4 + n^8



             Let x^4+a^4+b^4=(x^2-k)^2.........................................(1)

             Then, x^2=-1/2(a^4+b^4-k^2)/k.....................................(2)

             Substitute k= a^2+b^2 to (2),then x^2=a^2b^2/(a^2+b^2)............(3)

             Substitute a = m^2-n^2 and b = 2mn to (3), then x = 2mn(m^2-n^2)/(m^2+n^2).

             Hence we obtain above parametric solution.


   
    Case 2:  x^4 + y^4 + z^4 = w^2(New solution)
    
           
             x = -4m^4n^12 + 128m^6n^10 + 6m^8n^8 + 128m^10n^6 - 4m^12n^4 + n^16 + m^16

             y = 4(-m^2 + n^2)(m^2 + n^2)(n^10 - m^2n^8 + 14m^4n^6 - 14m^6n^4 + m^8n^2 - m^10)mn

             z = -8m^2n^2(m^2 + n^2)(n^10 - m^2n^8 + 14m^4n^6 - 14m^6n^4 + m^8n^2 - m^10)

             w = n^32 + 120m^4n^28 - 256m^6n^26 + 2332m^8n^24 + 768m^10n^22 + 5960m^12n^20 - 512m^14n^18 + 48710m^16n^16
                -512m^18n^14 + 5960m^20n^12 + 768m^22n^10 + 2332m^24n^8 - 256m^26n^6 + 120m^28n^4 + m^32
       


            (x0x+1)^4+(y0x)^4+(z0x)^4=(w0x^2+kx+1)^2........................(1)

            Let (x0,y0,z0,w0)=(2mn(m^2-n^2), (m^2-n^2)(m^2+n^2), 2mn(m^2+n^2), m^8+14m^4n^4+n^8) 
       
            k =-16m^3n^3(-3m^2n^4+n^6+3m^4n^2-m^6)/(m^8+14m^4n^4+n^8)

            x = -4(n^10-m^2n^8+14m^4n^6-14m^6n^4+m^8n^2-m^10)mn
                  /(n^16-8m^2n^14+12m^4n^12+8m^6n^10+230m^8n^8+8m^10n^6+12m^12n^4-8m^14n^2+m^16)

            Substitute k and x to (1), hence we obtain above parametric solution.






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