(m^4+1)X^4 + (m^4+1)Y^4 = 2Z^4 + 2W^4

Existence of solution for diophantine equation ax^4 + by^4 + cz^4 + dw^4 = 0 are known if abcd is square number.
Furthermore, condition of a + b + c + d = 0 or having one known solution is necessary to have a rational solution.
Anyway, we show some parametric solutions of ax^4 + by^4 + cz^4 + dw^4 = 0 with abcd is square number.

General solutions of ax^4 + by^4 + cz^4 + dw^4 = 0 are given below.
ax^4 + by^4 + cz^4 + dw^4 = 0
ax^4 + by^4 + cz^4 + dw^4 = 0 Part 2

Parametric solutions of x^4 + hy^4 = z^4 + ht^4 are given below.
x^4 + hy^4 = z^4 + ht^4
Many numeric solutions of x^4 + hy^4 = z^4 + ht^4 are given below.
x^4 + hy^4 = z^4 + ht^4

We show diophantine equation (m^4+1)X^4 + (m^4+1)Y^4 = 2Z^4 + 2W^4 has infinitely many integer solutions.
m is arbitrary.

This equation is related to (m^4+n^4)X^4 + (m^4+n^4)Y^4 = 2Z^4 + 2W^4.
m,n are arbitrary.


(m^4+1)X^4 + (m^4+1)Y^4 = 2Z^4 + 2W^4.......................................................(1)

We use an identity p(t+1)^4+p(t)^4=(t^2+at+b)^2+(ct^2+dt+e)^2,..............................(2)
with (a,b,c,e,p)=(d+2, 1/2d+1, 1+d, 1/2d, 1+d+1/2d^2).

So, we look for the integer solutions {Z^2 = t^2+(d+2)t+1/2d+1, W^2 = (1+d)t^2+dt+1/2d}.....(3)

By parameterizing the first equation and substituting the result to second equation, then we obtain quartic equation below.

Let d=m^2-1, then
u^2 = k^4m^2+4k^3-2k^2m^2+(-8m^4-4)k+4m^6+5m^2..............................................(4)

This quartic equation is birationally equivalent to an elliptic curve below.

Y^2-4YX/m+(8m+16m^5)Y = X^3-2(m^4+2)X^2/(m^2)+(-20m^4-16m^8)X+104m^6+32m^10+80m^2

It has a point P(X,Y)=( 2(m^4+2)/m^2, -16(-1+m^8)/m^3 ).

Hence we get 2P(X,Y)=( (m^8+4m^4+1)/(m^2), -(3m^8-m^4-3+m^12)/(m^3) ).

This point P is of infinite order, and the multiples nP, n = 2, 3, ...give infinitely many points.

This quartic equation has infinitely many parametric solutions below.

n=2:
X = m^4+4m^2-1
Y = m^4-4m^2-1
Z = 3m^4+1
W = (m^4+3)m

n=3:
X = m^12+12m^10-19m^8+40m^6+19m^4+12m^2-1
Y = m^12-12m^10-19m^8-40m^6+19m^4-12m^2-1
Z = 5m^12-27m^8-41m^4-1
W = m(m^12+41m^8+27m^4-5)
.
.
etc.

Hence we can obtain infinitely many integer solutions for equation (1).

Example:

m=2: 17X^4 + 17Y^4 = 2Z^4 + 2W^4:  [X,Y,Z,W]=[31, 1, 49, 38],[14431, 15361, 12911, 30038],...etc.