4(m^2-1)X^4 + (1-m^2)(m^2+3)Y^4 = -(m^2-1)^2Z^4 + 4(m^2+3)W^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.
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 4(m^2-1)X^4 + (1-m^2)(m^2+3)Y^4 = -(m^2-1)^2Z^4 + 4(m^2+3)W^4 has infinitely many integer solutions.
d,m are arbitrary.

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

We use an identity p(t+1)^4+nq(t)^4=np(t^2+at+b)^2+q(ct^2+dt+e)^2,.......................(2)
with (a,b,c,d,e,p)=(2/n, 1/n, 0, 2, 1, qn/(-1+n)).

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

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

u^2 = -2n^2k^3-2(n-2)nk^2+2n^2k+2(n-2)n..................................................(4)

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

Y^2 = X^3-2(n-2)nX^2-4n^4X+8(n-2)n^5

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

Hence we get 2P(X,Y)=( 1/16(m^12+4m^10+7m^8+7m^4-4m^2+1)/(m^2(m^2+1)^2), -1/64(20m^6-16m^4+1-46m^8+m^2-26m^10-8m^12+5m^16+4m^14+m^18)/(m^3(m^2+1)^3) ).

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

This quartic equation has infinitely many parametric solutions below.

h=2:
X = m^6+3m^4-m^2+1
Y = (m^3-m^2+m+1)(m^3+m^2+m-1)
Z = m^6+m^4-5m^2-1
W = 2m(m^2+1)(m-1)(m+1)

h=3:
X = m^24+12m^22+34m^20+12m^18+15m^16+216m^14+188m^12-264m^10-49m^8+28m^6+66m^4-4m^2+1
Y = (m^12-2m^11+4m^10-2m^9+5m^8+12m^7+8m^6+4m^5-5m^4-10m^3+4m^2-2m-1)(m^12+2m^11+4m^10+2m^9+5m^8-12m^7+8m^6-4m^5-5m^4+10m^3+4m^2+2m-1)
Z = m^24+4m^22-14m^20-124m^18-241m^16+8m^14+220m^12+104m^10+335m^8-12m^6-46m^4+20m^2+1
W = 4m(m^6+m^4-5m^2-1)(m^6+3m^4-m^2+1)(m^3+m^2+m-1)(m^3-m^2+m+1)(-1+m)(m+1)(m^2+1)
.
.
etc.

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

Example:

m=2: :  12X^4-21Y^4=-9Z^4+28W^4: [X,Y,Z,W]=[109, 91, 59, 60],[110918161, 85854961, 28202639, 70226520],...etc.