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

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

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

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

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

Let c=m^2, then
u^2 = m^2(-d+m^2)^2k^4+(4m^6-12dm^4+12m^2d^2-4d^3)k^3+(4dm^4-2m^2d^2-2m^6)k^2+(-12m^6+20dm^4-12m^2d^2+4d^3)k+5m^2d^2+9m^6-10dm^4............(4)

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

Y^2-4(-2dm^2+d^2+m^4)YX/m+(24m^9-64dm^7+64m^5d^2-32m^3d^3+8md^4)Y = X^3-2(2d^4-8m^2d^3+13m^4d^2-10dm^6+3m^8)X^2/(m^2)+(80m^6d^3-20m^4d^4-136m^8d^2+112m^10d-36m^12)X+80m^2d^8-640d^7m^4+2344d^6m^6-5104d^5m^8+7192d^4m^10-6688d^3m^12+3992d^2m^14-1392dm^16+216m^18
It has a point P(X,Y)=( 2(2d^4-8m^2d^3+13m^4d^2-10dm^6+3m^8)/(m^2), 16(d^5-6m^2d^4+15m^4d^3-20m^6d^2+14m^8d-4m^10)d/(m^3) ).

Hence we get 2P(X,Y)=( (d^4-4m^2d^3+10m^4d^2-12dm^6+6m^8)/(m^2), d(3d^5-18m^2d^4+46m^4d^3-64m^6d^2+48m^8d-16m^10)/(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 = 4m^4-2dm^2-d^2
Y = 4m^4-6dm^2+d^2
Z = 4m^4-2dm^2+d^2
W = m(4m^4-6dm^2+3d^2)

n=3:
X = 64m^12-224m^10d+320m^8d^2-216m^6d^3+64m^4d^4-6d^5m^2-d^6
Y = 64m^12-160m^10d+160m^8d^2-104m^6d^3+56m^4d^4-18d^5m^2+d^6
Z = 64m^12-224dm^10+288d^2m^8-184d^3m^6+56d^4m^4-6d^5m^2+d^6
W = m(64m^12-160dm^10+128d^2m^8-8d^3m^6-48d^4m^4+30d^5m^2-5d^6)
.
.
etc.

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

Example:

(d,m)=(2,2): 10X^4 + 10Y^4 = 4Z^4 + W^4:  [X,Y,Z,W]=[11, 5, 13, 14], [563, 893, 277, 1646],...etc.






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