When will a parametric solution exist for a Diophantine equation?

Question

Many Diophantine equations with infinite solutions I've seen have parametric solution. Example: $$a=m^2-n^2$$ $$b=2mn$$ $$c=m^2+n^2$$ Implies: $$a^2+b^2=c^2$$ So pythagorean triples can be generated with arbitrary $$m>n>0$$ I specifically have in mind this equation: $$ab(a+b)(a-b)=cd(c+d)(c-d)$$ So does there exist some f such that: $$f(r, s, t)=(a,b,c,d)$$

Answer 1

Equation, ab(a+b)(a-b)=cd(c+d)(c-d) has parametrization and is given below,

a=(k)(5k-1)

b=(3k-1)(7k-2)

c=(2k-1)(3k-1)

d=(5k-1)(4k-1)

for k=3 we have,

(a,b,c,d)=(21,76,20,77)

Answer 2

Your equation can be expressed in the form, $$ab(a^2-b^2)=cd(c^2-d^2)\tag1$$

More generally, $$ab(a^2+hb^2)=cd(c^2+hd^2)\tag2$$

This is the expanded version of the well-known,

$$(a + b \sqrt{h})^4 + (c - d \sqrt{h})^4 = (a - b \sqrt{h})^4 + (c + d \sqrt{h})^4\tag3$$

Yours was the case $\color{blue}{h=-1}$ so, in effect, you are looking for solutions in the complex numbers to $(3)$. We borrow the method used by Euler. Let, $$a,\;b,\;c,\;d = x,\; p y,\; q x,\; y$$

to get, $$(p - q^3) x^2 = -h (p^3 - q) y^2$$

implying $$(p - q^3)(p^3 - q) =-h\,z^2\tag4$$

This is an elliptic curve and if you find one rational point, you can find infinitely many.

I. For $h=1$, so $(3)$ is in the integers:

$$q = p + \frac{3p (1 - p^2)^3}{4p^6 + p^4 + 10p^2 + 1}$$

II. For $h=-1$, so $(3)$ is in the Gaussian integers:

$$q = -p - \frac{3p (1 + p^2)^3}{-4p^6 + p^4 - 10p^2 + 1}$$

From these initial formulas, one can then find infinitely more.

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Example. Let $h=-1$ and $p=2$, then, $$a,\;b,\;c,\;d = 186,\; 178,\; 128,\; 89$$

which solves your $(1)$ and implying, $$(186 + 178\,i)^4 + (128 - 89 \,i)^4 = (186 - 178 \,i)^4 + (128 + 89 \,i)^4$$