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: $$f(r, s, t)=(a,b,c,d)$$

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    $\begingroup$ as I said on your earlier question, there are usually infinite families of solutions of homogeneous polynomial equations such as this. As the other answer (Tito) showed, one may sometimes parametrize all rational solutions. The trickier thing is to find all integer primitive solutions; here, $\gcd(a,b,c,d) = 1,$ and be able to prove that you really have described all primitive integer solutions. $\endgroup$ – Will Jagy Oct 20 '16 at 2:13
  • $\begingroup$ Does this come from the question of "congruent numbers"? I think I've seen that term $ab(a^2-b^2)$ in that context. $\endgroup$ – Gottfried Helms Oct 20 '16 at 9:04
  • $\begingroup$ @GottfriedHelms Yah. Looking for a surjective parametrization of equal congrua (plural of congruum or "congruent number"). The smallest example is a=5, b=2, c=6, d=1 for the congruum 840. $\endgroup$ – Christian Woll Oct 20 '16 at 15:41
  • $\begingroup$ Christian - I see (I've just fiddled with this myself) Isn't that problem of congruent numbers one of the "open problems"? If so, then I suspect, that it shall be at least very difficult to make progress in your question... (Maybe I'm wrong - I'm not expert in this, just screened some small articles and essays in the last days) $\endgroup$ – Gottfried Helms Oct 20 '16 at 15:46

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





for k=3 we have,


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    $\begingroup$ You could use MathJax to format your formula. $\endgroup$ – Yujie Zha Jun 26 '17 at 4:12

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.

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

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

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