Diophantine equations

Question

Can I rigorously generate all possible solutions to the following system of Diophantine equations:

$x^2+y^2+z^2= a^2 +b^2+c^2+d^2=t^2$

Answer 1

For the system of equations.

$$\left\{\begin{aligned}&X^2+Y^2+Z^2=T^2\\&A^2+B^2+C^2+D^2=T^2\end{aligned}\right.$$

If we do the replacement.

$$p=r^2+n^2+q^2+j^2-i^2-v^2$$

$$k=r^2+n^2+q^2+j^2+i^2-v^2-2(r+n+q-v)i$$

$$s=r^2+n^2+q^2+j^2+v^2-i^2-2(r+n+q-i)v$$

$$b=2(i+v-r-n-q)j$$

$$t=r^2+i^2+v^2-n^2-q^2-j^2+2(n+q-i-v)r$$

$$c=q^2+i^2+v^2-r^2-n^2-j^2+2(n+r-i-v)q$$

$$a=n^2+i^2+v^2-r^2-q^2-j^2+2(q+r-i-v)n$$

Then the solution will have the form:

$$X=2ps$$

$$Y=2ks$$

$$Z=p^2+k^2-s^2$$

$$T=p^2+k^2+s^2=t^2+c^2+a^2+b^2$$

$$A=2at$$

$$B=2bt$$

$$C=2ct$$

$$D=a^2+b^2+c^2-t^2$$

Answer 2

Answer in progress

---

Let's first consider solving the equation $$x^2+y^2+z^2=t^2.$$ Let $0\le x \le y \le z \le t$. Let us introduce dummy variables $p$ and $q$ and substitute them as follows: $$(x-p)^2+(y+p)^2+(z+q)^2=(t+q)^2.$$ Substituting two dummy variables allows us to express one of the variables in terms of the other, when expanding and simplifying with the condition that $x^2+y^2+z^2=t^2$. Given the inequalities, this particular substitution will guarantee small sizes of $p$ and $q$ terms since a lot will cancel out. Expand this and we find that it is true if: $p^2-p(x-y)\pm q(z\mp t)=0$ Now let us make this simpler by substituting $q=p^2+p$. This allows us to cancel out a $p$ factor and reduce to a first-degree polynomial. This also allows to connect a coefficient to a $p$ value so if we want to set the left hand side to $0$, it is sufficient that the coefficient is $0$ where the coefficient is not in terms of $p$. It is also sufficient to consider just one case of $(\pm, \mp)$, so let's pick the $(+,-)$. We now have: $$-x+y+z-t + p(1+z-t)=0$$ where $1+z-t$ is the coefficient. Let $t=1$ and $z=0$. So $-x+y=1$. Now let $y=1$ and $x=0$ to achieve this condition. Substitute it all back into the original equation, rewrite $(-p)^2=p^2$ and voila: $$p^2+(p+1)^2+(p^2+p)^2=(p^2+p+1)^2$$ Now substitute $p$ with $p/q$ and multiply through by $q^4$, so we have: $$(pq)^2+(pq+q^2)^2+(p^2+pq)^2=(p^2+pq+q^2)^2$$

---

Fun curio, this identity can be further generalised as follows:

If $ac-bd=1$ then for all $p$ and $q$, $$\big(p^2(a^2+ab+b^2)+pq+q^2(c^2+cd+d^2)\big)^2=$$ $$\big(ap^2(a+b)+pq+dq(c+d)\big)^2+$$ $$\big(bp^2(a+b)+pq+cp^2(c+d)\big)^2+$$ $$(abp^2-pq+cdq^2)^2$$

If you want, see if you can find this generalisation from the one above, which is just the special case $(a,b,c,d)=(1,0,1,0)$. Anyways, back to topic.

---

Let us now use the Brahmagupta-Fibonacci identity to finish this off. We can use the famous Pythagorean identity: $$(x^2-y^2)^2+(2xy)^2=(x^2+y^2)^2$$ and BF: $$(m a^2+ n b^2)(m c^2 + n d^2) = (m a c\pm n b d)^2 + m n (a d\mp bc)^2$$ Let $m=1$ and $n=1$. Then substitute $a\mapsto a^2-b^2$, $b\mapsto 2ab$, $c\mapsto c^2-d^2$ and $d\mapsto 2cd$. By Pythagoras's identity, this will transform the BF equality to expressing $(a^2+b^2)^2(c^2+d^2)^2$ as the sum of two squares. Finally, substitute $p=a^2+b^2$ and $q=c^2+d^2$ to effectively write $(pq)^2$ as the sum of two squares. The next part we require is then writing $$(p^2+pq)^2+(q^2+pq)^2=(p+q)^2(p^2+q^2)$$ as a sum of two different non-trivial squares, but I'll work on this part when I have more time.