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In the world of Diophantine analysis, there are single equation problems (e.g., “Find all solutions $(x,y)$ such that $x^3=y^2+1$”), double equation problems (e.g., “Find all numbers $p$ such that $p+1$ and $p^2+1$ are both the double of a square”), and problems involving three or more equations which share one or more of the variables.

What, if any, are the principal methods of [non-trivially] “upgrading” one type to a “higher” type? Can every single equation problem be [non-trivally] turned into a multiple-equation problem?

Note: By “non-trivially”, I mean not just rational transformations like writing an odd number $x$ as $2u+1$. Rather, I mean taking a single equation which [apparently] provides 'exactly $N$' units of information about the numbers/variables involved and turning it into two or more equations where the total amount of information is greater than '$N$'.

Kieren MacMillan
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For some single equations, you can in fact split it and solve it as an intersection of two quadric surfaces. So you now have two equations to solve. But the upside is if there is a rational point, then there may be infinitely many.

This can be done for,

$$a^4+b^4+c^4 = d^4\tag1$$

as Elkies did in this post. For example, express the above as,

$$(p + r)^4 + (p - r)^4 + s^4 = q^4\tag2$$ $$2p^4 + 12p^2r^2 + 2r^4 + s^4 = q^4\tag{2b}$$

Then it can be split into two quadrics with extra parameter $m$,

$$-(3 m^2 - 8m + 6) p^2 + 2 (m^2 - 2) p q - 2 m q^2 = (m^2 + 2) r^2\tag3$$ $$ -4 (m^2 - 2) p^2 + 8 m p q + (m^2 - 2) q^2 = (m^2 + 2) s^2\tag4$$

To see the intersection of $(3),(4)$, simply eliminate $m$ between the two using resultants, then factor and one recovers $(2b)$.

Similarly, it can also be done for,

$$a^4+b^4+c^4+d^4 = (a+b+c+d)^4$$

as Jacobi and Madden did here.

Or for,

$$a^6+b^6+c^6 = d^2$$

as Bremner and Ulas showed here. But the trick is to find those quadric surfaces which is no easy feat.

Tito Piezas III
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