I am trying to understand the proof in Carmichaels book Diophantine Analysis but I have got stuck at one point in the proof where $w_1$ and $w_2$ are introduced.
The theorem it is proving is that the system of diophantine equations:
- $x^2 + y^2 = z^2$
- $y^2 + z^2 = t^2$
cannot simultaneously be satisfied.
The system is algebraically seen equivalent to
- $t^2 + x^2 = 2z^2$
- $t^2 - x^2 = 2y^2$
and this is what will be worked on. We are just considering the case where the numbers are pairwise relatively prime. That implies that $t,x$ are both odd (they cannot be both even). Furthermore $t > x$ so define $t = x + 2 \alpha$.
Clearly the first equation $(x + 2\alpha)^2 + x^2 = 2 z^2$ is equivalent to $(x + \alpha)^2 + \alpha^2 = z^2$ so by the characterization of primitive Pythagorean triples there exist relatively prime $m,n$ such that $\{x+\alpha,\alpha\} = \{2mn,m^2-n^2\}.$
Now the second equation $t^2 - x^2 = 4 \alpha (x + \alpha) = 8 m n (m^2 - n^2) = 2 y^2$ tells us that $y^2 = 2^2 m n (m^2 - n^2)$ by coprimality and unique factorization it follows that each of those terms are squares so define $u^2 = m$, $v^2 = n$ and $w^2 = m^2 - n^2 = (u^2 - v^2)(u^2 + v^2)$.
It is now said that from the previous equation either
- $u^2 + v^2 = 2 {w_1}^2$, $u^2 - v^2 = 2 {w_2}^2$
or
- $u^2 + v^2 = w_1^2$, $u^2 - v^2 = w_2^2$
but $w_1$ and $w_2$ have not been defined and I cannot figure out what they are supposed to be. Any ideas what this last part could mean?
For completeness, if the first case occurs we have our descent and if the second case occurs $w_1^2 + w_2^2 = 2 u^2$, $w_1^2 - w_2^2 = 2 v^2$ gives the descent. Which finishes the proof.