Write the Pythagorean equation $a^2 + b^2 = c^2$ under the form $N(z) =c^2$ , where $z = a + ib$ in $Z[i]$ and N denotes the norm map of $Q(i)/Q$ . Since the norm is multiplicative, it is sufficient to solve the same problem where c is replaced by a prime p factor of c. In your example(s), you take such a p to be congruent to 1 mod 4, so that it is totally decomposed in $Z[i]$, and the usual proof shows that $p$ (hence $p^2$) is indeed a norm. Process as you did, and we are through.
Let me propose another way to find all the "$c$ stuck triples", as you call them, by exploiting more heartily the arithmetic of the Gaussian integers. The parametrization of the integer triples verifying $a^2 + b^2 = c^2$ in $Z$ is classically known, but here is a « Galois solution ». Let us begin with $N(z) = 1$, with $z$ in $Q(i)$ . An element $X + iY$ in $Q(i)$ has norm 1 iff it has the form $(x +iy)/(x-iy)$ (Hilbert 90 !), equivalently iff $X = (x^2 - y^2)/(x^2 + y^2) , Y = 2xy/(x^2 + y^2)$. Since these expressions are homogeneous in x, y, we can take x, y in $Z$. Coming back to the original Pythagorean equation, we get the usual parametrization of the Pythagorean triples : $a = m^2 – n^2 , b = 2mn , c = m^2 + n^2 $. If we fix c, all the « c stuck triples » (a,b,c) will be given as above, starting from (m,n) such that $c = m^2 + n^2$ . More precisely, the quotient $(x+iy)/(x-iy)$ as above, with $x^2 + y^2 = c$, will yield all the "c stuck triples". These are parametrized as follows : $x + iy = t. (m + in)$, with $N(t) = 1$.
EDIT Particular case : if $t$ is in $Z [i]$, then $t$ is a unit (invertible), hence a power of $i$, and we recover the « manipulations » in your example(s). General case : as before, $t = \bar s /s$, with $s$ in $Z[i]$, hence $s. (x + iy) = \bar s . (m + in) (*)$, which means that $s$ divides $\bar s . (m + in)$ in $Z[i]$. Let us consider the prime factors $\pi$ of $s$. Denoting by $p$ the prime number under $\pi$, ramification theory in quadratic fields distinguishes three cases : (1) $p$ is inert ; (2) $p$ is totally ramified ; (3) $p$ is totally split (this corresponds to $p\equiv 1\pmod 4$). In the first two cases, there is only one prime ideal over $p$, hence $\pi$ and $\bar \pi$ differ by a unit, and we can simplify in $s$ and $\bar s$ . In the third case, $\pi $ and $\bar \pi$ are coprime, hence $\pi$ divides $(m + in)$ (and $p$ divides $c$). Conversely, taking for $s$ such a $\pi$, we get (x + iy) as in (*) . Conclusion : if no prime factor of $c$ is congruent to 1 mod 4, we are in the particular case ; if there is at least one prime factor of $c$ congruent to 1 mod 4, the solutions are more complicated.
