In a proof I‘m working on, I’m trying to characterize a prime $\phi$ which divides three numbers: one of the form $2x^2+1$, one of the form $2y^4+1$, and one of the form $32z^4+1$. Using various divisibility and congruence arguments, I have it down to
$$\phi \mid 3 \cdot 11 \cdot 641 \cdot 4241 \cdot 18059.$$
The answer I‘m ultimately trying to prove is $32z^4+1=33$ (i.e., $z=1$).
Are there any congruence or reciprocity arguments that would eliminate $641$, $4241$, or $18059$? A brute force search [of relatively small numbers] suggests that $641$ and $4241$ might not divide numbers of the form $32z^4+1$, but that could easily be shown false by searching on larger $z$.
EDIT: Here are some more details.
I have two positive integers, $p$ and $y$, which satisfy $(2p^2+1) \mid (2y^4+1)$. I have two other positive integers, $n$ and $m$, and I can characterize both $p$ and $y$ in terms of $n$ and $m$. I’ve found that if a prime $\phi \mid \gcd(2p^2+1,2y^4+1)$, then $\phi \mid (32m^4+1)$. Using that fact and other relationships I have between the various numbers involved, I’ve reduced the possibilities for $\phi$ to $3,11,641,4241,18059$. I want to prove $32m^4+1=33$, i.e., $m=1$. I’m hoping I can eliminate some or all of the possibilities for $\phi$ other than $3$ and $11$, and then use power arguments to find $32m^4+1=33$.
EDIT #2: As it turns out, I was able to take another path and prove $\phi \mid 3 \cdot 11 \cdot 257 \cdot 1171$, so I don’t need the answer to this question in order to move forward on my work. But I’m leaving it up/open, because I am interested in hearing comments and answers!