Using quadratic residues and/or reciprocity to prove relative primality?

Question

I have odd positive integers $q$ and $y$, with $3q^2 < y^2 < 4q^2$, such that the following are true:

\begin{align} (q^2+9) &\mid (y^2+5)(y^2+29) \\[0.25em] (q^2+2) &\mid (y^2+1)(y^2+5) \\[0.25em] (y^2+29) &\mid (q^2+9)(q^2+50). \end{align}

I’m trying to prove that $\gcd(q^2+9,y^2+29)=2$, so that $(q^2+9) \mid (y^2+5)$.

QUESTION #1: What can be done with what I have to obtain the conclusion I’m looking for?

QUESTION #2: What other information about $q$ and/or $y$ would be enough to reach the desired conclusion?

Answer 1

from your original quartic relation, we take new variables as the squares of yours, so the new relation is now quadratic. Your problem reduces to showing that the largest square in https://oeis.org/A001653 is 169, and this was done.

Apparently Ljunggren shows that 169 is the last square term.

Should be in Mordell's book (1969). YES, page 271

Note how the right hand column below alternates squares and double squares. That relation is $v_{j+2} = 6 v_{j+1} - v_j + 2.$ The left column is squares only for $1,169,$ that relation is $u_{j+2} = 6 u_{j+1} - u_j$ Let's see, the strict repetition of each value is what happens in "Vieta Jumping" when symmetry is weakened. We can draw a stairway between integral points in the first quadrant, start at 1,0 then up then right then up then right

Tue Jun 23 12:24:11 PDT 2020

      1      0
      1      1
      5      1
      5      8
     29      8
     29     49
    169     49
    169    288
    985    288
    985   1681
   5741   1681
   5741   9800
  33461   9800
Tue Jun 23 12:24:26 PDT 2020