What does it mean when a non-integer rational fraction is a square modulo an integer? Let’s say that $p$ and $q$ are coprime integers such that $p/q$ is a non-integer rational fraction in lowest terms. Now let’s say that $r$ is a positive integer such that $p/q$ is a square modulo $r$.
What does that even mean?
Note: In the best case scenario [vis-à-vis my current mathematical work], it would imply $r=1$. What I have is $7/6$ is a square modulo [an indeterminate/unknown integer] $r$.

EDIT (more details): I have a system of four equations in four
  indeterminate integers $x$, $y$, $p$, and $q$. Manipulating the
  system, I found that $7k^2 \equiv 6m^2\!\pmod{3q-4p}$ for some
  integers $k$ and $m$ [actually functions of the four unknown
  integers]. I'm trying to determine what I can say about $3q-4p$, which
  I'm hoping to prove equals $1$.



EDIT #2 (even more details): I can obtain a fairly large number of such conclusions, e.g. $-14/3$ and $9/2$ and $7/6$ are all squares modulo this indeterminate integer $3q - 4p$. Perhaps there is a way to use two or more of these results to force $3q - 4p = 1$? For example, is it kosher to say
  $$
\frac{-14}{3} \times \frac{9}{2} = -21
$$
  is a square modulo $3q-4p$, because each one is a square modulo $3q-4p$? Better yet, is
  $$
\frac{-14}{3} \times \frac{7}{6} = -\frac{49}{9} = -\biggl(\frac{7}{3}\biggr)^2
$$
  a square modulo $3q-4p$? Because then $-1$ is a square, so all primes dividing $3q-4p$ are of the form $4n+1$, I think?

 A: If $q$ is relatively prime to $r$, then division by $q$ is well-defined mod $r$ and the statement is simply saying that $p/q$ is a square in $\mathbb{Z}/r\mathbb{Z}$.
Probably the most reasonable interpretation of that statement would then say that $p/q$ cannot be a square modulo $r$ when $q$ is not relatively prime to $r$, because as the term "modulo $r$" indicates an element of that ring, and $p/q$ is not an element of that ring.
However, even if $q$ and $r$ are not relatively prime, it's possible there's a more general interpretation, perhaps using $r$-adic numbers. Rather than speculate about this, one would want more context for your statement.
A: If $q$ and $r$ are coprime, division by $q$ is well-defined in the ring $\mathbb Z/r \mathbb Z$ of integers mod $r$, i.e. there is some integer $x$ such that $p \equiv q x \mod r$.  So this is saying that there is such an $x$ that is the square of an integer.
For example,  $7/6 \equiv  5^2 \mod 11$, so $7/6$ is a square $\mod 11$.
The integers $r$ such that $7/6$ is a square mod $r$ are the divisors of $7 - 6 y^2$ for integers $y$.  The first few are $1, 7, 11, 13, 17, 19, 29, 41, 47, 53, 61, 77, 79, 89, 91$.
EDIT: Yes, the squares coprime to $r$ mod $r$ form a subgroup of the multiplicative group of $\mathbb Z/r \mathbb Z$.   $-1$ is a square mod $r$ (and therefore mod any prime factor of $r$) iff all odd prime factors of $r$ are congruent to $1 \mod 4$ and there is at most one factor of $2$: see OEIS sequence A008784.  However, be careful: $-14/3$, $7/6$, and $9/2$ are all squares mod $7$ but $-1$ is not.
A: We use fraction notation $\,x = a/b\,$ only when there exists a unique solution $\,x\,$ to  $\,b x  = a.\,$ 
Modulo $\,n,\,$ this is true iff $\,b\,$ is coprime to $\,n\,$ since then by Bezout $\, bj+nk = 1\,$ for some integers $\,j,k.\,$ Thus reducing this equation modulo $\,n\,$ we obtain $\,bj\equiv 1\pmod{n},\,$ i.e. $\, j\equiv b^{-1}.\,$ Scaling $\, bx\equiv a\,$ by $\,b^{-1}$ we obtain the unique solution $\,x\equiv ab^{-1},\,$ i.e. $\,a/b \equiv ab^{-1}.$
Conversely if $\,b\,$ is not coprime to $\,n,\,$ a unique solution of $\,bx\equiv a\,$ need not exist, i.e. there can be no solutions, or more than one solution, so $\,x \equiv a/b\pmod n\,$ is not well-defined.  
For example, mod $\rm 10,$ $\rm\:4\,x\equiv 2\:$ has solutions $\rm\:x\equiv 3,8,\:$ so the "fraction" $\rm\:x \equiv 2/4\pmod{10}\,$ cannot designate a unique solution of $\,4x\equiv 2.\,$ Indeed, the solution is $\rm\:x\equiv 1/2\equiv 3\pmod 5,\,$ which requires canceling $\,2\,$ from the modulus too, because $\rm\:10\mid 4x-2\iff5\mid 2x-1.\,$ 
Further $\,x \equiv 1/2\pmod{10}\,$ has no solution since $\,10\mid 2x-1\,\Rightarrow 10n = 2x-1\,$ hence $\,1 = 2x-10n\,$ is even, contradiction. See here for more.
Thus, when the fraction is well-defined, your question reduces to asking about when integers are squares mod $\,n,\,$ a classical question that is solvable by quadratic reciprocity and a generalized Euler criterion.
