<s>[INST] Solving for an unknown $c$ in relation to a ratio of gamma functions I have been working with ratios of gamma functions and I am surprised how difficult it is to make even elementary conclusions.  I am hoping it is just the learning curve.
Consider the following problem.  Find a positive real $c$ for an integer $x > 1$ such that:
$$\frac{\Gamma(2x+3-c)}{\Gamma(2x+1)} = x^2$$
Here's my reasoning for why I am confident that a solution exists for each integer $x$.
$\dfrac{\Gamma(2x+3)}{\Gamma(2x+1)} = \dfrac{(2x+2)!}{(2x)!}=(2x+2)(2x+1) = 4x^2 + 6x + 2 > x^2$
I am completely at a loss how to tackle what appears to me to be such a simple use of the Gamma function.
I would greatly appreciate if someone can either solve this problem for some $x > 1$ or help me to understand the methods that could be applied to this problem to provide an estimate for $c$ with, ideally, an upper and lower bound.
For the estimate, I am looking for something more interesting than the trivial:
$$0 < c < 1$$
 [/INST] One may use Newton's method to solve $\ln \Gamma(2x+3-c) = 2\ln x + \ln \Gamma(2x+1)$ to find $c$ (see [2]): 
Choose the initial $c_0$,
$$c_{k+1} = c_k - \frac{\ln \Gamma(2x+3-c_k) - 2\ln x - \ln \Gamma(2x+1)}{-\Psi(2x+3-c_k)}, \ k=0, 1, 2, \cdots$$
where $\Psi(x) = (\ln \Gamma(x))'$ is the digamma function.
(Remark: $c \mapsto \ln \Gamma(2x+3-c)$ is a convex function.)
For bounds, one may use good bounds for the gamma function. For example, in the following, we give some simple bounds.
First, we deal with the case when $x > 2$. Clearly, $0 < c < 1$.
The equation is written as
$$\frac{\Gamma(2x+3-c)}{\Gamma(2x+3)} = \frac{x^2}{(2x+1)(2x+2)}$$
or
$$\frac{\Gamma(2x+3)}{\Gamma(2x+3-c)} = \frac{(2x+1)(2x+2)}{x^2}. \tag{1}$$
Recall Gautschi's inequality [1]: for $y > 0$ and $s \in (0, 1)$,
$$y^{1-s} < \frac{\Gamma(y+1)}{\Gamma(y+s)} < (y+1)^{1-s}.$$
By letting $y = 2x + 2$ and $s = 1-c$ in Gautschi's inequality, we have
$$(2x+2)^c < \frac{\Gamma(2x+3)}{\Gamma(2x+3-c)} < (2x+3)^c.\tag{2}$$
From (1) and (2), we have
$$(2x+2)^c < \frac{(2x+1)(2x+2)}{x^2} < (2x+3)^c$$
which gives
$$\frac{\ln (2x+1) + \ln (2x+2) - 2\ln x}{\ln (2x+3)}  < c < \frac{\ln (2x+1) + \ln (2x+2) - 2\ln x}{\ln (2x+2)}.$$
Second, we deal with the case when $x = 2$. Clearly, $1 < c < 2$.
The equation is written as
$$\frac{\Gamma(2x+3-c)}{\Gamma(2x+2)} = \frac{x^2}{2x+1}$$
or
$$\frac{\Gamma(2x+2)}{\Gamma(2x+3-c)} = \frac{2x+1}{x^2}. \tag{3}$$
Using Gautschi's inequality, we get 
$$1 + \frac{\ln 5 - 2\ln 2}{\ln 6} < c < 2 - \frac{2\ln 2}{\ln 5}.$$
Reference
[1] https://en.wikipedia.org/wiki/Gautschi%27s_inequality
[2] Folitse Komla Amenyou, "Properties and Computation of the Inverse of the Gamma function".
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