<s>[INST] Find the sum of all positive integers $n$ such that $\frac{n+7}{25n+7}$ is kingly 
Say a rational number is $kingly$ if it is the square of another rational number. Find the sum of all positive integers $n$ such that $\frac{n+7}{25n+7}$ is kingly.

If $\dfrac{n+7}{25n+7}$ is kingly, then we can write $\dfrac{n+7}{25n+7} = \dfrac{p^2}{q^2}$ for some $p,q \in \mathbb{Z}^+$ and $\gcd(p,q) = 1$. Rearranging the equation gives $$q^2(n+7) = p^2(25n+7).$$ 
How can I continue?
 [/INST] Option 1: $n+7$ and $25n+7$ are both perfect squares
Not possible: 
All perfect squares are equivalent to $0,1$ or $4$ modulo $5$
$25n+7\equiv 2\pmod 5$
Option 2: $(n+7)$ divides $(n+25)$ and the ratio is a perfect square.
$\lim_\limits {n\to \infty} \frac {n+7}{25n+7}= \frac 1{25}$
There are 3 possible values that $\frac {n+7}{25n+7}$ could take on where the ratio is a perfect square. 
$\frac {n+7}{25n+7} = \frac 14 \implies n =1$
$\frac {n+7}{25n+7} = \frac 1{9} \implies 16 n = 56$ and $n$ is not an integer.
$\frac {n+7}{25n+7} = \frac 1{16} \implies 9 n = 105$ and $n$ is not an integer.
Indeed I have overlooked something.
$n = 20 \implies \frac {27}{507} = \frac {9}{169}$
$p^2(25n+7) = q^2(n+7)\\
(25p^2 - q^2) n = 7(q^2-p^2)$
$(5p+q) | 7(q+p)(q-p)$ and $5p-q |7(q+p)(q-p)$
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