<s>[INST] $a^2-b^2$ will be a perfect square if the ratio of $a$ and $b$ is 
$a^2-b^2$ will be a perfect square if the ratio of $a$ and $b$ is

Options are: A) $1:2$,  B) $3:4$, C) $5:4$, D) $4:5$, E) None of these 
So my first observation was that $|a|$ has to be greater than $|b|$. So only options that suffice this condition are $C)$ and $E)$. Also, if $C)$ was correct, then we get $a=5k$; $b=4k$, $25k^2-16k^2=9k^2$. Which is giving us a perfect square. But isn't there any rigorous way also?
 [/INST] A) $1^2-2^2=-3$,
B) $3^2-4^2=-7$,
C) $5^2-4^2=9$,
D) $4^2-5^2=-9$.
This rules out A), B) and D).
C) implies $(5k)^2-(4k)^2=(3k)^2$, so that the property holds for all $(a,b)=k(5,4)$.

If you don't want to use the multiple choices, let
$$a^2-b^2=c^2.$$
You can factor as
$$(a+b)(a-b)=c^2=pq$$
and solve with
$$a=\frac{p+q}2,b=\frac{p-q}2.$$
The ratio is 
$$\frac ab=\frac{p+q}{p-q},$$ where $pq$ is a perfect square.
For instance, $p=c^2,q=1$ for any $c$ gives
$$\frac ab=\frac{c^2+1}{c^2-1}$$among which
$$\frac ab=\frac{9+1}{9-1}=\frac54.$$
And there are other possibilities...
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