$a^2-b^2$ will be a perfect square if the ratio of $a$ and $b$ is

Question

$a^2-b^2$ 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?

Answer 1

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 $(a,b)=k(5,4)$.

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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...

Answer 2

In a multiple choice question it can often be quicker to answer it via trial and error as you know one of the choices must be correct. If we ignore the multiple choice part then are looking at: $a^2-b^2=x^2$. This is simplify Pythagoras' Theorem rewritten so if you know solutions to that you have additional solutions. A general solution to Pythagoras is for any $a$ and $b$ with $a>b>0$ then you can construct the following three numbers which will satisfy Pythagoras' Theorem:

$$(a^2-b^2)^2+(2ab)^2=(a^2+b^2)^2$$

Each number could also be multiplied by a constant.

Update: To use express these as a ratio rewrite it to the form you needed:

$$(a^2+b^2)^2-(a^2-b^2)^2=(2ab)^2$$ or $$(a^2+b^2)^2-(2ab)^2=(a^2-b^2)^2$$ Hence you want a ratio of: $$a^2+b^2\,:\,a^2-b^2$$ or $$a^2+b^2\,:\,2ab$$ for any integers $a>b>0$.

Note: Your solution has $a=2$ and $b=1$ as is the second form of the ratio.