This particular problem is more a brainteaser than mathematics.
A little mathematical insight will tell you the bear's walk must
have started at the North Pole. But that's all the math there is in the
problem.
There is no mathematical reason why the bear must be white;
someone could have airlifted a grizzly bear to the North Pole and
let it loose to walk this path.
But since the color of a bear whose walk starts at the North Pole
is the "brain teaser" part of the problem,
you are intended to think through the "most plausible"
events that could have occurred, and the idea that the bear was one that
was likely to be found naturally near the North Pole seems to
be a better fit to the story than the possibility of any other
kind of bear appearing there.
These reasons why the color of the bear is determined (or not)
involve kinds of real-world reasoning that are sometimes called
"using common sense" or "exercising judgment"
and are notoriously difficult to express in mathematical logic.
The difficulty of this sort of thing has fueled a lot of the research
that was done in artificial intelligence over the last sixty years or so.
As far as I'm aware there is still no general solution to the problem.
Now let's consider questions in which
the options to be chosen among are susceptible to mathematical logic
(unlike the color of a bear found at the North Pole).
First, consider your example:
If $a<b$, is $a^2 < b^2$ or is $b^2 < a^2$?
A somewhat more general format of this question, allowing for any finite
number of unknown quantities and any finite number of choices (two or more), is,
"Given $P(x_1, \ldots, x_n)$, which is true: $Q_1(x_1, \ldots, x_n)$, $Q_2(x_1, \ldots, x_n),\ldots$,
$Q_{k-1}(x_1, \ldots, x_n)$, or $Q_k(x_1, \ldots, x_n)$?"
It seems to me you also mean to require the following two conditions:
The formulas $Q_1,\ldots,Q_k$ are mutually exclusive, that is,
no two formulas $Q_i(x_1, \ldots, x_n)$ and $Q_j(x_1, \ldots, x_n)$,
where $i\neq j$, can be true simultaneously for any choice of values of
$x_1, \ldots, x_n$ such that $P(x_1, \ldots, x_n)$ is true.
$Q_1,\ldots,Q_k$ are exhaustive, that is,
for every choice of values of $x_1, \ldots, x_n$,
if $P(x_1, \ldots, x_n)$ then
at least one of the formulas
$Q_1(x_1, \ldots, x_n),\ldots$, $Q_k(x_1, \ldots, x_n)$ is true.
This format assumes an implicit quantification over the variables
$x_1, \ldots, x_n$, namely, $Q(x_1, \ldots, x_n)$, and
an answer to the question (if there is one)
depends on which one of the $k$ following statements is true:
\begin{align}
(\forall (x_1, \ldots, x_n))
& (P(x_1, \ldots, x_n)\implies Q_1(x_1, \ldots, x_n)), \tag{$\phi_1$}\\
& \vdots \\
(\forall (x_1, \ldots, x_n))
& (P(x_1, \ldots, x_n)\implies Q_k(x_1, \ldots, x_n)). \tag{$\phi_k$}
\end{align}
For $i=1\ldots,k$, if $\phi_i$ is true then the answer is $Q_i$.
If no $\phi_i$ is true then the problem is indeterminate.
The example above also assumes (somewhat ambiguously) a domain
for the variables; in context, the domain would usually be whatever
things we had most recently defined that could be compared using $<$,
whose squares existed, and whose squares could be compared using $<$.
Out of context, we have to choose a suitable domain arbitrarily,
so I'll choose the integers.
(Like the color of the bear, that particular choice is an exercise
in judgment.)
This leads to consideration of the following two statements:
\begin{align}
\newcommand{Z}{\in \mathbb Z}
(\forall (a,b)) & ((a \Z \land b \Z \land a < b) \implies a^2 < b^2)
\tag{$\phi_1$}\\
(\forall (a,b)) & ((a \Z \land b \Z \land a < b) \implies b^2 < a^2)
\tag{$\phi_2$}\\
\end{align}
In this particular example, it's easy enough to show
that the statements $a^2 < b^2$ and $b^2 < a^2$ are mutually exclusive
(because $<$ is antisymmetric),
and not much harder to show that these statements are exhaustive
(because if $x$ and $y$ are integers, one of $x<y$, $x=y$, or $x>y$
must be true, and the condition $a<b$ rules out the case $a^2=b^2$),
so this is indeed a problem in the desired format.
To show that the answer to the problem is indeterminate,
we have to disprove both $\phi_1$ and $\phi_2$.
Disproofs of statements of this form are typically done by
counterexample; for example, a disproof of
$$
(\forall (a,b)) ((a \Z \land b \Z \land a < b) \implies a^2 < b^2) \tag{$\phi_1$}
$$
is just a proof of
$$
\lnot(\forall (a,b)) ((a \Z \land b \Z \land a < b) \implies a^2 < b^2),
\tag{$\lnot\phi_1$}
$$
which is equivalent to
$$
(\exists(a,b)) ((a \Z \land b \Z \land a < b) \land \lnot(a^2 < b^2)).
$$
By far the easiest way to prove this last formula
(and therefore disprove $\phi_1$) is by presenting a "witness"
such as $a=-2$, $b=-1$, which proves the last formula
because $-2\Z$, $-1\Z$, $-2<-1$, and $(-2)^2 \not< (-1)^2$.
Similarly, to disprove $\phi_2$ we can simply present the witness
$a=1$, $b=2$.
For a more general problem of this format, with $k$ alternatives,
you just need $k$ witnesses, not necessarily all different.
The "exercise in judgment" above was that I did not choose $\mathbb N$
as the domain because the idea of this example in the first place
was to show how something could be indeterminate (which would
not be the case here if the domain were $\mathbb N$);
I did not choose $\mathbb Q$ or $\mathbb R$ because any proof of
indeterminacy for $\mathbb Z$ can be extended to those domains as well;
and I did not choose any other domain because that would have been weird.
