Functional Equations, Linear Symmetry, Group Theory

I am taking a topics in group theory course, and a small section of the course concerns functional equations (called linear symmetries in this course).

the question is as follows;

Find all linear symmetries of the equation $ x^{2}-y^{2}=1$, find product, inverse and neutral element

any page or video on functional equations does not explore their properties as a group. This professor seems to be marrying the ideas of group theory with functional equations.

attempted solution:

I am brand new at solving functional equations so I begin by stating the question

$$ f(x^{2})-f(y^{2})=f(1) $$

find $f$ that satisfies this equation.

Ill begin by setting $f(1)=c$ since it is always a constant. Additionally, ill fix $y=1$

This gives

$$f(x^{2})-f(1)=a \implies f(x^{2})-a=a \implies f(x^{2})=2a $$

$x^2$ is always positive, so may I can assume that $f(x^2)$ is also positive, and as a result $2c$ and $c$ must also be positive.

This previous attempt does not give us any pertinent information so I will attempt another substitution.

given $ f(x^{2})-f(y^{2})=f(1)$, let $ y=x-1 $, which gives

$$f(x^{2})-f(x-1)^2=f(x^2)-f(x^2-2x)=f(x^2)-f(x^2)+2f(x)=f(1)$$

therefore,

$$2f(x)=f(1) \implies f(x)=\frac{f(1)}{2}$$

This is so far my attempt at a solution, although I don't believe my result for $f(x)$ is all solutions for $f(x)$.

Could I get a nudge (or two) in the correct direction, and could you help me understand the notation to display functional equations as mappings of groups from $\mathbb R \to \mathbb R$?

Thanks.

Solution 1:

The functional equation $ f \big( x ^ 2 \big) - f \big( y ^ 2 \big) = f ( 1 ) $ has only one solution, and that is the constant zero function. To see that, just let $ y = x $ to get $ f ( 1 ) = 0 $, and then let $ y = 1 $ to find out that $ f \big( x ^ 2 \big) = 0 $, which means that $ f $ is identically zero on the domain of the objects of the square form (and that's all you can get from the equation, since the arguments appearing in the scope of $ f $ in the equation are all squares; if domain of $ f $ is intended to contain non-square points, the equation says nothing about the value of $ f $ at those points, and $ f $ can be arbitrarily defined at those point and identically zero on the squares, and it still satisfies the equation). If $ x $ and $ y $ are intended to be real numbers, this mean that $ f $ is identically zero on nonnegative real numbers, and if they are intended to be complex numbers, that means $ f $ is identically zero on the whole plane, and if they are intended to be integers then $ f $ is identically zero on perfect squares, and so on.

But aside from that, I think you've misunderstood the original equations. It seems to me that "symmetries of the equation $ x ^ 2 - y ^ 2 = 1 $" have nothing to do with the functional equation. That's something that you have to check with the source of the problem, but as far as I can see, those "symmetries" are intended to mean transformations on "$ x ^ 2 - y ^ 2 = 1 $" as a symbolic formula, so that the result doesn't change as an equation about numbers or whatever is intended for $ x $ and $ y $. As an example, the transformation $ x \mapsto - x , y \mapsto y $ is such a symmetry, as it changes the expression to $ ( - x ) ^ 2 - y ^ 2 = 1 $, and by formal rules of algebra of numbers, that is equivalent to the original equation. Another examples are $ x \mapsto x , y \mapsto y $ (identity), $ x \mapsto x , y \mapsto - y $ and $ x \mapsto - x , y \mapsto - y $. In this sense, these symmetries form a group. The operation is the composition of transformations, which is associative; as an example, the composition of $ x \mapsto - x , y \mapsto - y $ and $ x \mapsto - x , y \mapsto y $ is $ x \mapsto - ( - x ) , y \mapsto - ( y ) $, or equivalently $ x \mapsto x , y \mapsto - y $. The neutral element is the identity transformation, which in composition with any transformation, results in the same transformation. And the inverse of each of the transformations above is itself; i.e. the composition of each of them with itself results in the identity transformation.

I don't know whether the above transformations are all of the symmetries of the given equation or not. If they are, you've found the group you wanted, and found out that in fact it is isomorphic to Klein four-group. It seems that characterizing this group may not be trivial at all. Moreover, you may find symmetries which don't have any inverse, and thus the "symmetries" may not form a group, but just a monoid.

EDIT: Another thing that came to my mind is that the question may have asked about the geometrical symmetries of the locus of a point such that its coordinates satisfy the equation $ x ^ 2 - y ^ 2 = 1 $. This makes sense, too, and in fact in this case, characterizing the group of symmetries is not that difficault.

EDIT: I realized that the two groups I mentioned are in fact the same; not only they are isomorphic, but also they are conceptually the same thing. They are two different ways of looking at the same notion; one from algebraic and formal point of view, and the other from pictorial and geometric point of view. That said, it's easy to see that in fact the group we're looking for is just the four-element group mentioned above (when we consider the points on a plane, and not for example in a three dimensional space, in which infinitely many new symmetries arise; that's also true from the algebraic point of view, when we consider the equation as an expression of just the two variables $ x $ and $ y $, and not another one not appearing in the equation).