A Nice and Easy Functional Equation 

Why can’t we consider a family of functions of kind f(x) = 1 if P(x) f(x) = -1 otherwise ?

If f is defined from Z to Z then f (x) = 1 for evens and -1 for odds will work. Otherwise, as in the video.

The last step wasn't correct When you got f(x)^2=1 You can not jump and say f(x)=1 for all x Or f(x)=-1 for all x you can only say that [f(x)=1 or -1] for all x Meaning thatsome values of f are 1 and some are -1 He needed to add one more step to fix it Set x to 2y This shows that if the function was from Z to Z, you can not say 2y contains all the numbers, so you'd have more solutions like it was mentioned in other comments

Kind of interesting from an algebraic perspective. Setting x, y = 0, you have f(0) = 1. So you have f(x-x) = f(x)f(x) -> f(x) = -1 or 1. Lastly, you have f(0-y) = f(0)f(y) -> f(-y) = f(y) -> f(x+y) = f(x)f(-y) = f(x)f(y). Considering the domain to just be some additive (abelian) group G, this means we have a symmetric homomorphism into the trivial multiplicative group {1, -1}. The kernel K is a normal subgroup, and the symmetry tells us that it has equivalent left and right cosets i.e. g + K = K - g -> K = g + K - g = g + (g + K) = 2g + K -> ... -> 2g is in K for all g in G. What does this say about the domain? One is that any "even" element, i.e. an element expressible as the sum of 2 elements, maps to 1. If the domain were the reals or rationals, this would be the whole set. If this were the integers, this would mean every even integer maps to 1, but what about the odd ones? f(2n+1) = f(2n)f(1) = 1*f(1) = f(1) -> every odd integer maps to the same thing. This gives us one of 2 options: f(x) = 1 (this includes the zero function since we're just identically mapping to an identity), or f(n) = (-1)^n

How do you know if value of the function is not 1 for some values and -1 for other values?

For y = 0 we have f(x) = f(x)f(0), so f(0) = 1. For y = x/2 we have f(x - x/2) = f(x/2) = f (x)f(x/2) which gives f(x) = 1.

You haven't adequately addressed the possibility that f is not a constant function.

I'm sure that any standard approach will show f(x) is constant, and the values 0 and 1 follow quickly.

Take x=2y => f(y) = f(2y)f(y) => f(2y)=1 OR f(y)=0 So only 2 possible solutions : f(x)=0 or f(x)=1

But how do you prove the function doesn’t jump from the values 1 and -1

f(x-y) = f(x)f(y) Set y = 0: f(x) = f(x)f(0) If f(x) = 0 for all x, then f(x-y) = f(x)f(y) becomes 0 = (0)(0), which works, so a solution has been found. If there exists some x such that f(x) =/= 0, then f(0) = 1. Consider x free, y = x. Then f(x - y) = f(x)f(y) => f(0) = f(x)^2 => 1 = f(x)^2. So, f(x) = 1 or -1. Note that the problem did not require continuity, so it is possible that f(x) = 1 for some values of x, but -1 for other values. Next, consider x = 0, y free. Substituting into our equation, we get f(0-y) = f(0)f(y) => f(-y) = 1 f(y) => f(-y) = f(y). Next, consider x = a, y = -b for some a and b. Substituting into the equation, we get f(a - (-b)) = f(a)f(-b) => f(a+b) = f(a)f(b). But that's our Cauchy equation, which you touched upon. So, we have f(x) = k^x for some k, and we know that f(1) = k. And since f(x) is always 1 or -1, then f(1) = 1 or -1. So, we can have f(x) = 1 or f(x) = (-1)^x. And here is where we run into a problem. Syber did not specify the domain of the problem. If the problem is Z -> Z, then (-1)^x is valid. If it's real -> real, then (-1)^x is not valid. And if it's complex -> complex then (-1)^x is valid... except we get f(1/2) = i, which breaks the part where f(x) can only be 1 or -1. So, if the problem is integers -> integers, then f(x) = 0, 1, or (-1)^x. If we are working over R -> R or C -> C, then f(x) = 0 or 1.

Here is my rigorous solution (point out any flaws if you see them, I will read your comment but I might not reply) From f(x - y) = f(x)f(y) we can plug in y = x and get f(0) = f(x)² for all real numbers. If the function f exists for all real numbers, then f(0) exists and is a constant. plugging in x = 0 and y = 0, we get f(0)² - f(0) = 0 which means f(0) = 1 or f(0) = 0. Case 1: f(0) = 0. If f(0) = 0, then f(x)² = 0 for all real numbers x, and this implies f(x) = 0 for all real numbers x. Let's test this solution out: f(x - y) = f(x)f(y) 0 = 0*0. It works. f(x) = 0 is a solution. Case 2: f(0) = 1. Since f(x)² = f(0) = 1 for all real numbers x, then f(x) = 1 or f(x) = -1 for any given number x. (That is to say, it is possible that, for example, f(2) = 1 and f(3) = -1). However, plugging in x = 2y, we get f(y) = f(2y)f(y); and since f(y) = -1 or f(y) = 1 then f(y) is nonzero. This implies f(2y) = 1 for all real numbers y; which means that f(x) = 1 for all real numbers x. Let's test this solution out: f(x - y) = f(x)f(y) 1 = 1*1 It works. f(x) = 1 is a solution. No assumptions have been made throughout this solution. There must be no other possible solutions. The solutions are: 1. f(x) = 0 for all real numbers x; OR 2. f(x) = 1 for all real numbers x.

beta(x,y)=gamma(x)gamma(y)/gamma(x+y) if that was f(x+y) on the left, this might be a cool path to follow. Unfortunately I don't see how to transform it.

f(1)=f(1)f(0) f(0)=1 f(1)=f(1)f(2) f(2)=1 f(2)=f(3)f(1) f(x)=1

I'd like to see some functional equations where the solution is a little more complicated

Awesome :)

f(x) =sqrtf(0)

The teacher must be Turkish.

f(x) could be -1 as well.

What is f(y) ?

Plug in y=0 to get f(x) = f(x)f(0). So either f(x)=0 or f(0)=1. In the latter case, plug in y=x to get 1 = f(0) = f(x)^2. So f(x)=1 or f(x)=-1. If f is continuous, we must have f = 0 or f = 1