How Can We Solve the Functional Equation f(f(x+y) - f(x-y)) = xy?

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The functional equation f(f(x+y) - f(x-y)) = xy has been analyzed with key findings indicating that f(0) = 0 and f(f(2x)) = x^2 for x = y. The discussion suggests that the function f(x) may involve an improper exponent, specifically f(x) = x^√2 / A, where A is a scaling factor. Additionally, derivatives of the function have been explored, revealing that the first derivatives yield specific relationships between x and y, while second derivatives are consistently zero.

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steve B. 98
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I'm trying to solve this problem from a high school math competition:
Find all functions f : R → R such that, f(f(x+y)-f(x-y))=xy, for all real x,y.
Any ideas of how to approach it.
I have found that f(0)=0, if x=y f(f(2x))=x^2
 
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Some thoughts:
I would note that the domain and range of the function must be all real numbers. This should eliminate many of the trig functions and exponentials.
Based on what you have, it seems like f(x) might incorporate some improper exponent...
Say...##f(x) = \frac{x^\sqrt {2} }{A}## where A scales 2 to 1 over two iterations.
I am not 100% sure how you would expand this out for the sums, and I don't think that function is defined for all x,y in the real numbers.

Another option,
Think of the derivatives:
##\frac{\partial}{\partial x} f ( f( x+ y) - f(x-y) ) = \frac{\partial}{\partial x} xy ##
##f ' ( f( x+ y) - f(x-y) ) * (f'(x+y)-f'(x-y)) = y ##
##\frac{\partial}{\partial y} f ( f( x+ y) - f(x-y) ) = \frac{\partial}{\partial y} xy ##
##f ' ( f( x+ y) - f(x-y) ) * (f'(x+y)+f'(x-y)) = x ##
And second derivatives are all zero.
 

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