An introduction to functional equation
How many functions f ( x ) from reals to reals are there such that f ( 0 ) = 0 and for all nonzero reals x , y we have f ( x y ) = f ( x ) f ( y ) − 2
The answer is 2.
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2 solutions
Btw, P ( x , 0 ) works just as well as P ( x , 1 ) in this case.
x, y are supposed to be nonzero reals.
Let n = f ( 1 ) ∈ R . Set y = 1 to get, after rearranging, f ( x ) = n − 1 2 . Hence the function is a real nonzero constant. Setting f ( x ) = a ∈ R ∀ x ∈ R gives the quadratic a 2 − a − 2 = 0 ⟹ ( a − 2 ) ( a + 1 ) = 0 ⟹ a = − 1 , 2 .
Notice how your solution is exactly identical to mine.
But this contradicts the first statement, f(0) = 0. Since f(x) = a ~= 0
Let P ( x , y ) be the statement f ( x y ) = f ( x ) f ( y ) − 2 .
P ( x , 1 ) ⟹ f ( x ) ( 1 − f ( 1 ) ) = − 2
⟹ f ( x ) = 1 − f ( 1 ) − 2 = c , c ∈ R , c = 0 , ∀ x ∈ R
⟹ c = c 2 − 2 ⟺ ( c − 2 ) ( c + 1 ) = 0 ⟺ ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ f ( x ) = 2 or f ( x ) = − 1