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Question Number 3654 by prakash jain last updated on 17/Dec/15

f (x f (y) + x) = x y + f (x)

f : R \to R

f (x) = ?

Commented by RasheedSindhi last updated on 18/Dec/15

f (x f (y) + x) = x y + f (x) \dots . (i)

L e t y = 0

f (x f (0) + x) = x (0) + f (x)

f (x) = f (x f (0) + f (x)) \dots \dots . . (i i)

f (0) = f (0 . f (0) + f (0))

f (0) = f (f (0))

x ⇋ y i n (i)

f (y f (x) + y) = x y + f (y) \dots \dots . (i i i)

y = x i n (i)

f (x f (x) + x) = x^{2} + f (x)

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

f (1) = f (f (1) + 1) - 1

f (1) + 1 = f (f (1) + 1)

Commented by prakash jain last updated on 18/Dec/15

based on above

f (x) = x is one solution, are there any other

solutions ?

Commented by prakash jain last updated on 18/Dec/15

Let us say f (x) \neq x and f (y) = 0 f o r s o m e y .

f (x) = {x f}^{- 1} (y) + f (x)

f^{- 1} (y) = 0

s o i f f (x) \neq x t h e n e i t h e r

f (x) \neq 0 for all x \in R or f (0) = 0 .