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# Understand the problem

Z be the set of all integers.
Determine all f : Z -> Z such that, for all integers a, b

$f(2a) + 2f(b) = f(f(a+b)) \rightarrow [1]$

##### Source of the problem
IMO 2019 Problem 1
##### Topic
Functional Equation
5 out of 10
##### Suggested Book
Functional Equation and How to Solve Them

Do you really need a hint? Try it first!
This problem needs some patience to play around with the variables to see the beauty of symmetry emerge. Ready for it?

• Prove $f(f(b)) = 2f(b) + f(0) \forall b$ in Z $\rightarrow [2]$ .
• Prove $f(2a) = 2f(a) - f(0) \forall a$ in Z
• Prove $f(2a) + 2f(a) = f(f(2a)) \forall a$ in Z.
• Prove $f(0) + 2f(2a) = f(f(2a)) \forall a$ in Z
• Hence prove $f(2a) = 2f(a) - f(0) \forall a$ in Z $\rightarrow [3]$
• Hence prove $f(a) + f(b) = f(a+b) + f(0) \forall a$ in Z $\rightarrow [4]$ using [1], [2], [3]
• Prove $f(n) = n(f(1) - f(0)) + f(0) \forall n$ in Z $\rightarrow [5]$ using induction in [4].
• Use [5] in [2] to prove that $f(1) - f(0) = 2$ or $f(n) = 0 \forall n$ in Z.
• Hence prove that the solution of the functional equation is $f(n) = 2n + f(0) \forall n$ in Z or $f(n) = 0 \forall n$ in Z

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