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Let Z be the set of integers. Determine all functions f : Z → Z such that, for all integers a and b,
f(2a) + 2f(b) = f(f(a + b)

Let a be zero such that

f(0)+2f(b)=f(f(b))……(i)

Let b be x and assume that it is true for all integers

f(0)+2f(x)=f(f(x))……(ii)

Then,let a be one such that

f(2)+2f(b)=f(f(b+1))……(iii)

Since x is true for all integers, replace all x in equation (ii) with (b+1) to get

f(0)+2f(b+1)=f(f(b+1))……(iv)

Equate equation (iii) and (iv)

f(0)+2f(b+1)=f(2)+2f(b)

This is equal to

2f(b+1)-2f(b)=f(2)-f(0)

f(b+1)-f(b)=(f(2)-f(0))/2

LHS is difference of consecutive terms, RHS is a constant

Therefore f(x) is an arithmetic progression.

f(x)=mx+k

Using equation (i)

m(2a)+k+2(mb+k)=f(m(a+b)+k)

Which further simplifies to

2m(a+b)+3k=m²(a+b)+mk+k

Comparing coefficients of (a+b)

2m=m²

And also

3k=mk+k

It is easy to see that there are only two solutions

m=0 or 2

So if m=0,k=0 and if m=2,k can be any integer

Therefore,the answers are

f(x)=0 or f(x)=2x+n