Question

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Here is a question

Let Z be the set of integers. Determine all functions f : Z gives Z such that, for all integers a and b,


f(2a) + 2f(b) = f(f(a+b).

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Answer 1

Answer :

• f(x) = 0
• f(x) = 2x + n for some integer n ∈ Z

Let us consider the :

$\sf \implies f(2a) + 2f(b) = f(f(a + b))\quad[\bf\: Equation \: 1 \:]$

Now,

Now,If a = 0, then:

$\implies \sf f(0) + 2f(b) = f(f(b)) \quad[ \: Equation \: 1 \: ]$

The above is true for all integers b.

Let's assume a = 1.

$\sf \implies f(2) + 2f(b) = f(f(b + 1))$

Using Equation 2 to evaluate f(f(b + 1)),

→ f(2) + 2f(b) = f(0) + 2f(b + 1)

→ (f(2) - f(0))/2 = f(b + 1) - f(b)

The equation is true for all integers b,

Hence the function f is an arithmetic progression.

So,

→ f(x) = mx + n

Now we substitute this into the original situation 1

→ f(2a) + 2f(b) = f(f(a + b))

→ (m(2a) + n) + 2(mb + n) = f(m(a + b) + n)

→ 2m(a + b) + 3n = m2(a + b) + (mn + n)

Now The coefficients of a + b & of a + b are identified.

So,

• 2m = m2
• 3n = mn + n

Now

There are two solutions to 2m = m2.

Let's consider them separate.

• m = 0

→ 3n = n

→ n = 0

• m = 2

→ 3n = 3n

→ n ∈ Z

Hence, We separate solutions:

• f(x) = 0
• f(x) = 0f(x) = 2x + n for some integer n ∈ Z

Answer 2

Answer :

f(x) = 0

f(x) = 2x + n for some integer n ∈ Z

Let us consider the :

Now,

Now,If a = 0, then:

The above is true for all integers b.

Let's assume a = 1.

Using Equation 2 to evaluate f(f(b + 1)),

→ f(2) + 2f(b) = f(0) + 2f(b + 1)

→ (f(2) - f(0))/2 = f(b + 1) - f(b)

The equation is true for all integers b,

Hence the function f is an arithmetic progression.

So,

→ f(x) = mx + n

Now we substitute this into the original situation 1

→ f(2a) + 2f(b) = f(f(a + b))

→ (m(2a) + n) + 2(mb + n) = f(m(a + b) + n)

→ 2m(a + b) + 3n = m2(a + b) + (mn + n)

Now The coefficients of a + b & of a + b are identified.

So,

2m = m2

3n = mn + n

Now

There are two solutions to 2m = m2.

Let's consider them separate.

m = 0

→ 3n = n

→ n = 0

m = 2

→ 3n = 3n

→ n ∈ Z

Hence, We separate solutions:

f(x) = 0

f(x) = 0f(x) = 2x + n for some integer n ∈ Z