-2

I found this problem in an old book of math, here is my try:

$$P(a;a) \mapsto ff(a)=2f(a)$$ $$P(a;f(b)) \mapsto f(a+f(b))=f(a)+2f(b)$$ $$P(0;b) \mapsto f(0)=0$$ $$P(0;b) \mapsto f(f(b)-b)=f(b)...(1)$$ $$P(b-f(b);b)\mapsto f(b-f(b))=-f(b)..(2)$$ From (1) and (2), f is odd

I really would be thankful if someone could give me the full solution please, thanks in advance !

Maxwol
  • 161
  • 1
    I believe there is a typo on the third line: shouldn't it start $P(0;0)$? – Michael Cohen Aug 28 '21 at 13:00
  • 2
    @Maxwol This is a nice functional equation, but your post is poorly written. To avoid it being downvoted/closed, could you improve it so that it matches the quality guidelines ? – SolubleFish Aug 28 '21 at 14:08
  • Could you help me to prove it plz – Maxwol Aug 28 '21 at 14:16
  • 1
    To start with, you should probably write full sentences to : - explain briefly where the equation comes from ; - detail the reasoning in (what I believe to be) your proof attempt ; - clearly ask your question (do you want us to check your proof, give a hint, provide a full solution, etc.). You can take a look at what the OP here looks like. – SolubleFish Aug 28 '21 at 14:22

1 Answers1

3

As you've seen, by setting $a=b$ we have

$$ f(f(a))=f(a-b+f(b))=f(a)+f(b)=2f(a) $$

And by setting $b=f(b')$ for some $b'\in\mathbb Z$ we get (by using the last result)

$$ f(a+f(b'))=f(a-b+f(b))=f(a)+f(b)=f(a)+f(f(b'))=f(a)+2f(b') $$

And by setting $a=0$ we get

$$ 2f(b')=f(f(b'))=f(0)+2f(b') $$

Thus $f(0)=0$.

Now let $a=f(1)$, we will show that $f(n)=na$ for all $n\in\mathbb Z$, first by induction to show it for all $n\geq 0$:

  • The base case is $f(0)=0=0a$
  • Now suppose $f(n)=na$ for some $0\leq n\in\mathbb Z$, then we have $$ f(n+1)+na=f(n+1)+f(n)=f(n+1-n+f(n))=f(1+f(n))=f(1)+2f(n)=a+2na $$ So $f(n+1)=(n+1)a$.

Finally, for all $0 > n\in\mathbb Z$ we have that $-n, -2n$ are positive so $$f(n)-4na=f(n)+2f(-2n)=f(n+f(-2n))=f((-n)-(-2n)+f(-2n))=f(-n)+f(-2n)=-na-2na=-3na$$ And we get $f(n)=na$.

Now to show $a\in\{0,2\}$, note that we have: $$ a^2=f(a)=f(f(1))=2f(1)=2a $$

Mor A.
  • 821