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Find $ f ( 2021 ) $ if the function $ f : \mathbb R \to \mathbb R $ satisfies the following Conditions:

  • $ f ( 1 ) = 1 $;
  • $ f ( x + y ) = f ( x ) + f ( y ) $, $ \forall x , y \in \mathbb R $;
  • $ f \left( \frac 1 x \right) = \frac { f ( x ) } { x ^ 2 } $, $ \forall x \in \mathbb R \setminus \{ 0 \} $.

I tried it to solve it using the first two conditions:

$ f(1)=1 $ is given. $ \quad $ (using the first condition)

So, if we take $ x = y = 1 $, we will have $ f ( 2 ) = f ( 1 + 1 ) = f ( 1 ) + f ( 1 ) = 1 + 1 = 2 $. $ \quad $ (using the second condition)

Now, if $ x = 2 $ and $ y = 1 $, we have $ f ( 3 ) = f ( 2 + 1 ) = f ( 2 ) + f ( 1 ) = 2 + 1 = 3 $. $ \quad $ (using the second condition and $ f ( 2 ) = 1 $ from above)

The same pattern will give us $ f ( 4 ) = 4 $ and if we apply it generally it will give us $ f ( x ) = x $ for $ x \in \mathbb Z _ + $.

As $ 2021 \in \mathbb Z _ + $, we get $ f ( 2021 ) = 2021 $.

The question seems to be interesting. But I found the above reasoning to be easy, and I'm not sure my solution is correct or not.

Mohsen Shahriari
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Atique Ahmed
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    It seems we don't need $f\left(\frac 1x\right)$ and $f(n)=n, n\in\mathbb Z^+$ – lone student Apr 08 '21 at 12:26
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    It looks right to me. – saulspatz Apr 08 '21 at 12:26
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    Is this problem from an ongoing contest? – Albus Dumbledore Apr 08 '21 at 12:28
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    @Albus this problem is trivial and problematic – lone student Apr 08 '21 at 12:30
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    @lonestudent absolutely! but the 2021 always raises a radar – Albus Dumbledore Apr 08 '21 at 12:31
  • @AlbusDumbledore No Sir its a simple problem send to me by my junior friend – Atique Ahmed Apr 08 '21 at 12:36
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    @lonestudent Sir what type of radar sir? – Atique Ahmed Apr 08 '21 at 12:36
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    @lulu Sir i am new here If u downvote thats will be bad for me – Atique Ahmed Apr 08 '21 at 12:46
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    @lulu and 2nd thing, before posting this question, i searched it 1st and when i don't found any clue about this then i post it – Atique Ahmed Apr 08 '21 at 12:48
  • @lulu see my https://math.meta.stackexchange.com/questions/33425/how-likely-is-it-that-a-question-with-2021-is-from-a-current-contest new today – Will Jagy Apr 08 '21 at 17:03
  • You haven't answered the question: Is it from an ongoing contest? If it is, this is a cheating attempt, whether by you or your so-called 'friend'. – user21820 Apr 18 '21 at 12:02
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    @user21820 I asked him where he get this question. He said he give some online test on some website. He got that question and he has just 10 min to answer 10 mcq's . And he also tell me that he tries by him self and two or three days after that test he send to me this question . Thats the story bro... And he is not my friend he is my junior and i post this question because it seems interesting but i found it easy and i am not sure that my solution is correct or not. untill now i never ever participate in any mathematical contest. – Atique Ahmed Apr 19 '21 at 00:08
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    @user21820 and bro i never cheated even when i was a student. So sir i request you that try before using such a harsh words. 1st try to search.. And one thing that i just forget to tell you. He(Junior) said that was not any competative test. – Atique Ahmed Apr 19 '21 at 00:17
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    @AtiqueAhmed: I did not claim that you cheated; please read my comments and carefully note the "If". Thank you for clarifying the source and circumstances of this question. Please also realize that there are a huge number of cheaters using Math SE to cheat, so you have to understand why many of us want to be careful not to allow it to happen. – user21820 Apr 19 '21 at 07:25

1 Answers1

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Your argument is correct, as is already noted in several comments above. But I want to add some remarks that I think would be helpful to you. First of all, the functional equation $$ f ( x + y ) = f ( x ) + f ( y ) $$ is a well-known one. It's named Cauchy's functional equation after the French mathematician Augustin-Louis Cauchy, and functions that satisfy it are called additive functions. You can take a look at the post with the title "Overview of basic facts about Cauchy functional equation" to find out about the properties of the functions that satisfy it (I won't go into some details here, as you can find them there). The most relevant one for us here is that for an additive function $ f : \mathbb R \to \mathbb R $, a rational number $ q $ and a real number $ x $, we have $$ f ( q x ) = q f ( x ) \text . $$ What you've proven is part of this fact. So, if the question had asked for the value of $ f \left( \frac { 2021 } 2 \right) $ or $ f ( - 2021 ) $, we could find them using the above fact, and we would have $ f \left( \frac { 2021 } 2 \right) = \frac { 2021 } 2 f ( 1 ) $ and $ f ( - 2021 ) = - 2021 f ( 1 ) $.

But the important thing to know is that there are additive functions that have very wild behaviors, and we can't always get $ f ( x ) = x f ( 1 ) $ for arbitrary (not necessarily rational) $ x $. For example, if the question had asked for $ f \left( \sqrt { 2021 } \right) $, you couldn't find it only having $ f ( 1 ) $ and the fact that $ f $ is additive. That's where the third condition comes into play. For an additive function $ f $ that also satisfies $ f \left( \frac 1 x \right) = \frac { f ( x ) } { x ^ 2 } $, we always have $ f ( x ) = x f ( 1 ) $, whatever $ x $ is. To see why, you can for example take a look at "Cauchy equation $ f ( x + y ) = f ( x ) + f ( y ) $ with an additional condition $ f \left( \frac 1 x \right) = \frac { f ( x ) } { x ^ 2 } $" or "Confusing functional equation problem (application of Cauchy equation when $ f \left( \frac 1 x \right) = \frac { f ( x ) } { x ^ 2 } $)".

Mohsen Shahriari
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