A BDMO functional equation problem.

Question

The problem is as follows.

$f:\mathbb{Z}\rightarrow \mathbb{Z}$

$f(f(x+y))$= $f(x^2$$)$+$f(y^2)$

$f(f(2020))$=$1010$

Find $f(2025)$

I approached by substituting $f(2020)$= $\alpha$

Then I substituted that$f(\alpha)$=1010

I tried to figure out that if I can find $f(45^2)$=$f(2025)$

Then I tried to find out if I can find the degree of $f(x)$.

Is it possible to find an $\beta$ such that,$f(\beta)$=1010

If $x^2$=$y^2$ then the equation becomes $f(f(2x))$ = $2f(x^2)$. surely,$f(2x)$ $\not=$$f(x^2)$.

I don't actually have any clue now.

Answer 1

Choosing $y=-x$, we have that $2f(x^2) = f(f(0))$, in other words $f$ is constant on square numbers. From the initial condition we can deduce a bunch of things, one of them is that $f(1010^2) = 505$. So $505$ is the value on all squares, including $f(2025) = f(45^2)$.

Answer 2

preferred_anon already gave the answer, but since it hasn't been "Green-ticked", I guess I will try my luck expanding his answer.

We know$f$ is a function from a Domain of integers to a Range of integers. We also know that when the input is equal to any squared number, the output remains the same; as proved by preferred_anon. Here, we are asked to find $f(2025)$ where the input is a square. In a nutshell, this question is basically asking you what number comes out when a perfect square goes in. So let's look at the other part of the question for more clues.

We know that

$$f(f(x+y)) = f(x^2)+f(y^2)$$

$$f(f(2020))=1010$$

Notice that I can use the second equation in the first one if $x=y=1010$, as $$2020=1010+1010$$

Substituting in the first equation, I get

$$f(1010+1010)=f(1010^2)+f(1010^2)$$ $$f(2020)=2f(1010^2)$$

Substituting the second equation and flipping LHS and RHS, I get

$$2f(1010^2)=1010$$ $$f(1010^2)=505$$

Or in other words, as we saw earlier,

$$f(Any Square Number) = 505$$

Therefore $f(2025)=505$, hence proved