### Functional Equations 101

Let's get to the math:

Let there be two sets $X$ and $Y$. A function from $X$ to $Y$ denoted as $f \colon X \to Y$ is assigning a value in $Y$ for every element in $X$. We say that $X$ is the domain of the function $f$ and $Y$ is the range.

A function $f \colon X \to y$ is said to be injective if $f(x) = f(x^{\prime}) \implies x = x^{\prime}$ To put it in a more abstract way, if there is some $a \in Y$ then there is at most one $b \in X$ such that $f(b) = a$ holds true.

A function is said to be surjective when for any $a \in Y$ there is at least one $b \in X$ such that $f(b) = a$ holds true.

A function is bijective if for every $a \in Y$ there is exactly one $b \in X$ such that $f(b) = x$. Bijective functions are basically functions which are both injective and surjective.

Bonus: A function $f \colon X \to X$ is known as an involution if $f(f(x)) = x \; \forall x \in X$
As an exercise, the readers should try to prove that every function that is an involution is a bijective function as well.

Let's look at some problems now:

Problem 1: (IMO 2019) Let $\mathbb{Z}$ be the set of integers. Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that, for all integers $a$ and $b$, $$f(2a)+2f(b)=f(f(a+b)).$$

Solution: The answers are $f(x)=0$ and $f(x)=2x+d$ for some $d$.
Note that $$f(2)+2f(b) = f(f(b+1)) = f(0)+2f(b+1) \implies f(b+1)-f(b)=\frac{f(2)-f(0)}{2}.$$ This means $f$ is linear, so let $f(x)=cx+d$. We have that $$c(2a)+d+2(cb+d)=c(c(a+b)+d)+d \iff c(2a+2b)+2d = c^2(a+b)+cd,$$ which is true if the coefficients of the monomials line up, or $2c=c^2$ and $cd=2d$. From the first equation, $c=0$ or $c=2$. $c=0$ gives that $d=0$ and the solution $f(x)=0$ and $c=2$ gives that $d$ can be anything and the solution $f(x)=2x+d$.

Problem 2:: (JMO 2021) Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that for positive integers $a$ and $b,$ $$f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2$$

Solution: We claim that $f(x)=1$ is the only solution. Since $f(1^2)=f(1)^2$, this means that $f(1)=1$. We also have $f(1^2+1^2)=f(1)f(1)=1$, so $f(2)=1$. Suppose that there exists another solution. Then, we must have $f(k)\neq1$ for some $k$. Consider the smallest value of $k$. Then, $f(x)=1$ for all $x<k$. We will prove that this is impossible.

Case 1: $k$ is even
Then, substituting $a=2x$ and $b=x^2-1$, we get

$$f((2x)^2+(x^2-1)^2)=f(2x)f(x^2-1)$$
$$f((x^2+1)^2)=f(2x)f(x^2-1)$$
$$f(x^2+1)^2=f(2x)f(x^2-1)$$
$$f(x)^2f(1)^2=f(2x)f(x^2-1)$$
$$f(x)^2=f(2x)f(x^2-1)$$
Now, let $x=\frac k2$. Since $k\geq4$, we have $x\geq2$, so $a,b>0$. Since $x<k$, this gives
$$f(k)f(x^2-1)=1,$$ so since $f(k)$ is a positive integer, this means that $f(k)=1$, which is a contradiction.

Case 2: $k$ is odd
Substituting $a=2x(x+1)$ and $b=(x+1)^2-x^2=2x+1$, we get

$$f((2x(x+1))^2+((x+1)^2-x^2)^2)=f(2x(x+1))f(2x+1)$$
$$f(((x+1)^2+x^2)^2)=f(2x(x+1))f(2x+1)$$
$$f((x+1)^2+x^2)^2=f(2x(x+1))f(2x+1)$$
$$f(x+1)^2f(x)^2=f(2x(x+1))f(2x+1)$$
Now, let $x=\frac{k-1}2$. Since $k\geq3$, we have $x\geq1$, so $x+1<2x+1$ and $a,b>0$. Therefore, we have $$f(2x(x+1))f(k)=1,$$ so $f(k)=1$.

Since $f(x)=1$ obviously works, this means that the only function that satisfies the conditions is $\boxed{f(x)=1}$ for all $x\in\mathbb N$.

Problem 3: (IMO 2010) Find all function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all $x,y\in\mathbb{R}$ the following equality holds:
$$f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor$$ where $\left\lfloor a\right\rfloor$ is greatest integer not greater than $a.$

Solution: Let $x=1$. We can see that $$f(y)=f(1)\lfloor f(y)\rfloor.$$ If $f(1)=0$, we know that all $f(y)=0$, which we will check later. Plugging the equation into the original equation, $$f(1)f(\lfloor x\rfloor y)=f(x)f(y).$$ However, if we switched the order of $x$ and $y$, we would get that $$f(\lfloor x\rfloor y)=f(\lfloor y\rfloor x).\; (1)$$ Now, for all $y$ such that $y>0$, plug $$\left(\frac{y}{\lfloor y\rfloor+1},\lfloor y\rfloor +1\right)$$ into (1). We get that $$f(0)=f\left(\left(\lfloor y\rfloor+1\right) x\right)=f(y).$$ This means that $f(y)$ is constant for all nonnegative $y$. For all $y$ such that $y<0$, plug $$\left(\frac{y}{\lfloor y\rfloor-1},\lfloor y\rfloor -1\right)$$ into (1). We get that $$f(0)=f\left(\left(\lfloor y\rfloor-1\right) x\right)=f(y).$$ This means that $f(y)$ is constant for all nonpositive $y$. Therefore, we can conclude that $f(x)$ is constant. The previous $f(x)=0$ also follows this, so just let $f(x)=C$. We have that $C=C\lfloor C\rfloor$, so $C=0$ or $\lfloor C\rfloor=1$. Both of these clearly work, so we have our answer: $f(x)=C$ where $C$ can either be $0$ or have a floor of $1$.

Problem 4: (IMO 2008) Find all functions $f: (0, \infty) \mapsto (0, \infty)$ (so $f$ is a function from the positive real numbers) such that
$$\frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}$$
holds for all positive real numbers $w,x,y,z,$ satisfying $wx = yz$

Solution: We claim the only solutions are $f(x)=x$ and $f(x)=\frac1x$.
Plugging in $w=x=y=z$, we get
$$\frac{2f(w)^2}{2f(w^2)}=\frac{2w^2}{2w^2}=1$$
$$f(w)^2=f(w^2)$$
It follows that $f(1)=1$.
In addition, if $wx=yz$, then
$$\frac {f(w^2) + f(x^2)}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}.$$
Let $ab=cd$. Plugging in $w=\sqrt{a}$, $x=\sqrt{b}$, $y=\sqrt{c}$, $z=\sqrt{d}$, we get
$$\frac {f(a) + f(b)}{f(c) + f(d) } = \frac {a + b}{c + d}.$$
Let $ab=c^2$. Then, we have
$$\frac{f(a)+f(b)}{2f(c)}=\frac{a+b}{2c}$$
$$\frac{f(a)+f(b)}{a+b}=\frac{f(c)}{c}$$
Let $a=1$, $c=\sqrt{b}$, and $b\neq 1$. We get
$$\frac{f(1)+f(b)}{1+b}=\frac{f(\sqrt{b})}{\sqrt{b}}$$
$$\frac{1+f(b)}{1+b}=\frac{f(\sqrt{b})}{\sqrt{b}}$$
$$\frac{(1+f(b))^2}{(1+b)^2}=\frac{f(b)}{b}$$
This is a quadratic in $f(b)$. We can check that $f(b)=\frac1b$ and $f(b)=b$ both satisfy this equation. Since $b\neq 1$, these are distinct, so these are the only solutions, so there are no other possible values of $f(b)$.

Let $c\neq 1$. There are 2 cases.

Case 1: $f(c)=c$
Then, for all $a, b$ such that $ab=c^2$, we have
$$\frac{f(a)+f(b)}{a+b}=\frac{f(c)}{c}=1$$
If $f(a)=a$, then $f(b)=b$.
Assume towards a contradiction that $f(a)=\frac1a$ and $f(b)=\frac1b$. Then,
$$\frac1a+\frac1b=a+b$$
$$\frac{a+b}{ab}=a+b$$
$$ab=1$$
Then, $c=1$, a contradiction. So, we have $f(a)=a$ and $f(b)=b$. So, $f(x)\equiv x$.

Case 2: $f(c)=\frac1c$
Then, for all $a, b$ such that $ab=c^2$, we have
$$\frac{f(a)+f(b)}{a+b}=\frac{f(c)}{c}=\frac1{c^2}=\frac1{ab}$$
$$f(a)+f(b)=\frac{a+b}{ab}=\frac1a+\frac1b$$
If $f(a)=\frac1a$, then $f(b)=\frac1b$. Assume towards a contradiction $f(a)=a$ and $f(b)=b$. Then,
$$a+b=\frac1a+\frac1b$$
$$ab=1$$
Then, $c=1$, a contradiction. so, we have $f(a)=\frac1a$ and $f(b)=\frac1b$. So, $f(x)\equiv \frac1x$.

Problem 5: (Shortlist 2018) Let $\mathbb{Q}_{>0}$ denote the set of all positive rational numbers. Determine all functions $f:\mathbb{Q}_{>0}\to \mathbb{Q}_{>0}$ satisfying $$f(x^2f(y)^2)=f(x)^2f(y)$$ for all $x,y\in\mathbb{Q}_{>0}$

Solution: The answer is $\boxed{f\equiv 1}$, which obviously works.
Let $P(x,y)$ be the assertion. Suppose that $f(1)=c$. Then $P(1,1)$ gives that $f(c^2)=c^3$ and $P(1,c^2)$ gives $f(c^6)=c^5$. But also $P(c^2, 1)$ gives $f(c^6)=c^7$ which implies that $f(1)=1$. Then $P(1, x)$ and $P(x, 1)$ give
$$f(f(x)^2)=f(x)$$
$$f(x^2)=f(x)^2$$
or that $f(x)=f(f(x)^2)=f(f(x))^2$. Then $f(x)$ is a $2^n$th power for arbitrarily large $n$, implying that $\boxed{f\equiv 1}$, and we are done.

Problem 6: (EGMO 2021) Find all functions $f:\mathbb{Q}\to\mathbb{Q}$ such that the equation
$$f(xf(x)+y) = f(y) + x^2$$ holds for all rational numbers $x$ and $y$.

Solution: Let $P(x,y)$ denote the assertion that $$f(xf(x)+y) = f(y) + x^2.$$

Lemma 1: $f$ is surjective.
Proof: $P(x,0): f(xf(x))=f(0)+x^2$.

$P(x,-xf(x)): f(0)=f(-xf(x))+x^2$, so $f(-xf(x))=f(0)-x^2$.

Thus, by induction, $f(0)+kx^2$ for any integer $k$ is in the image of $f$.

But note that $\frac{p}{q}=pq\left(\frac{1}{q}\right)^2$, so $f(0)+\frac{p}{q}$ is in the image of $f$, for any integers $p,q$, where $q\ne 0$.

Thus, any rational number is in the image of $f$. $\blacksquare$

Lemma 2: $f(0)=0$.
Suppose $f(k)=0$ (a such $k$ must exist as $f$ is surjective).

$P(k,x): f(x)=f(x)+k^2\implies k=0$.

Lemma 3: $f(1)=\{-1,1\}$.
$P(1,0): f(f(1))=1$.

$P(f(1),0): f(f(1))=f(1)^2$.

This implies $f(1)^2=1$, which proves our claim.

Case 1: $f(1)=1$.
Then $P(1,x): f(x+1)=f(x)+1\implies f(x)=x\forall x\in\mathbb{Z}$.

We claim that $f(kxf(x))=kx^2$ for all positive integers $k$.
Proof: We proceed by induction. Setting $y=0$ gives $f(xf(x))=x^2$. Now suppose for all positive integers $l$ less than $k$, $f(lxf(x))=lx^2$.

$P(x,(k-1)xf(x)): f(kxf(x))=(k-1)x^2+x^2=kx^2$, which completes the induction.

For any rational $x$, we take a positive integer $k$ so that $kxf(x)$ is an integer (this is obviously always possible).

So $kxf(x)=kx^2\implies xf(x)=x^2\implies \boxed{f(x)=x\forall x\in\mathbb{Q}}$.

Case 2: $f(1)=-1$.
$P(1,x): f(x-1)=f(x)+1$.

So for all integers $n$, $f(n)=-n$.

We claim that $kxf(x)=kx^2$ for all positive integers $k$.
Proof: Same as above case.

For any given $x$, take a positive integer $k$ so that $kxf(x)$ is an integer.

Then $-kxf(x)=kx^2\implies -xf(x)=x^2\implies \boxed{f(x)=-x\forall x\in\mathbb{Q}}$

Bratin.

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