Functional Equations with Finitely Many Ratios

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Problem 1. Determine all functions $𝑓 : ℝ \to ℝ$ such that

$𝑓 (𝑥 + 𝑦) + 𝑓 (𝑥 - 𝑦) = 2 𝑓 (𝑥) + 2 𝑓 (𝑦)$
$𝑓 (𝑥 𝑦) = 𝑓 (𝑥) 𝑓 (𝑦)$
$𝑓 (1) = 1$
$\forall 𝑥 \in ℝ, \exists 𝑛 \in {1, 2, \dots, 10}$ such that $𝑓 (𝑥) = 𝑛 𝑥 2$ .

Solution. It’s not hard to see that the final answer is $𝑓 (𝑥) = 𝑥 2$ . Next we prove it.
Let $𝑥 = 0$ , $𝑦 = 0$ , then $2 𝑓 (0) = 4 𝑓 (0)$ . Hence, $𝑓 (0) = 0$ . Let $𝑥 = 0$ , we obtain $𝑓 (𝑦) + 𝑓 ( - 𝑦) = 2 𝑓 (𝑦)$ . Thus, $𝑓 (𝑦) = 𝑓 ( - 𝑦)$ . Therefore $𝑓 ( - 1) = 1$ .
Then we will show that $𝑓 (𝑛) = 𝑛 2 \forall 𝑛 \in ℤ$ . By $𝑓 (𝑦) = 𝑓 ( - 𝑦)$ , it suffices to show that $𝑓 (𝑛) = 𝑛 2 \forall 𝑛 \in ℕ$ . Since $𝑓 (0) = 0$ and $𝑓 (1) = 12$ are already known, by induction, we obtain $𝑓 (𝑛) = 𝑓 (𝑛 - 1 + 1) = - 𝑓 (𝑛 - 2) + 2 𝑓 (𝑛 - 1) + 2 𝑓 (1) = - (𝑛 - 2) 2 + 2 (𝑛 - 1) 2 + 2 = 𝑛 2$ .
Next, suppose $𝑓 (𝑥) = 𝜑 (𝑥) 𝑥 2$ . We stipulate that $𝜑 (0) = 1$ , since $𝑓 (0) = 0$ . Then we obtain $𝜑 : ℝ \to {1, 2, \dots, 10}$ . We study the case in which $𝑥 \neq 0$ , since we already know that $𝑓 (0) = 0$ . By property 1, we obtain $𝜑 (𝑥 + 𝑦) (𝑥 + 𝑦) 2 + 𝜑 (𝑥 - 𝑦) (𝑥 - 𝑦) 2 = 2 𝜑 (𝑥) 𝑥 2 + 2 𝜑 (𝑦) 𝑦 2$ . Simplify this expression to obtain $(𝜑 (𝑥 + 𝑦) + 𝜑 (𝑥 - 𝑦) - 2 𝜑 (𝑥)) 𝑥 2 + (𝜑 (𝑥 + 𝑦) + 𝜑 (𝑥 - 𝑦) - 2 𝜑 (𝑦)) 𝑦 2 + 2 (𝜑 (𝑥 + 𝑦) - 𝜑 (𝑥 - 𝑦)) 𝑥 𝑦 = 0$ . Then we will show that $\exists 𝑦 \in ℤ$ such that $𝜑 (𝑥) = 𝜑 (𝑦)$ . Let $𝑎 = 𝜑 (𝑥 + 𝑦) + 𝜑 (𝑥 - 𝑦) - 2 𝜑 (𝑥)$ , $𝑏 = 𝜑 (𝑥 + 𝑦) + 𝜑 (𝑥 - 𝑦) - 2 𝜑 (𝑦)$ , $𝑐 = 2 𝜑 (𝑥 + 𝑦) - 2 𝜑 (𝑥 - 𝑦)$ , then $𝑎 𝑥 2 + 𝑏 𝑦 2 + 𝑐 𝑥 𝑦 = 0$ . Since $𝜑$ $𝜑$ is a bounded integer-valued function, we know that there are only finitely many possibilities of $(𝑎, 𝑏, 𝑐)$ $(𝑎, 𝑏, 𝑐)$ . Thus, this equation gives at most finitely many solutions of $𝑦$ $𝑦$ satisfying this condition for a given $𝑥$ $𝑥$ when $𝑎$ $𝑎$ , $𝑏$ $𝑏$ , $𝑐$ $𝑐$ is not all zero. But $𝑦 \in ℤ$ can take infinitely many values. Thus, $\exists 𝑦$ $\exists 𝑦$ such that $𝑎 = 𝑏 = 𝑐 = 0$ for a given $𝑥$ $𝑥$ . In this case, we obtain $𝜑 (𝑥 + 𝑦) + 𝜑 (𝑥 - 𝑦) = 2 𝜑 (𝑥) = 2 𝜑 (𝑦)$ , which implies $𝜑 (𝑥) = 𝜑 (𝑦)$ .

Last, since $\forall 𝑥 \in ℝ, \exists 𝑦 \in ℤ, 𝜑 (𝑥) = 𝜑 (𝑦)$ and $𝜑 (𝑥) = 1 \forall 𝑥 \in ℤ$ , we obtain $𝜑 (𝑥) = 1 \forall 𝑥 \in ℝ$ . Thus, $𝑓 (𝑥) = 𝑥 2$ .

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#show: tylenotes.with(
  title: "Functional Equations with Finitely Many Ratios",
  date: "2026-10-30",
  tags: ("functional equation", "finite sets", "lean"),
  summary: "A technique for solving a functional equation with finitely many ratios",
)

#problem[
  Determine all functions $f: RR->RR$ such that
  + $f(x+y) + f(x-y) = 2f(x) + 2f(y)$
  + $f(x y) = f(x) f(y)$
  + $f(1) = 1$
  + $forall x in RR, exists n in {1,2,dots,10}$ such that $f(x) = n x^2$.
]

#solution[
  It's not hard to see that the final answer is $f(x) = x^2$.
  Next we prove it. \
  Let $x = 0$, $y = 0$, then $2f(0) = 4f(0)$. Hence, $f(0) = 0$.
  Let $x = 0$, we obtain $f(y) + f(-y) = 2f(y)$.
  Thus, $f(y) = f(-y)$. Therefore $f(-1) = 1$. \
  Then we will show that $f(n) = n^2 thick forall n in ZZ$.
  By $f(y) = f(-y)$,
  it suffices to show that $f(n) = n^2 thick forall n in NN$.
  Since $f(0) = 0$ and $f(1)=1^2$ are already known,
  by induction, we obtain
  $f(n) = f(n-1 + 1) = -f(n-2) + 2f(n-1) + 2f(1)
  = -(n-2)^2 + 2(n-1)^2 + 2 = n^2$. \
  Next, suppose $f(x) = phi(x)x^2$.
  We stipulate that $phi(0) = 1$, since $f(0) = 0$.
  Then we obtain $phi: RR->{1,2,dots,10}$.
  We study the case in which $x eq.not 0$,
  since we already know that $f(0) = 0$.
  By property 1, we obtain 
  $phi(x+y)(x+y)^2 + phi(x-y)(x-y)^2 = 2phi(x)x^2 + 2phi(y)y^2$.
  Simplify this expression to obtain
  $(phi(x+y)+phi(x-y)-2phi(x))x^2+(phi(x+y)+phi(x-y)-2phi(y))y^2+
  2(phi(x+y)-phi(x-y))x y = 0$.
  Then we will show that
  $exists y in ZZ$ such that $phi(x)=phi(y)$.
  Let $a = phi(x+y)+phi(x-y)-2phi(x)$,
  $b = phi(x+y)+phi(x-y)-2phi(y)$,
  $c = 2phi(x+y)-2phi(x-y)$,
  then $a x^2 + b y^2 + c x y = 0$.
  Since $phi$ is a bounded integer-valued function,
  we know that there are only finitely
  many possibilities of $(a,b,c)$.
  Thus, this equation gives at most finitely many solutions
  of $y$ satisfying this condition for a given $x$ when 
  $a$, $b$, $c$ is not all zero.
  But $y in ZZ$ can take infinitely many values.
  Thus, $exists y$ such that $a=b=c=0$ for a given $x$.
  In this case,
  we obtain $phi(x+y)+phi(x-y)=2phi(x)=2phi(y)$,
  which implies $phi(x)=phi(y)$.

  Last, since $forall x in RR, exists y in ZZ, phi(x)=phi(y)$
  and $phi(x)=1 thick forall x in ZZ$,
  we obtain $phi(x)=1 thick forall x in RR$.
  Thus, $f(x)=x^2$.
]