Find all functions $f :\Bbb{ R}\to \Bbb{Z}$ such that $$\left( f(f(y) - x) \right)^2+ f(x)^2 + f(y)^2 = f(y) \cdot \left( 1 + 2f(f(y)) \right),$$for all $x, y \in \Bbb{R}.$
Problem
Source:
Tags: algebra, functional equation
Tintarn
01.05.2016 23:47
As usual, denote $P(x,y)$ the assertion that $f(f(y)-x)^2+f(x)^2+f(y)^2=f(y) \cdot (1+2f(f(y)))$. Let $a=f(0)$.
$P(0,0)$ implies $a-a^2=(f(a)-a)^2 \ge 0$. Hence $a=0$ or $a=1$.
If $a=0$, then $P(x,0)$ implies $f(-x)^2+f(x)^2=0$ and hence $\boxed{f(x)=0}$ for all $x$ which is indeed a solution.
If $a=1$, then $P(x,0)$ implies $f(1-x)^2+f(x)^2=2$ and hence $f(x)^2=1$ for all $x$.
So $P(x,y)$ simplifies to $f(y) \cdot (1+2f(f(y)))=3$ for all $y$ and hence $1+2f(f(y))=\pm 3$ which implies $f(f(y))=1$ and hence $\boxed{f(x)=1}$ for all $x$ which is indeed another solution.
BobThePotato
02.05.2016 00:15
(never mind this doesn't work)
bobthesmartypants
02.05.2016 00:20
No, the function $f(x)=\lfloor x\rfloor$ is a non-constant function that maps reals to integers...
KMO1
03.05.2016 10:35
Tintarn wrote: If $a=1$, then $P(x,0)$ implies $f(1-x)^2+f(x)^2=2$ and hence $f(x)^2=1$ for all $x$. But $P(x,0)$ implies $f(1-x)^2+f(x)^2=2f(1)$
shinichiman
03.05.2016 10:46
KMO1 wrote: But $P(x,0)$ implies $f(1-x)^2+f(x)^2=2f(1)$ We can find $f(1)=1$ by doing the next step $P(0,0)$ implies $f(1)^2+1=2f(1)$ so $f(1)=1$. Hence, $f(1-x)^2+f(x)^2=2f(1)=2$.
Khazix
10.05.2016 01:25
Tintarn wrote:
... which implies $f(f(y))=1$ and hence $\boxed{f(x)=1}$ for all $x$ which is indeed another solution.
How does $f(f(y)) = 1$ imply that $f(x) = 1$ for all x? Couldn't $f(y) = -1$, and then $f(f(y)) = 1$ for some $y$?
lebathanh
19.08.2016 15:59
Khazix wrote: Tintarn wrote:
... which implies $f(f(y))=1$ and hence $\boxed{f(x)=1}$ for all $x$ which is indeed another solution.
How does $f(f(y)) = 1$ imply that $f(x) = 1$ for all x? Couldn't $f(y) = -1$, and then $f(f(y)) = 1$ for some $y$? I can explain for you : from euation f(y)(1+f(f(y))) =3 and f(f(y))=1 ==> f(y)=1
MathStudent2002
28.02.2017 07:34
We claim the only solutions are $\boxed{f(x) = 0, f(x)=1}$. Plugging them in yields $0=0$ and $3=3$, respectively, so it is clear that they both work. Now we shall show that they are the only solutions.
Let $P(x,y)$ denote the assertion.
Observe that if $f(n) = 0$ for any $n$, then use $P(x,n)$ to get $f(-x)^2 + f(x)^2 = 0$, which implies that $f(x), f(-x)$ are both $0$ since $f(x)^2+f(-x)^2 \ge 0$, with equality when they are both zero. Hence $f(x) = 0$ for all $x$.
Use $P(0,0)$ to conclude that $f(f(0))^2 + f(0)^2 + f(0)^2 = f(0)+2f(f(0))f(0)$. Hence we see that $f(0)-f(0)^2 = (f(f(0))-f(0))^2 \ge 0$, so $f(0) = 0$ or $1$ If $f(0) = 0$, then $f(x) = 0$ for all $x$ for reasons stated above.
Now, we assume $f(0) = 1$. Observe that we have $f(f(0)) - f(0) = 0$, since $0 = f(0)-f(0)^2 = (f(f(0)) - f(0))^2$. so, $f(1) = 1$.
By $P(0,y)$, we conclude that $f(f(y))^2 + 1 + f(y)^2 = f(y) + 2f(f(y))f(y)$, so $f(y)-1 = (f(f(y))-f(y))^2 \ge 0$, so $f(y) \ge 1$ for all $y$.
Now, by $P(x,0)$, we conclude that $f(1-x)^2 + f(x)^2 + 1 = 1+2f(f(0)) = 1+2f(1) = 3$, so $f(1-x)^2 + f(x)^2 = 2$. But $f(1-x), f(x)$ are integers, so $f(1-x)^2 = 1, f(x)^2 = 1$. But we also know that $f(1-x), f(x) \ge 0$, hence we can conclude that $f(1-x) = f(x) = 1$ for all $x$.
Thus $f(x)$ is identically zero or identically 1, so we are done. $\blacksquare$
ZETA_in_olympiad
30.07.2022 15:59
Let $P(x,y)$ denote the given assertion. We claim that all solutions are $f\equiv 1$ and $f\equiv 0.$ Note that $P(0,0)$ implies $f(0)-f(0)^2=(f(f(0))-f(0))^2\geq 0$ hence $f(0)\in \{0,1\}.$ Obviously $f\equiv 0$ if $f(0)=0.$ Consider $f(0)=1.$ Then $P(0,0)$ gives $f(1)=0.$ Now $P(x,0)$ implies $f(x)^2+f(1-x)^2=2$ since working in $\mathbb{Z}$ we have $f(x)=\pm 1.$ Finally $P(x,y)$ implies $f(f(y))\in \{-2,1\}$ and thus $f(f(y))=1$ and we are done.
whencence
06.12.2024 21:20
Let \( P(x,y) \) denote the original assertion. \( P(0,0) \) gives \[ f(f(0))^2 + f(0)^2 + f(0)^2 = f(0)(1+f(f(0))). \]For simplicity, let \( f(0) = a \) and \( f(f(0)) = b \). Then the last equation becomes \[ b^2 - 2ab + 2a^2 - a = 0. \]Treat this as a quadratic equation in \( b \). Taking the determinant \( D = 4a^2 - 4(2a^2 - a) = 4a(1 - a) \). Since \( D \geq 0 \) and \( a(1 - a) \leq \frac{1}{4} \), we obtain \( a(1 - a) = 0 \). Hence, let's explore 2 cases: $\textbf{Case 1:}$ \( a = 0 \). In this case, \( f(0) = 0 \). Then \( P(x,0) \) gives \[ f(-x)^2 + f(x)^2 = 0 \quad \text{for all } x \in \mathbb{R}. \]This gives us a solution of \( \boxed{f(x) = 0} \) for all \( x \in \mathbb{R} \). $\textbf{Case 2:}$ \( a = 1 \). Then the quadratic equation above gives \( b = 1 \). So, \( f(0) = 1 \) and \( f(1) = 1 \). Again, \( P(x,0) \) gives \[ f(1 - x)^2 + f(x)^2 + 1 = 3 \quad \text{for all } x \in \mathbb{R}. \]Since we are in \( \mathbb{Z} \), we get \( \boxed{f(x) = 1} \) for all \( x \in \mathbb{R} \).