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#1 h Apr 6, 2025, 6:08 PM • 2 Y

Y by PikaPika999, Fishheadtailbody

Find all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying the equation
$(f(x+y))^2= f(x^2) + f(2xf(y) + y^2), \quad \forall x, y \in \mathbb{R}.$

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#2 h Apr 6, 2025, 7:31 PM • 2 Y

Y by PikaPika999, hanzo.ei

This is actually very difficult.

It is clear that all three functions f(x) = 0 , f(x) = 2 and f(x) = x satisfy the equation.
Let P(x, y) represent the statement.
P(0,0) gives f(0)^2 = 2f(0) , so f(0) = 0 or 2 .
Let first discuss when f(0) = 2 .
P(x,0) gives f(x)^2 = f(x^2) + f(4x) ,
while P(0,x) gives f(x)^2 = f(0) + f(x^2) .
We got f(4x) = f(0) . We have found our constant solutions.

Now, f(0) = 0 .
P(x,0) gives f(x)^2 = f(x^2) .
We force (x+y)^2 = 2xf(y)+y^2 ,
P(2f(y)-2y,y) gives f((2f(y)-2y))^2 = 0 after simplification.
We hope to know something about the zero set of f , ideally, it only contains 0 .
If f(r) = 0 for some real number r \neq 0 , then so is r^2 .
P(x,r) gives f(x+r)^2 = f(x)^2 .
This can be inducted to all nr where n \in \mathcal{Z}, f(x+nr)^2 = f(x)^2 .
By P(x,nr), 2xf(nr) +n^2r^2 is also zero of the function.
If one of the nr is not zero of the function, 2xf(nr) can run through all real numbers, meaning f(x) = 0, which we already found.
Assume that there is a real number s that f(s) \neq 0 .

Maybe we can consider P(x+y,z) and P(x+z,y) .
Their LHS are the same, comparing the RHS, we have
f((x+y)^2)+f(2(x+y)f(z)+z^2) = f((x+z)^2)+f(2(x+z)f(y)+y^2) .

I still need some time to proceed. Maybe next time or someone can bump me. Also, new user cannot pose LaTeX.

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#3 h Apr 6, 2025, 10:55 PM • 2 Y

Y by Funcshun840, hanzo.ei

Denote $P(x,y)$ the assertion of the given F.E.
$P(0,0)$ gives $f(0)^2=2f(0)$ so $f(0)=0$ or $f(0)=2$ .
Case 1: $f(0)=2$ .
In this case $P(x,0)$ gives that $f(x)^2=f(x^2)+f(4x)$ and $P(0,x)$ gives $f(x)^2=2+f(x^2)$ which finishes as it gives $f(4x)=2$ for all reals $x$ and by shifting this gives $f(t)=2$ for all reals $t$ .
Case 2: $f(0)=0$ .
In this case $P(x,0)$ gives $f(x)^2=f(x^2)$ , now notice we want to have $x^2+2xy=2xf(y)$ which can be done when $x=0$ or $x=2f(y)-2y$ , in the later case we get that $f(2f(y)-2y)=0$ for all reals $y$ . Now suppose there existed $c \ne 0$ such that $f(c)=0$ , let $S$ the set of zeroes of $f$ then clearly $-c \in S$ and thus $\pm c^{2^n} \in S$ for all integers $n$ (we can wlog $c>0$ btw so dw).
Now $P(x,-2x)$ gives that $f(2x(f(-2x)+2x))=0$ which gives that $f(t(f(t)-t))=0$ for all reals $t$ . Now $P(x,c)$ gives that $f(x+c)^2=f(x)^2$ , now notice we had that $f(x)>0$ for all $x>0$ and thus for $x>0$ this implies that $f(x+c)=f(x)$ therefore in fact $f(x)=f(x+kc)$ for all positive integers $k$ and positive reals $x$ , this gives $nc \in S$ for all integers $n$ .
In fact we have that $S$ is closed under addition, also if $f$ had period $T$ on positive reals then $f(T)=0$ must hold otherwise $f(x)=0$ for all positive reals $x>T^2$ and then clearly this is sufficient to get it for all reals eventually, but we can also notice that $P(x+c,y)-P(x,y)$ gives that $f(2xf(y)+y^2)=f(2xf(y)+y^2+cf(y))$ for all reals $x,y$ now if $y \in S$ this is trivially true otherwise if $y \not \in S$ then we have $f(t)=f(t+cf(y))$ for all reals $t$ and the given $y$ which shows that for some $c \in S$ we have $cf(y) \in S$ for $y \not \in S$ but since it is trivial on the othercase we can just say $cf(y) \in S$ for any $c \in S$ and $y$ real.
So we have that if $c \in S$ then $-c \in S$ , and also $S$ is closed under addition and under squaring and taking square root for positive values, and also the latest thing we got.
Now we could have $2xy+y^2=c$ for some $c \in S$ then $f \left(\frac{(c-y^2)f(y)}{y}+y^2 \right)=0$ for all $y \ne 0$ .
Now $f(1)=f(1)^2$ as well so $f(1)=0$ or $f(1)=1$ , if $f(1)=0$ then $1 \in S$ and thus $f(n)=0$ for all integers $n$ and we also have $f(f(x))=0$ for all reals $x$ and thus $f(2f(x))=0$ for all reals $x$ , however remember that that zeroes of $f$ are additive so $2x=2f(x)+2x-2f(x)$ is also in $S$ for all reals $x$ and thus $f$ is zero everywhere, so now suppose $f(1)=1$ .
now it shows that for any real $x$ we have $f(x)=f(x+c)$ for any $c \in S$ and thus now consider $f(-1)^2=1$ , if it were $1$ then $f(1)=f(-1)$ and thus $f(4)=0$ which shows $f(4f(x))=0$ for all reals $x$ and then a similar thing can be applied were we find $4x \in S$ for all reals $x$ and thus $f(x)=0$ everywhere which here is a contradiction, so instead we must have $f(-1)=-1$ .
Now we must recall the thing found above, the $f \left(\frac{(c-y^2)f(y)}{y}+y^2 \right)=0$ for all $y \ne 0$ . Since $y^2-yf(y)$ for all reals $y \ne 0$ is in $A$ we NOW can jump and conclude that $f \left(\frac{cf(y)}{y} \right)=0$ for all $c \in S$ and $y \ne 0$ .
And to finish with style we now use full additiviness given in order to find out that if $c,d \in S$ then $c^2, d^2 \in S$ but also $(c+d)^2 \in S$ and thus $(c+d)^2-c^2-d^2=2cd \in S$ which comes really close from showing we have multiplicativiness on $S$ .
Now $P \left(\frac{c}{2}, x \right)$ gives that $f \left(\frac{c}{2}+x \right)^2=f \left(\frac{c}{2} \right)^2+f(x)^2$ and then setting $x=-\frac{c}{2}$ here for $c \in S$ gives that $f \left(\frac{c}{2} \right)=0$ and therefore if $c \in S$ then $0.5c \in S$ and this combined with the previous thing give that $S$ is multiplicative.
Now consider $2xz+z^2=2xy+y^2$ which holds on the case of $y \ne z$ when $2x=-y-z$ which gives that $f(y^2-(y+z)f(y))=f(z^2-(y+z)f(z))$ for all reals $y,z$ (cause obviously it is true when $y=z$ ) and due to the periodicness with elements of $S$ found previously this gives that $f(-zf(y))=f(-yf(z))$ for all reals $y,z$ , from $z=-1$ we can conclude that $f(y)=f(f(y))$ , however because $S$ is multiplicative we have that that $cx=cx-cf(x)+cf(x) \in S$ for all reals $x$ and therefore from here we can conclude $f$ is zero everywhere unless $S$ is only $0$ .
So now if $f$ was injective at $0$ then $f(x)=x$ for all reals $x$ which works. Since all solutions have been found, we are done :cool:

:cool:

.

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#4 h May 15, 2025, 4:44 AM

Y by

hanzo.ei wrote:

Find all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying the equation
$(f(x+y))^2= f(x^2) + f(2xf(y) + y^2), \quad \forall x, y \in \mathbb{R}.$

funny solution using the obscure fact that any real number is expressible in a form like $f(a)+f(b)+f(c)-a-b-c$ .

Let $P(x,y)$ be the assertion $f(x+y)^2=f\left(x^2\right)+f\left(2xf(y)+y^2\right)$ . Let $g(x)=f(x)-x$ , obviously $\boxed{f(x)=x}$ is a solution, otherwise we'll look for solutions such that there exists some $j\in\mathbb R$ with $g(j)\ne0$ . Let $S=\{x\in\mathbb R\mid f(x)=0\}$ .

$P(0,0)\Rightarrow f(0)\in\{0,2\}$
If $f(0)=2$ , we have:
$P(x,0)\Rightarrow f(x)^2=f\left(x^2\right)+f(4x)$
$P(0,x)\Rightarrow f(x)^2=f\left(x^2\right)+2$
So $f(4x)=2$ for all $x$ , we get the solution $\boxed{f(x)=2}$ for all $x$ which indeed fits.
Otherwise, consider the case $f(0)=0$ .

$P(x,0)\Rightarrow f\left(x^2\right)=f(x)^2$
Note now that $S$ is closed under addition, since if $u,v\in S$ then $f\left(u^2\right)=f\left(v^2\right)=0$ as well so $P(u,v)\Rightarrow f(u+v)=0$ . It's also closed under subtraction, since if $u\in S$ then $f(-u)^2=f\left(u^2\right)=f(u)=0$ . Writing in terms of $g$ , $P(x,y)$ becomes:
$g((x+y)^2)=g\left(x^2\right)+g\left(2xf(y)+y^2\right)+2xg(y),$ and taking $y=j$ we find that every real number $k$ is expressible as $k=2g(a)-2g(b)-2g(c)$ for some $a,b,c\in\mathbb R$ depending on $k$ .
$P(2g(x),x)\Rightarrow 2g(x)\in S$ , so by closure under subtraction we have $k=2g(a)-2g(b)-2g(c)\in S$ for any $k\in\mathbb R$ , that is, $\boxed{f(x)=0}$ for all $x$ which fits.

This post has been edited 2 times. Last edited by jasperE3, May 15, 2025, 4:47 AM

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