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#1 h Sep 23, 2021, 6:05 PM

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Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfying:
$f(x^2+y+f(y))=f^2(x)$ for all real numbers $x$ and $y$

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#2 h Sep 24, 2021, 7:48 AM • 3 Y

Y by teomihai, KazukoHashima, AlexCenteno2007

KazukoHashima wrote:

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfying:
$f(x^2+y+f(y))=f^2(x)$ for all real numbers $x$ and $y$

Let $P(x,y)$ be the assertion $f(x^2+y+f(y))=f(x)^2$

If $f(y)<0$ for some $y$ , $P(\sqrt{-f(y)},y)$ $\implies$ $f(y)=f\left(\sqrt{-f(y)}\right)^2\ge 0$ , contradiction.
So $f(x)\ge 0$ $\forall x$

Comparing $P(x,y)$ with $P(-x,y)$ , we get $f(x)^2=f(-x)^2$ and so $f(-x)=f(x)$ (since both are nonnegative).

Comparing $P(x,y)$ with $P(x,-y)$ , we get $f(x^2+f(y)-y)=f(x^2+f(y)+y)$
And so assertion $Q(x,y)$ : $f(x)=f(x+2y)$ $\forall x\ge f(y)-y$ $\forall y\ge 0$

Let $A=f(\frac 12)-\frac 12$ : $Q(x,\frac 12)$ $\implies$ $f(x)=f(x+1)$ $\forall x\ge A$
And so $f(x)=f(x+n)$ $\forall x\ge A$ and $\forall n\in\mathbb Z_{\ge 0}$

Let $u>v>A$ and $B=f(\frac {u-v}2)-\frac{u-v}2$ : $Q(x,\frac{u-v}2)$ $\implies$ $f(x)=f(x+u-v)$ $\forall x\ge B$

Let then $n\in\mathbb Z_{\ge 0}$ such that $v+n\ge B$ : $f(v+n)=f(v+n+u-v)=f(u+n)$
But $f(u+n)=f(u)$ and $f(v+n)=f(v)$

And so $f(u)=f(v)$ $\forall u>v>A$

Let then $x>y\ge 0$ :
$P(x,A+1)$ $\implies$ $f(x)^2=f(x^2+A+1+f(A+1))$
$P(y,A+1)$ $\implies$ $f(y)^2=f(y^2+A+1+f(A+1))$
And since $x^2+A+1+f(A+1)>y^2+A+1+f(A+1)>A$ , we have $f(x^2+A+1+f(A+1))=f(y^2+A+1+f(A+1))$

And so $f(x)^2=f(y)^2$ $\forall x>y\ge 0$
So $f(x)=f(y)$ $\forall x>y\ge 0$ (since both are nonnegative)
So $f(x)=f(y)$ $\forall x,y$ (since $f(x)$ is even)

And so solutions
$\boxed{\text{S1 : }f(x)=0\quad\forall x}$ and $\boxed{\text{S2 : }f(x)=1\quad\forall x}$
Which indeed both fit.

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