Let $P(x,y)$ be the assertion $f(y+f(x))^2=f(y)^2+f(8xy)+8x^2f(x^2)$
$P(0,y); f(y+f(0))^2=f(y)^2+f(0)$
We can easily see that $f(0)=0$, since if $f(0) \neq 0$, we can prove by induction that
$f(nf(0))^2=f(0)^2+nf(0)$
when $n$ is an integer
and we can choose $n$ so that $nf(0)<-f(0)^2$ so that make $f(nf(0))^2=f(0)^2+nf(0)<0$, a contradiction!
So $f(0)=0$
$P(x,0); f(f(x))^2=8x^2f(x^2) \quad (1)$
$f(x^2) \ge 0$
We see that for $x>0$, $f(x)\ge 0$
We can rewrite $P(x,y)$ as $f(y+f(x))^2=f(y)^2+f(8xy)+f(f(x))^2$
If there are $x_0 \neq 0$ and $f(x_0)=0$ then $f(y)^2=f(y)^2+f(8x_0y)$
or $f(8x_0y)=0$
then we get for any $x$, $f(x)=0$ then it is finished
So suppose for $x \neq 0$, $f(x) \neq 0$, in particular when $x>0, f(x)>0$
Substitute $y$ with $-f(x)$ in $P(x,y)$
$0=f(-f(x))^2+f(f(x))^2+f(-8xf(x))$
for $x<0$, $f(x) \neq 0$, $f(f(x)) \neq 0$
$f(-8xf(x))<0$
$-8xf(x)<0$
$f(x)<0$
for $x<0, f(x)<0$
Let $x \neq 0$
From $(1); f(f(x))^2=f(f(-x))^2$ but $f(f(x))$ and $f(f(-x))$ have different signs
so $f(f(-x))+f(f(x))=0$
$P(f(x),y); \quad f(y+f(f(x)))^2=f(y)^2+f(8f(x)y)+f(f(f(x)))^2$
$P(f(-x),y+f(f(x)));\quad f(y)^2=f(y+f(f(x)))^2+f(8f(-x)(y+f(f(x))))+f(f(f(-x)))^2$
Add the two equations together
$f(8f(x)y)+f(8f(-x)(y+f(f(x))))+f(f(f(x)))^2+f(f(f(-x)))^2=0$
Substitute $x$ with $-x$ and $y$ with $y+f(f(x))$
$f(8f(x)(y+2f(f(x))))+f(8f(-x)(y+f(f(x))))+f(f(f(x)))^2+f(f(f(-x)))^2=0$
$f(8f(x)y)=f(8f(x)(y+2f(f(x))))$
Set $y=-f(f(x))$
$f(-8f(x)f(f(x)))=f(8f(x)f(f(x)))$
but they should have different signs so, a contradiction
therefore $f(x)=0$ for all $x$
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