Find all functions $f:\Bbb R\mapsto \Bbb R$ satisfying $$f(x+y)f(x-y)=2x+f(x^2-y^2)$$
I let $a=x+y$ and $b=x-y$ and noticed that $f(a)f(b)=a+b+f(ab)$. It's obvious now that $f(n)=n+1$ satisfies, for all real $n$.
How do I show this more rigorously, and that no other functions satisfy?
Below is my slightly more rigorous approach.
I let $y=0$ so $f(x)^2=2x+f(x^2)$ and let $f(x)=p+q$ for real $p$ and $q$. We see that $$2pq+p^2+q^2=2x+f(x^2).$$ Since $2pq$ and $2x$ have the same integer coefficient, letting $p=kx$ and $q=\frac 1k$ for some $k$ implies $pq=x$ so the $2x$ terms cancel out, thus $f(x)=kx+\frac 1k$ or $$k^2x^2+\frac 1{k^2}=kx^2+\frac 1k$$ for all $x$. Since this must be valid for all $x$, it must be valid for $x=1$, so substituting that neatly removes the $x$ variable. Trivially $k\neq 0$ and upon solving for $k$, since $f:\Bbb R\mapsto\Bbb R$ then $k$ must be real, and we can thus show that $k=1$ can only satisfy, yielding $f(x)=x+1$.
However, although this approach is more rigorous, it assumes $pq=x$ out of elegant consequence. So I tried a different approach:
Let $x=y=0$, then we can see $f(0)^2=f(0)$ thus $f(0)\in\{0,1\}$. Now let $x=y$ and we have $$f(2x)f(0) = 2x+f(0).$$ This must be true for all $x$ but plugging $f(0)=0$ wouldn't make the statement true for all $x$, so $f(0)=1$ and therefore $f(2x)=2x+1$ or, through $2x\mapsto x$, we have $f(x)=x+1$.
This approach looks good to me, but I don't know how to show no other functions exist to satisfy the problem. Since $f(x+m)-f(x)=m$ which is constant for any $x$, I thought of employing this property to attempt showing that $f$ is linear. But I think I could only pull this off if $f:\Bbb Z\mapsto \Bbb Z$ instead.
Any thoughts?
Thanks.
This is much simpler. Put $x=y=0$ to see that $f(0)^{2}=f(0)$. Hence $f(0)=0$ or $1$. But if $f(0)=0$ we get a contradiction by putting $y=x$ so we must have $f(0)=1$. Now put $x=y$ to get $f(2x)=2x+1$ for all $x$. Hence $f(x)\equiv x+1$.