Find $g(x)$ from functional equation involving integrals 
If $g(x)$ is continuous function in $[0,\infty)$ satisfying
  $g(1)=1$ and $$\int _{0}^{x}{2xg^{2}(t)\,dt} =\left(\int _{ 0 }^{ x }{
2g(x-t)\,dt} \right)^2$$Find $g(x)$.

I differentiated both sides w.r.t $x$. But what to do next? I'm stuck!
 A: I don't know how to find all solutions, but it so happens that
$$g(x):=x^{1+\sqrt{2}}\qquad(x\geq0)$$
satisfies the conditions.
How I got that? I remarked that the "Ansatz" $g(x):=x^\alpha$ would lead to a homogeneous identity in $x$. It then only remained to determine the "characteristic value" $\alpha$.
A: Here's some progress.
A simple change of variable shows that the original equality may be rewritten as $\displaystyle \forall x\geq 0, x\int _{0}^{x}{g^{2}(t)\,dt} =2\left(\int _{ 0 }^{ x }{g(t)\,dt} \right)^2$
Differentiating both sides of this equality yields $$\forall x\geq 0, x\int_0^x{g^2(t)} dt +(xg(x))^2=4xg(x)\int_0^x{g(t)} dt$$
Combining both equalities we have $$2\left(\int _{ 0 }^{ x }{g(t)\,dt} \right)^2 +(xg(x))^2=4xg(x)\int_0^x{g(t)} dt$$
Setting $F(x) = \int_0^x g(t)dt$, we have $$\forall x\geq 0, 2F^2(x)+x^2(F'(x))^2=4xF(x)F'(x)$$
which luckily factors as $$ (xF'(x)-(2-\sqrt2)F(x))(xF'(x)-(2+\sqrt2)F(x))=0$$
It would be nice to infer that either $\forall x, F'(x)-(2-\sqrt2)F(x)=0$ or $\forall x, F'(x)-(2+\sqrt2)F(x)=0$ which gives Christian Blatter's solution, but that seems difficult.
