I'll write my attempt to solve this question any (???) mark means i'm not sure that it is correct.
So first i assumed that A and B is densely defined.(It is not written in the queestion but in order to $A^*$ to be well defined we need it but maybe it can be done without assuming this(??)
Proof $\Leftarrow$ :
Assume $A^*A=B^*B$. By this 2 links $A^*A$ is self-adjoint(1), $A^*A$ is self-adjoint(2) we know form domain of $D(A)=\mathfrak{Q}(A^*A)=\mathfrak{Q}(B^*B)=D(B)$, using definitions of $A^*$ and $B^*$ we obtain (where $x\in D(A^*) $ or $ D(B^*)$)
$$ (A^*x,y)=(x,Ay) \enspace \forall y \in D(A) $$
$$ (B^*x,y)=(x,By) \enspace \forall y \in D(B) $$
now choose $x\in D(A^*A)$ and $y=x$ then we have
$$(A^*Ax,x)=(Ax,Ax) \quad \text{also} \quad (A^*Ax,x)=(B^*Bx,x)=(Bx,Bx)$$
So this shows $(Bx,Bx)=(Ax,Ax) \enspace \forall x\in D(A^*A)=D(B^*B)$ But we want this to be true for all $z\in D(A)=D(B)$. To do this(??) let $z\in D(A)$ then it can be approximated via $a_n\in D(A^*A)$ with graph norm because $D(A)$ is closed with graph norm ($\|x\|+\|Ax\|$) and $D(A^*A)$ is dense in $D(A)$(With graph norm of course)
Now to see $(Az,Az)=(Bz,Bz)$ (??)
$$(Az,Az)=\lim_{n\rightarrow \infty}(Aa_n,Aa_n)=\lim_{n\rightarrow \infty}(Ba_n,Ba_n)=(Bz,Bz)$$
Proof $\Rightarrow$ :
Now we assume otherwise.Let $x\in D(A^*A)$ then because $A^*A$ and $B^*B$ is self-adjoint $Ran(A^*A+1)=\mathfrak{H}$ ($\mathfrak{H}$=Whole hilbert space) then,
$$\langle(A^*A+1)x,y\rangle=\langle x,y \rangle + \langle Ax,Ay \rangle \enspace \forall y\in D(A)$$Also $\exists z \in D(B^*B)$ s.t. $(A^*A+1)x=(B^*B+1) z $
I also claimed (???) $\langle x,x \rangle + \langle Ax,Ax \rangle = \langle x,x \rangle + \langle Bx,Bx \rangle $ implies $\langle x,y \rangle + \langle Ax,Ay \rangle = \langle x,y \rangle + \langle Bx,By \rangle $ (By polarization of quadratic form or what not(??) )
Which gives me
$$\langle x,y \rangle +\langle Ax,Ay \rangle = \langle x,y \rangle +\langle Bx,By \rangle=\langle z,y \rangle +\langle B^*Bz,y \rangle \enspace \forall y\in D(B)$$
After this i claimed $z=x$(??) therefore $A^*Ax=B^*Bx$ and because they are self-adjoint $A^*A\subseteq B^*B \Rightarrow A^*A=B^*B$
I wrote the proof because i wanted to show my work maybe it is true maybe it is false i don't know.
Because after thinking about one question for a long time I can't be sure of myself even if the proof is perfectly right
Thank you for your answers in advance :)
