Multiplication via squaring and addition Is it possible to define multiplication of two positive integers only using addition and squaring? Of course I have $5 \cdot 3 = 5 + 5 + 5$ but I would like something without saying do this $n$-times. 

Peano Arithmetic has the following two axioms:
  
  
*
  
*$x \cdot 0 = 0$
  
*$x \cdot y = x \cdot (y-1) + x$

So I could also write $3 \cdot 5 = 3 \cdot (5-1) + 3 = 3^2 + 3 + 3$ but again I don't "know" how often I need to apply the $2$nd axiom.
I tried a few things and noticed that one has:
$$2xy = (x+y)^2-x^2-y^2 \text{ and } 4xy = (x+y)^2-(x-y)^2$$ 
Close to $xy$ but still not what I am looking for. And actually this uses subtraction...
Edit: As clarified in the comments: I am asking how to define multiplication inside the structure $(\mathbb{N}, +, \cdot^2)$.
 A: Per your comment, the precise question you're asking is:

Is multiplication definable in the structure $(\mathbb{N}; +,\cdot^2)$?

The answer is yes: we have $z=x\cdot y$ iff $z+z+x^2+y^2=(x+y)^2$. 

This is a bit unsatisfying; can we do better?
Well, one natural hope would be for a specific term built out of $+$ and $\cdot^2$ which gives multiplication. E.g. raising to the fourth power isn't just definable, it's given by the term $(x^2)^2$. So we now ask:

Is the term $xy$ equivalent (in the obvious sense) to a term in the language $+,\cdot^2$? 

The answer to this new question is no. One way to see this is by taking derivatives. Suppose $t(x,y)$ is a term built out of $+$ and $\cdot^2$. Then when we write ${\partial\over\partial x}t(x,y)$ as a fully-cancelled-where-possible sum of monomials, every monomial in which $y$ occurs have even coefficient$^1$. But the monomial $xy$ itself doesn't have this property.$^2$ 

$^1$This takes proof, but it's a straightforward induction so I'll leave it to the reader.
$^2$OK fine, technically we need to prove that the fully-cancelled-sum-of-monomials form of a polynomial is unique, but meh - I'll leave it to the reader as well. Induction builds character.
