Prove that there exists just one function $ f:\mathbb{N}^2\longrightarrow\mathbb{N} $ which simultaneously satisfies: $ \text{(1)}\quad f(m,n)=f(n,m),\quad\forall m,n\in\mathbb{N} $ $ \text{(2)}\quad f(n,n)=n,\quad\forall n\in\mathbb{N} $ $ \text{(3)}\quad n>m\implies (n-m)f(m,n)=nf(m,n-m), \quad\forall m,n\in\mathbb{N} $