real analysis - Bijection from $mathbb{N}$ to $mathbb{N}$ s.t $phi(n)ne n^2$

Is it possible to find a bijection $\phi:\mathbb{N}\to \mathbb{N}$ such that $\forall n\in \mathbb{N} ,\quad\phi(n)\ne n^2$

If it is not, how to prove, $\forall \phi\in \mathcal{L(\mathbb{N})},\;\forall N\in \mathbb{N},\exists p>N, \text{s.t.}\quad\phi(p)=p^2$

(with $\mathcal{L(\mathbb{N})}$ the set of all bijections from $\mathbb{N}$ to $\mathbb{N}$

Answer

Yes. Put $\phi(1)=2, \phi(2)=1$ and $\phi(n)=n$ for all $n\geq 3$.