$P(1,1): f(1+f(1))=f(1)+2 >1+f(1)$ $P(2,1): f(1+f(2))=f(1)+3>1+f(2)$ since $f(1)+3>f(1)+2 \implies f(1+f(2))>f(1+f(1)) \implies $ if $f(m)>m$ then $f(M)>M$ for $M>m$ Thus, $f(1)+3>1+f(2) \implies f(1)+2>f(2) \implies f(1+f(1))>f(2) \implies 1+f(1)>2 \implies f(1)>1$ Using similar logic, we can find using $P(n,1)$ that $f(n)>n$. Thus, $f(n)+m+1 \geq n+f(m)+1 \implies f(n)-n \geq f(m)-m$ for all $n,m \in Z_+$ However, we can swap $m$ and $n$ to find that $f(m)-m \geq f(n)-n$. Thus, $f(m)=m+c$ and checking, we find $f(m)=m+1 \implies f(2006)=2007$