Let $P(m,n)$ be the assertion. $P(1,1)\implies 2f(1)\ge f(2)-2f(1)^2\ge 2f(1)^2\implies f(1)\ge f(1)^2\implies f(1)=1$ as $f(1)\in \mathbb{N}$ $P(m,1)\implies 2f(m)+f(m)^2+f(1)\ge f(m^2+1)\ge 2f(m)+f(m)^2+1$ $\implies (f(m)+1)^2\ge f(m^2+1)\ge (f(m)+1)^2\implies f(m^2+1)=(f(m)+1)^2$ $\implies f(m^2+1)=(f(m)+1)^2$. Take $m=1$ to get $f(2)=4$, Now,it is easy to get $f(5)=25$, $f(26)=676$.... As $ 2f(mn)\geq f(m^2+n^2)-f(m)^2-f(n)^2\geq 2f(m)f(n) $ From the LHS, RHS, we get, $f(mn)\ge f(m)f(n)\cdots (\spadesuit )$ $f(2.m^2)\ge f(2).f(m^2)\implies f(2m^2)\ge 4f(m^2)$ $P(m,m)\implies 2f(m^2)\ge f(2m^2)-2f(m)^2$ Multiplying this by $2$, we get, $\implies 4f(m^2)\ge 2f(2m^2)-4f(m)^2$ We have already proved $f(2m^2)\ge 4f(m^2)$ Therefore, $f(2m^2)\ge 2f(2m^2)-4f(m)^2\implies 4f(m)^2\ge f(2m^2)$ From $(\spadesuit )$, we have, $f(2m^2)\ge f(2)f(m^2)=4f(m)^2$ Hence, $f(m)^2\ge f(m^2)$ But $(\spadesuit )$ implies $ f(m^2)\ge f(m)^2$ Therefore, we have $f(m^2)=f(m)^2$ Considering $P(m,m)$, it is easy to see that $f(2m^2)=4f(m)^2$. Now, it is easy to see that $f(2^k)=2^{2k}\forall k\in\mathbb{N}$. in $(\spadesuit )$, $P(m,2m)\implies f(2m^2)=f(m)f(2m)$ But $f(2m^2)=4f(m)^2$ $\implies 4f(m)^2=f(m)f(2m)\implies f(2m)=4f(m)\forall m\in\mathbb{N}$ We have proved $(f(m)+1)^2=f(m^2+1)$ Take $m=3$ to get $(f(3)+1)^2=f(10)$ We have proved $f(5)=5$ and $f(2n)=4f(n)$, therefore $f(10)=4f(5)=100\implies f(10)=10$ $\implies (f(3)+1)^2=100\implies f(3)=9$ We have proved $f(1)=1,f(2)=4,f(3)=9,f(4)=16,f(5)=25$ We now claim $f(n)=n^2\forall n\in \mathbb{N}$ Let's say it is true till some positive integers $n$. We have already proved $f(2n)=4f(n)$, hence, if we know $f(n),$ we can find $f(2n)$. Let $m=2x+1$ be an odd integer. $(f(m)+1)^2=f(m^2+1)\implies (f(m)+1)^2=f(4x^2+4x+2)=2f(2x^2+2x+1)$ $=2f(x^2+(x+1)^2)$ Taking $P(x,x+1)$ in $ 2f(mn)\geq f(m^2+n^2)-f(m)^2-f(n)^2\geq 2f(m)f(n) $ We can find $f(x^2+(x+1)^2)$. $u_1=1,u_2=2, u_n=u_{n-1}^2+1$. We have proved that $f(m^2+1)=(f(m)+1)^2$ It is easy to show that for all the numbers in the sequence, $f(u_n)=u_n^2$. Taking $P(m,u_n)$ we can conclude $f(m^2+u_n^2)=(f(m)+u_n^2))^2\forall n\in\mathbb{N}$(All the numbers in this sequence) To complete the induction, we only have to prove the "odd integers case" Let $m=2x+1$ be an odd integer. Taking $n=1$, we get $u_n=1$ If we take $n=1$, it only remains to find the value of $f(x(x+1))$ Notice that taking even $n$ does not help, so we will take only odd $n'$s Taking $n=3$, we get $u_n=5$ We have proved $f(m^2+u_n^2)=(f(m)+u_n^2))^2\forall n\in\mathbb{N}$ Therefore $f(m^2+25)=(f(m)+25)^2$ Take $m=2x+1$, $f(4x^2+4x+26)=(f(2x+1)+25)^2$. $=2f(2x^2+2x+13)=2f((x+3)^2+(x-2)^2)$. Take $P(x+3,x-2)$ in the original inequality, it remains to find the value of $f(x^2+x-6)$. @Tima I think you understand now. Induction Complete. Therefore $f(n)=n^2\forall n\in\mathbb{N}\blacksquare$. BTW, what is SRMC?

Ashutoshmaths wrote: BTW, what is SRMC? Silk Road Mathematics Competition. Each March math students from Kyrgyzstan, Tajikistan and Kazakhstan participate in this olympiad in the city of Alma-ata. Other countries which located along the Great silk way: Uzbekistan, Turkey, Turkmenistan and Azerbaijan in absentia participates in this olympiad.

Ashutoshmaths wrote: Taking $P(x,x+1)$ in $ 2f(mn)\geq f(m^2+n^2)-f(m)^2-f(n)^2\geq 2f(m)f(n) $ We can find $f(x^2+(x+1)^2)$. Can you explain how you got $f(x^2+(x+1)^2)$ ?

Substitute $m=x$ and $n=x+1$ in the original inequality. You will see that $f(x^2+(x+1)^2)$ is between to equal squares, then you'll get $f(x^2+(x+1)^2)=(2x^2+2x+1)^2$.

Ashutoshmaths wrote: Substitute $m=x$ and $n=x+1$ in the original inequality. You will see that $f(x^2+(x+1)^2)$ is between to equal squares, then you'll get $f(x^2+(x+1)^2)=(2x^2+2x+1)^2$. why $f(x^2+x)=(x^2+x)^2$?

Complete the above solution. By inequality we can easily have f(m^2) = f(m)^2, f(2m) = 2f(m), f(m^2+1) = ((f(m)+1)^2 (*). Now by induction, we only need to prove for m odd. Consider m = 2x+1, let f(2x+1) = k by P(2x+1,2x-1) we have 4f(4x^2+1) >= (f(2x-1) + k)^2 so by (*) and inductive hypothesis we have (k-(2x+1)^2).A<=0 with A>0 by f:N->N so k<=(2x+1)^2 (1). Again, by P(2x+1,1): 2k=2f(2x+1) >= f(4x^2 + 4x + 2) - k^2 - 1 = 4f(x^2 + (x+1)^2) - k^2 - 1 hence (k+1)^2 >= 4B with B = f(x^2 + (x+1)^2). Finally, P(x,x+1) gives B >= (f(x) + f(x+1))^2 = (x^2 + (x+1)^2)^2 so k + 1 >= 2(2x^2 + 2x + 1) so k >= (2x + 1)^2 (2). Thus, by (1) and (2), we have the conclusion

Solution: Let $P(m,n)$ be the assertion. \[P\left( {1;1} \right) \Rightarrow f\left( 1 \right) = 1.\]\[P\left( {1,n} \right) \Rightarrow f\left( {{n^2} + 1} \right) = {\left[ {f\left( m \right) + 1} \right]^2},\,\,\forall n \in \,{\mathbb{Z}^ + } \Rightarrow f\left( 2 \right) = 4\]\[P\left( {n,n} \right) \Rightarrow 2f\left( {{n^2}} \right) \geqslant f\left( {2{n^2}} \right) - 2{f^2}\left( n \right) = f\left( {2.{n^2}} \right) - 2.f\left( {n.n} \right) \geqslant f\left( 2 \right).f\left( {{n^2}} \right) - 2.f\left( {{n^2}} \right) = 2f\left( {{n^2}} \right),\,\,\forall n \in \,{\mathbb{Z}^ + }\,\]\[ \Rightarrow f\left( {2{n^2}} \right) = 4f\left( {{n^2}} \right)\,;\,\,\,\,f\left( {{n^2}} \right) = {f^2}\left( n \right),\,\,\forall n \in \,{\mathbb{Z}^ + }\]\[P\left( {x,2{\text{x}}} \right) \Rightarrow 4{f^2}\left( x \right) = 4f\left( {{x^2}} \right) = f\left( {2{{\text{x}}^2}} \right) \geqslant f\left( {2{\text{x}}} \right).f\left( x \right) \geqslant 4{f^2}\left( x \right),\,\,\forall x \in {\mathbb{Z}^ + } \Rightarrow f\left( {2{\text{x}}} \right) = 4f\left( x \right),\,\,\forall x \in {Z^ + }\]\[P\left( {2,{n^2}} \right) \Rightarrow 2f\left( {{n^2}} \right).f\left( 2 \right) = 2f\left( {2{n^2}} \right) \geqslant f\left( {{n^4} + 4} \right) - {f^2}\left( {{n^2}} \right) - {f^2}\left( 2 \right) \geqslant 2f\left( {{n^2}} \right).f\left( 2 \right),\,\,\forall n \in {\mathbb{Z}^ + } \Rightarrow f\left( {{n^4} + 4} \right) = {\left[ {f\left( {{n^2}} \right) + 4} \right]^2},\,\,\forall n \in {\mathbb{Z}^ + }\]\[ \Rightarrow {f^2}\left( n \right) + 4 = \sqrt {f\left( {{n^4} + 4} \right)} = \sqrt {f\left( {\left( {{{\left( {n - 1} \right)}^2} + 1} \right).\left( {{{\left( {n + 1} \right)}^2} + 1} \right)} \right)} \geqslant \sqrt {f\left( {{{\left( {n - 1} \right)}^2} + 1} \right)} .\sqrt {f\left( {{{\left( {n + 1} \right)}^2} + 1} \right)} \geqslant \left[ {f\left( {n - 1} \right) + 1} \right]\left[ {f\left( {n + 1} \right) + 1} \right]\,\,,\,\,\forall n \in {\mathbb{Z}^ + }\,\,\,\,\,\,\,{\,^{(*)}}\]We will prove $f(n)=n^2$ by induction on $n$. At (*), put $n=2;3$ then $9\leqslant f(3) \leqslant 9$ so $f\left( n \right) = {n^2},\,\,\forall n = 1;2;3;4$ Assume that $f\left( i \right) = {i^2},\,\,\forall i \leqslant n - 1$ If $n$ is even so $f\left( n \right) = 4f\left( {\frac{n}{2}} \right) = {n^2}$ as $\frac{n}{2} \leqslant n - 1$ If $n$ is odd so $f\left( {n + 1} \right) = 4f\left( {\frac{{n + 1}}{2}} \right) = {\left( {n + 1} \right)^2}$. At (*) we have: \[\left\{ \begin{gathered} {f^2}\left( n \right) + 4 \geqslant \left[ {f\left( {n - 1} \right) + 1} \right]\left[ {f\left( {n + 1} \right) + 1} \right] \hfill \\ {f^2}\left( {n - 1} \right) + 4 \geqslant \left[ {f\left( {n - 2} \right) + 1} \right]\left[ {f\left( n \right) + 1} \right] \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} f\left( n \right) \geqslant {n^2} \hfill \\ f\left( n \right) \leqslant {n^2} \hfill \\ \end{gathered} \right. \Rightarrow f\left( n \right) = {n^2},\,\,\forall n \in {\mathbb{Z}^ + }\]