Iranian Mathematical Olympiad

Every polynomial in the form of $P(x,y)=A{x}^2+2Ay+Bx$ for some $A,B\in \mathbb{R}$.

Let $Q(a,b)=P(a,b^2)$ and define $Q_i(x,y)$ to be the homogeneous polynomial which consists of $i$-th degree coefficients of $Q(x,y)$. It is directly implied that $$Q_i(a,\sqrt{2bc})+Q_i(b,\sqrt{2ac})+Q_i(c,\sqrt{2ab})=Q_i(a+b+c,\sqrt{ab+bc+ac})$$ --- Let $A_i(a,b,c)$ denote the above equality. Plugging $c=0$ in this equality results in $$Q_i(a,0)+Q_i(b,0)+Q_i(0,\sqrt{2ab})=Q_i(a+b,\sqrt{ab}).$$ Define $x=a+b,y=\sqrt{ab}$. First of all, mind that $$Q_i(a,0)=a^i{Q}_i(1,0)$$ and $$Q_i(0,a)=a^i{Q}_i(0,1).$$ Secondly, the above equality concludes that $$Q_i(x,y)=(\sqrt{2y^2}{)}^n{Q}_i(0,1)+\left[{\left(\frac{x-\sqrt{x^2-4y^2}}{2}\right)}^i+{\left(\frac{x+\sqrt{x^2-4y^2}}{2}\right)}^i\right]Q_i(1,0)$$ Next, we state that if $i\ge 3$, $Q_i=0$. This means $P(x,y^2)$ is of degree at most 2 which then will result in $P(x,y)=A{x}^2+Bx+Cy+D$ and one can simply see that $D=0,C=2A$ and every polynomial of form $$P(x,y)=A{x}^2+Bx+2Ay$$ satisfies the problem. For the sake of contradiction, assume that $i\ge 3$. Note that $P(x,y^2)=Q(x,y)$, therefore every coefficient of $Q(x,y)$, and by extension, $Q_i(x,y)$, has $y$ to an even power. $$Q_i(x,y)=x^i{m}_0+x^{n-2}{y}^2{m}_2+m_4{x}^{n-4}{y}^4+\dots$$ By plugging this in $A_i(a,b,c)$ and evaluating the coefficient of $a^{n-1}b$, one can see the coefficient on left hand side to be zero, and on the right hand side $$\begin{array}{r}{m}_0(x+y+z{)}^i+m_2(xy+yz+xz)(x+y+z{)}^{i-2}+\dots \\ implies[x^{i-1}y]=\left(\genfrac{}{}{0ex}{}{i}{1}\right)m_0+\left(\genfrac{}{}{0ex}{}{i-2}{0}\right)m_2=0⟹m_2=-i{m}_0\end{array}$$ Furthermore, by evaluating the coefficient of $x^{i-2}yz$ we get $$\begin{array}{rl}[x^{i-2}yz]& =\left(\genfrac{}{}{0ex}{}{i-1}{1}\right)m_0+\left[1+\left(\genfrac{}{}{0ex}{}{i-2}{1}\right)+\left(\genfrac{}{}{0ex}{}{i-2}{1}\right)\right]m_2+\left(\genfrac{}{}{0ex}{}{2}{1}\right)m_4=2m_2\\ implies2m_4& =i{m}_0(2i-5)-i\times (i-1)m_0⟹m_4=\frac{i(i-4)}{2}{m}_0\end{array}$$ --- Moreover, let $$T(x,y)={\left(\frac{x-\sqrt{x^2-4y}}{2}\right)}^i+{\left(\frac{x+\sqrt{x^2-4y}}{2}\right)}^i.$$ Evaluate the coefficient of $x^i,x^{i-4}{y}^2$ in $T$. $$\begin{array}{rl}[x^i]& =T(1,0)=1,Q_i(x,y)\\ & =2^{\frac{i}{2}}{y}^i{Q}_i(0,1)+T(x,y^2)Q_i(1,0)\\ implies{m}_0& =Q_i(1,0)\end{array}$$ And (Evaluating coefficient in $Q_i$) $$\begin{array}{rl}[x^{i-2}{y}^2]& =\frac{1}{2}{T}_{yy}(1,0)=\frac{i(i-3)}{2}\\ implies{m}_4& =\frac{i(i-3)}{2}{Q}_i(1,0)\\ impliesa& =Q_i(1,0)=\frac{2m_4}{i(i-3)}=\frac{i(i-4)a}{i(i-3)}\end{array}$$ So either $i=3$, $i=4$ or $a=b=c=0$, otherwise we have a contradiction. In case of the latter two, $$\begin{array}{rl}{Q}_i(1,0)& =0\\ implies{Q}_i(x,y)& =2^{\frac{i}{2}}{y}^i{Q}_i(0,1)+0\\ implies{Q}_i(0,1)& =2^{\frac{i}{2}}{1}^i{Q}_i(0,1)\\ implies{Q}_i(0,1)(2^{\frac{i}{2}}-1)& =0\\ implies{Q}_i(0,1)& =0⟹Q_i(x,y)=0\end{array}$$ Therefore, the only remaining case is $i=3$, which is easily disproven. This concludes our proof. $\square$