51st IMO Shortlisted Problems

The equality $x^2+7=x^2+2^2+1^2+1^2+1^2$ shows that $n\le 5$. It remains to show that $x^2+7$ is not a sum of four (or less) squares of polynomials with rational coefficients. Suppose by way of contradiction that $x^2+7=f_1(x{)}^2+f_2(x{)}^2+f_3(x{)}^2+f_4(x{)}^2$, where the coefficients of polynomials $f_1,f_2,f_3$ and $f_4$ are rational (some of these polynomials may be zero). Clearly, the degrees of $f_1,f_2,f_3$ and $f_4$ are at most $1$. Thus $f_i(x)=a_{i}x+b_i$ for $i=1,2,3,4$ and some rationals $a_1,b_1,a_2,b_2,a_3,b_3,a_4,b_4$. It follows that $x^2+7={\sum }_{i=1}^4{\left(a_{i}x+b_i\right)}^2$ and hence $$\begin{array}{c}\sum _{i=1}^4{a}_i^2=1,\sum _{i=1}^4{a}_i{b}_i=0,\sum _{i=1}^4{b}_i^2=7.\end{array}\text{\hspace{1em}(1)}$$ Let $p_i=a_i+b_i$ and $q_i=a_i-b_i$ for $i=1,2,3,4$. Then $$\begin{array}{rl}\sum _{i=1}^4{p}_i^2& =\sum _{i=1}^4{a}_i^2+2\sum _{i=1}^4{a}_i{b}_i+\sum _{i=1}^4{b}_i^2=8,\\ \sum _{i=1}^4{q}_i^2& =\sum _{i=1}^4{a}_i^2-2\sum _{i=1}^4{a}_i{b}_i+\sum _{i=1}^4{b}_i^2=8\\ \text{ and }\sum _{i=1}^4{p}_i{q}_i& =\sum _{i=1}^4{a}_i^2-\sum _{i=1}^4{b}_i^2=-6,\end{array}$$ which means that there exist a solution in integers $x_1,y_1,x_2,y_2,x_3,y_3,x_4,y_4$ and $m>0$ of the system of equations $$\text{ (i) }\sum _{i=1}^4{x}_i^2=8m^2,\text{ (ii) }\sum _{i=1}^4{y}_i^2=8m^2,\text{ (iii) }\sum _{i=1}^4{x}_i{y}_i=-6m^2$$ We will show that such a solution does not exist. Assume the contrary and consider a solution with minimal $m$. Note that if an integer $x$ is odd then $x^2\equiv 1(\mathrm{mod}8)$. Otherwise (i.e., if $x$ is even) we have $x^2\equiv 0(\mathrm{mod}8)$ or $x^2\equiv 4(\mathrm{mod}8)$. Hence, by (i), we get that $x_1,x_2,x_3$ and $x_4$ are even. Similarly, by (ii), we get that $y_1,y_2,y_3$ and $y_4$ are even. Thus the LHS of (iii) is divisible by $4$ and $m$ is also even. It follows that $\left(\frac{x_1}{2},\frac{y_1}{2},\frac{x_2}{2},\frac{y_2}{2},\frac{x_3}{2},\frac{y_3}{2},\frac{x_4}{2},\frac{y_4}{2},\frac{m}{2}\right)$ is a solution of the system of equations (i), (ii) and (iii), which contradicts the minimality of $m$.

Solution 2:

We prove that $n\le 4$ is impossible. Define the numbers $a_i,b_i$ for $i=1,2,3,4$ as in the previous solution. By Euler's identity we have $$\begin{array}{rl}\left(a_1^2+a_2^2+a_3^2+a_4^2\right)\left(b_1^2+b_2^2+b_3^2+b_4^2\right)& ={\left(a_1{b}_1+a_2{b}_2+a_3{b}_3+a_4{b}_4\right)}^2+{\left(a_1{b}_2-a_2{b}_1+a_3{b}_4-a_4{b}_3\right)}^2\\ & +{\left(a_1{b}_3-a_3{b}_1+a_4{b}_2-a_2{b}_4\right)}^2+{\left(a_1{b}_4-a_4{b}_1+a_2{b}_3-a_3{b}_2\right)}^2\end{array}$$ So, using the relations (1) from the Solution 1 we get that $$\begin{array}{c}7={\left(\frac{m_1}{m}\right)}^2+{\left(\frac{m_2}{m}\right)}^2+{\left(\frac{m_3}{m}\right)}^2\end{array}\text{\hspace{1em}(2)}$$ where $$\begin{array}{rl}& \frac{m_1}{m}=a_1{b}_2-a_2{b}_1+a_3{b}_4-a_4{b}_3\\ & \frac{m_2}{m}=a_1{b}_3-a_3{b}_1+a_4{b}_2-a_2{b}_4\\ & \frac{m_3}{m}=a_1{b}_4-a_4{b}_1+a_2{b}_3-a_3{b}_2\end{array}$$ and $m_1,m_2,m_3\in \mathbb{Z},m\in \mathbb{N}$. Let $m$ be a minimum positive integer number for which (2) holds. Then $$8m^2=m_1^2+m_2^2+m_3^2+m^2$$ As in the previous solution, we get that $m_1,m_2,m_3,m$ are all even numbers. Then $\left(\frac{m_1}{2},\frac{m_2}{2},\frac{m_3}{2},\frac{m}{2}\right)$ is also a solution of (2) which contradicts the minimality of $m$. So, we have $n\ge 5$. The example with $n=5$ is already shown in Solution 1.