69th Belarusian Mathematical Olympiad

Answer: 3.

Note that the constant term of each $P_i$, $i=1,\dots ,7$, equals to zero: $$0=Q(0,\dots ,0)=P_1(0,\dots ,0{)}^2+P_2(0,\dots ,0{)}^2+\cdots +P_7(0,\dots ,0{)}^2,$$ whence each $P_i(0,\dots ,0)=0$. Moreover, the degree of each $P_i$ does not exceed 1. Indeed, if the degree of some $P_i$ is greater than 1, then $P_i$ contains the monomial of the form $a\cdot x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}\dots x_7^{{\alpha }_7}$, where ${\alpha }_1+\cdots +{\alpha }_7>1$, but in this case the coefficient of $Q$ at $x_1^{2{\alpha }_1}{x}_2^{2{\alpha }_2}\dots x_7^{2{\alpha }_7}$ is not less than $a^2>0$, since the coefficients of all other summands of $P_1^2+P_2^2+\cdots +P_7^2$ at $x_1^{2{\alpha }_1}{x}_2^{2{\alpha }_2}\dots x_7^{2{\alpha }_7}$ are nonnegative. And this contradicts to $\mathrm{deg}Q=2$.

Since each $P_i$ has the degree not exceeding 1 and no constant term, we can write $P_i=a_{i1}{x}_1+a_{i2}{x}_2+\cdots +a_{i7}{x}_7$.

Consider the equality $3=Q(1,0,0,\dots ,0)$, it can be written as $$P_1(1,0,0,\dots ,0{)}^2+P_2(1,0,0,\dots ,0{)}^2+\cdots +P_7(1,0,0,\dots ,0{)}^2=a_{11}^2+a_{21}^2+\cdots +a_{71}^2.$$ The coefficients $a_{11},a_{21},\dots ,a_{71}$ are nonnegative integers, therefore, exactly three of them are equal 1, and the rest are equal 0. Similarly for each $i=1,\dots ,7$ the coefficient at $x_i$ of exactly three polynomials equals 1 and the coefficient at $x_i$ of the rest four polynomials equals 0. Hence the sum of all coefficients of polynomials $P_1,P_2,\dots ,P_7$ equals $3\cdot 7=21$. Therefore, $$P_1(1,\dots ,1)+P_2(1,\dots ,1)+\cdots +P_7(1,\dots ,1)=3\cdot 7=21.$$ Note that the sum of seven numbers $P_1(1,\dots ,1),\dots ,P_7(1,\dots ,1)$ equals to 21 wherein the sum of their squares equals to $Q(1,\dots ,1)=63$, and $\sqrt{\frac{63}{7}}=3=\frac{21}{7}$. Hence the arithmetic and geometric means of these numbers are equal. Therefore, all these numbers are equal i.e. $P_1(1,\dots ,1)=P_2(1,\dots ,1)=\cdots =P_7(1,\dots ,1)=21/7=3$.