Problems from Ukrainian Authors

Yes, you can.

Number the columns from left to right and the rows from top to bottom with the numbers $1,2,\dots ,2022$. Write in the cell $(r,c)$, i.e. the one in the $r$-th row and $c$-th column, the number $x^{2(r+c)}$, where the number $x$ is selected later.

Consider an arbitrary rectangle $R$, which lies in rows with numbers $s,s+1,\dots ,t$ and columns with numbers $u,u+1,\dots ,v$. By construction, the sum of all numbers in the rectangle $R$ is equal to $$(x^{2s}+x^{2s+2}+\cdots +x^{2t})\cdot (x^{2u}+x^{2u+2}+\cdots +x^{2v})=x^{2s+2u}\cdot \frac{x^{2a}-1}{x^2-1}\cdot \frac{x^{2b}-1}{x^2-1},$$ where $a=t-s+1$ and $b=v-u+1$. It is immediately clear that if $a=b$, the sum of numbers in $R$ is the perfect square of any real number $x$. Now it remains to find such an $x$ that no matter how $a\ne b$, where $a,b\in \{2,3,\dots ,2021\}$, the number $(x^{2a}-1)(x^{2b}-1)$ would not be a perfect square.

Proving. Let $P(x)=x^{2n}+a_{2n-1}{x}^{2n-1}+\cdots +a_{1}x+a_0$. We can immediately see that starting from some point $P(x)$ is positive, because its highest coefficient is $1$. In the following, we will consider only $x$ from this point onwards. First, let's find the following polynomial $Q(x)=x^n+b_{n-1}{x}^{n-1}+\cdots +b_{1}x+b_0$ with rational coefficients such that the degree of the polynomial $P(x)-Q(x{)}^2$ is less than $n$. To do this, it is enough to restore the coefficients one by one $b_{n-1},\dots ,b_1,b_0$, using identities: $$\sum _{k=n}^{2n-1}\sum _{i+j=k}{b}_i{b}_j{x}^k=a_k{x}^k.$$ Thus, we can write the equality $P(x)=Q(x{)}^2+R(x)$, $\mathrm{deg}R<n$. Let's multiply both parts of this equality by $M^2$ so that the polynomials $M^{2}P(x)=P_1(x)$, $MQ(x)=H(x)$ and $M^{2}R(x)=S(x)$ become integer coefficients. We have that $P_1(x)=H^2(x)+S(x)$, where again $\mathrm{deg}S<n$ and all polynomials $P_1,H$ and $S$ have integer coefficients. Suppose that what you want to prove is not true. This would mean that $S(x)$ is not identically zero. We know that there are infinitely many $x$ such that $P_1(x)=M^{2}P(x)$ is a perfect square. We will see that since the coefficients of the polynomial $S(x)$ are integer and nonnegative and $S(x)$ is not exactly zero, starting from some point $S(x)\ne 0$, and therefore from the same point $H(x{)}^2\ne P_1(x)$. Then consider sufficiently large positive integers $t$, of which there are infinitely many, such that $P_1(t)=w^2$, where $w$ is a positive integer and $H(t{)}^2\ne P_1(t)$. We see $$|P_1(t)-H^2(t)|\ge 2w-1\Rightarrow |S(t)|\ge 2w-1=2\sqrt{P(t)}-1\ge 2t^n-1>t^n.$$ But since $\mathrm{deg}S<n$, the inequality $S(t)>t^n$ can only hold for a finite number of different $t$, which leads to a contradiction.

The lemma implies that for any given $a$ and $b$, there are only a finite number of $x$ such that $(x^{2a}-1)(x^{2b}-1)$ is a perfect square. Indeed, otherwise, there would be a polynomial $Q(x)$ with rational coefficients such that $(x^{2a}-1)(x^{2b}-1)=Q^2(x)$. But we can see that there is a complex root $e^{\frac{2\pi i}{a}}$ on the left-hand side, which has a multiplicity of $1$, and this is where we get the contradiction. Alternatively, we can see that $(x^{2a}-1{)}^{\prime }=2a{x}^{2a-1}$, which has a GCD of $1$ with $(x^{2a}-1)$, and therefore $x^{2a}-1$ has no roots of multiplicity. Therefore, for any given $a$ and $b$, there are only a finite number of $x$ such that $(x^{2a}-1)(x^{2b}-1)$ is a perfect square. And since there are a finite number of ways to choose a pair $(a,b)$, it means that there is an $x_0$, for which $(x_0^{2a}-1)(x_0^{2b}-1)$ is not a perfect square for any positive integers $a$ and $b$ not greater than $2019$. Therefore, we can put $x=x_0$ and the statement is proved.