This is a known problem about tiling a rectangle of $ 1 \times x $,$x$ is irrational with $n$ squares with length of their sides $s_1,s_2,\ldots , s_n$. You have to prov this is not possible. I want to try a proof similar but different than the one in https://kam.mff.cuni.cz/~matousek/stml-53-matousek-1.pdf Miniature 12.
Proof by contradiction - lets assume that such tiling exists. Let $ R/Q $ be a vector space of the real numbers over the field of the rational numbers. Let $V$ be a subspace of $ R/Q $ = $span\{1, x, s_1^2,s_2^2 \ldots ,s_n^2\} $ where $s_i$ is the length of the side of square $i$. Let $B = \{1, x, b_1,b_2 \ldots ,b_m\} $ be a basis for $V$. We define a linear function $f : V \to V$ that satisfies $ f(1)=1, f(x)=-1,f(b_i) = b_i$. We clearly see that $nullity(f) = 1$ and $ 1+x $ is the only basis vector for the kernel of $f$.
We know that $x = \sum_{i=1}^n S_i^2 $. Apply $f$ to both sides and use the linearity of $f$ and have $f(x) = -1 = \sum_{i=1}^n f(S_i^2)$. If $S_i^2 \in Q$ then $f(S_i^2) \in Q $ because $f(1) = 1$. We can move all the rational terms $f(S_i^2)$ with $S_i^2\in Q$ to the left side and have $$ k= \sum_{ S_i^2 \not \in Q} f(S_i^2) $$ where $k$ a rational number.
$ f(x) = -1 \implies f(-kx) = k = \sum_{ S_i^2 \not \in Q} f(S_i^2) = f(\sum_{ S_i^2 \not \in Q} S_i^2 ) $ and we get $$ \sum_{ S_i^2 \not \in Q} S_i^2 = -kx + \alpha(1+x) $$ for some $\alpha \in Q$ (recall that $f( \alpha(1+x))=0 $).
$x = \sum_{ S_i^2 \not \in Q} S_i^2 + \sum_{ S_i^2 \in Q} S_i^2 = \sum_{ S_i^2 \in Q} S_i^2 -kx + \alpha (1+x) \implies \sum_{ S_i^2 \in Q} S_i^2 = x(1 + k - \alpha) - \alpha$
but this is impossible because $x(1 + k - \alpha) - \alpha $ is irrational and we got a contradiction.
I hope to get feedback on the validity of this proof.
Edit: This proof is invalid because, as noted below by Calvin if 1+k−α=0 then there is no contradiction. I wonder if there any elegant algebraic way to prove it. The proof idea in https://kam.mff.cuni.cz/~matousek/stml-53-matousek-1.pdf is based on geometry (extending the edges of the squares ) and I wonder if there is a proof that is more algebraic.