# Is it possible to cover a $70\times70$ square with $24$ squares with side length $1,2,3\ldots24$?

Is it possible to cover a $70\times70$ square with $24$ squares with side length $1,2,3\ldots24$?

• When I encountered this problem in the past, the first thing that came to my mind was: How do you use the $1$x$1$ square in such a way that you don't require an additional $1$x$1$ to 'fill in the hole' created by that $1$x$1$. I'm not sure if the way in which I described it makes sense through text. There may be a counterexample to that as well. Commented Jul 23, 2014 at 6:25

Further notes: Consider the general problem: is it possible to tile an $L\times L$ square with $k$ squares of side length $1,2,3\ldots k$? It turns out, the only solution to $$1^2+2^2+3^2+\cdots+k^2=L^2$$ is $k=24,L=70$ (besides the obvious $k=L=1$). Here is a link to the proof by Watson in 1918 (it's in the first few pages) So this problem is unique, in a way.
• $1^2+2^2+3^2+...+24^2=70^2$ its true :) Commented Jul 23, 2014 at 6:30