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Here is another one. Prove that the power of $13$ can be writen as a sum of two squares. I will give two profs of it. First one is more involved and includes lemma $$(a^2+b^2)(x^2+y^2)= (ax+by)^2+(bx-ay)^2$$ yet second takes step 2 and it is much more elegant. proof Base: $n=1$, then $13 = 2^2+3^2$ and we are done. We know that $13^n = a^2+b^2$ ...


For what natural $n$ does there exist a square composed of $n$ squares? Example: 1, 4, and 6 are valid, but one cannot construct a square from 2, 3, or 5 squares. Proof:


A simple consequnce of: Postage Stamp Problem, which states that for any two relatively prime positive integers $m,n$, the greatest integer that cannot be written in the form $am + bn$ for nonnegative integers $a, b$ is $mn-m-n$, is that every natural number greater or equal to $mn-m-n+1$ can be writen in a form $am+bn$ for some $a,b$. And for some ...

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