In a Group $G$, $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ holds, $\forall a,b\in G$.Prove that the group $G$ is abelian. My approach was following: Let $a,b\in G$ Then, $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ holds. Now, $$a^3b^3=b^3a^3 \\ \implies aaabbb=bbbaaa\\ \implies a\cdot a^2\cdot b^2\cdot b=b...

In a Group $G$, $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ holds, $\forall a,b\in G$.Prove that the group $G$ is abelian. My approach was following: Let $a,b\in G$ Then, $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ holds. Now, $$a^3b^3=b^3a^3 \\ \implies aaabbb=bbbaaa\\ \implies a\cdot a^2\cdot b^2\cdot b=b...

Suppose a $m \times m$ square can be dissected into $7$ rectangles such that no two rectangles have a common interior point and the side lengths of the rectangles form the set $\{1,2,3,4,5,6,7,8,9,10,11,12,13,14\}$. Find the maximum value of $m$. I have calculated that the value is at most $22...

How would you like to calculate this one? Do you see a fast, neat way here? Ideas? $$\int_0^1 \frac{\log (x) \log \left(\frac{1}{2} \left(1+\sqrt{1-x^2}\right)\right)}{x} \, dx$$ Sharing solution is only optional.