Macedonian Mathematical Competitions

a) For $2520$ we have $2520 = 2^3 \cdot 3^2 \cdot 5 \cdot 7$. We notice that in order to get a square of a positive integer this number has to be multiplied with at least $2 \cdot 5 \cdot 7 = 70$. We obtain the product $2520 \cdot 70 = 420^2$ and the desired number is $70$.

b) We denote by $2k-1$ and $2k+1$, $k \in \mathbb{Z}$, the two consecutive odd integers. Then their sum is $(2k-1)+(2k+1)=2k-1+2k+1=4k$ which is divisible with $4$.