Consider this magic square:
$$\begin{array} {|r|r|}\hline a & b & c \\ \hline d & e & f \\ \hline j & h & i \\ \hline \end{array} $$ Where $a,b,c,d,e,f,j,h,i\in \mathbb N^*$ and the $\gcd$ of all nine elements is $1$. This magic square is A multiplicative one such that the product in each row, column and diagonal is equal to $P$
The first question: Prove that $e^3=P$
The second question: construct a multiplicative magic square with the divisors of $100$
My attempt:
It’s not actually an attempt, it is just an observation, since the $\gcd(a,b,c,d,e,f,j,h,i)=1$ And $$aei=cej=beh=def=P$$ $e\in\mathbb N^* \implies e\geq1$, so we can divde by it: $$ai=cj=bh=df$$ Wich mean that $a\mid cj,bh,df$ And $i\mid cj,bh,df$ and so on..., but what’s next?
$$e^3=\frac{aei\cdot beh\cdot ceg}{abc\cdot ghi}=\frac{P^3}{P^2}=P$$
The knight move strategy seems to work just as in the additive case:
$$\begin{matrix}50 &1&20\\4&10& 25\\5&100 &2\end{matrix}$$