Originally Posted by

**Soroban** Hello, hastings!

Are they any other restrictions to the numbers?

As stated, the problem is rather silly.

Since numbers can be repeated, the problem is trivial.

. . A solution is: .$\displaystyle \left[\begin{array}{ccc} n & n & n \\ n & n & n \\ n& n & n\end{array}\right]$ . for any $\displaystyle n \in I^+$

Using three different integers: $\displaystyle a,b,c$

. . a solution is: .$\displaystyle \left[\begin{array}{ccc}a & b & c \\ b & c & a \\ c & a & b\end{array}\right]$ . for any $\displaystyle a,b,c \in I^+$

Even if the numbers must be *consecutive* integers,

. . a solution is: .$\displaystyle \left[\begin{array}{ccc} a+7 &a+2 & a+3 \\ a & a+4 & a+8 \\ a+5 & a+2 & a+1 \end{array}\right]$ . for any $\displaystyle a \in I^+$