No. The middle cell can only be $2$, respectively $4$, but then $7$, respectively $5$, cannot be placed.

A similar idea: notice that all the prime sums have to be odd, so all the sums have to be between an even and an odd number. That forces the following configuration of evens and odds: OEO EOE OEO If the center number is $x$, then $x+2, x+4, x+6, x+8$ must all be prime. This is impossible since $x+4, x+6, x+8$ contains a multiple of 3 greater than 3.

yea, obviously the bonus problem this year

no the edges should be even numbers, so we cannot make every pairs prime. (if all the edges are not even numbers, there will e a pair of odd integers which sum is not a prime)

I thought you could show that evens have to be in the middle squares of the top and bottom rows and middle squares of left and right columns with a parity argument. From there, you could show that no odd number can be placed in the middle square, so contradiction. (There exist no solutions)

Well, we can use that every cell with the coordinates with the same parity contains the odd numbers, because two neighouring cells have different parity (their sum is greater than 2) and every odd number which we put in the middle gives the contradiction because of divisibility by 3.

chaotic_iak wrote: Is it possible to fill a $3 \times 3$ grid with each of the numbers $1,2,\ldots,9$ once each such that the sum of any two numbers sharing a side is prime? $\color{blue}\boxed{\textbf{Answer: No}}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Suppose we can complete the board: Note that $7$ can only have $4$ and $6$ as neighbors $\Rightarrow 7$ is in a corner, and the numbers $4$ and $6$ are adjacent to $7$ Next to $4$ and $6$ there can only be $1$ $\Rightarrow 1$ is in the center But $1$ can only have at most 3 neighbors $(\Rightarrow \Leftarrow)_\blacksquare$ $\color{blue}\rule{24cm}{0.3pt}$