Just Addition

$0$ cannot be used as this would require one of the equations to be $0 + n = n$ , and duplication is not allowed. So we are only working with eight of $1, 2, 3, 4, 5, 6, 7, 8, 9$ in any potential solution.

Now the sum $S$ of the numbers in the lower row must equal the sum of the numbers in the rightmost column, since each represents the sum of all the numbers in the remaining 4 boxes. This in turn implies that the sum of all 8 numbers is $3S$ , i.e., divisible by $3$ . The range of sums in general is $1 + 2 + ... + 7 + 8 = 36$ to $2 + 3 + .... + 8 + 9 = 44$ , so the valid sums we can work with are $36, 39$ and $42$ .

With a sum of $36$ we are using the integer $1$ through $8$ . The sums of the numbers in the outer row and column must each be $36/3 = 12$ , which can only be accomplished with the pairs $(4,8)$ and $(5,7)$ . For one of the equations we must then have $1 + 3 = 4$ ; let this be one of the rows. Then to get a vertical sum of $7$ , since $4$ has already been used, we must have in a column the equation $1 + 6 = 7$ . This forces $2$ into the last remaining box, giving us the solution

$\Large 1 + 3 = 4$

$\Large + \space \space +$

$\Large 6 + 2 = 8$

$\Large = \space =$

$\Large 7 \space \space \space \space \space 5$

With a sum of $39$ we now must include $9$ and drop $6$ . The sum of the numbers in the outer row and column must now be $39/3 = 13$ , which can be accomplished with the pairs $(4,9), (5,8)$ and $(6,7)$ . But since we dropped 6, the pairs we must use are $(4,9)$ and $(5,8)$ . Again, one equation must be $1 + 3 = 4$ , which we can let be one of the rows. Since $4$ is then already used, one of the vertical equations must be $3 + 2 = 5$ , forcing $7$ into the remaining box, giving us the solution

$\Large 2 + 7 = 9$

$\Large + \space \space +$

$\Large 3 + 1 = 4$

$\Large = \space =$

$\Large 5 \space \space \space \space \space 8$

Finally, for a sum of $42$ we must add back $6$ and drop $3$ . The outer row and column sums must now be $42/3 = 14$ , which can only be accomplished with the pairs $(5,9)$ and $(6,8)$ . This forces the placement of $7$ in the inner boxes, which in turn forces the placement of $1,2$ and then $4$ , giving us the third and final solution

$\Large 4 + 1 = 5$

$\Large + \space \space +$

$\Large 2 + 7 = 9$

$\large = \space \space =$

$\Large 6 \space \space \space \space \space 8$