Good nut  
 
Solution. 
 
It is obvious that column \'c\' (5, 8, 4, 1) and row \'2\' (9, 2, 4, 3) are only column and row which correspond to the initial conditions. So 5 + 8 + 4 + 1 = 9 + 2 + 4 + 3 = 18.  
Further observation hints that only \'a1\' or \'b1\' may be chosen as \'only one other number in the place.  
 
Choice of \'b1\' leave us with no options but to have  
 
\'a1\' + \'d1\' = 14 
\'b3\' + \'b4\' = 13 
 
However, we have only {5, 2, 5, 9, 8, 3, 4, 1} to satisfy them. 
 
Quick glance gives: 
 
\'a1\' or \'d1\' = 5 or 9 
and 
\'b3\' or \'b4\' = 5 or 8 
 
But due to \'a2\' = 9, \'a1\' can be only 5. So \'a1\' = 5, \'d1\' = 9. 
Same story goes for \'b3\' and \'b4\', so \'b3\' = 5, \'b4\' = 8. 
 
And we have {1, 2, 3, 4} left at our disposal which can be quickly arranged. 
 
So the answer is: 
 
3 8 5 2 
1 5 8 4 
9 2 4 3 
5 3 1 9

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