The puzzle is to place a different number in each of the ten squares so that the sum of the squares of any two adjacent numbers shall be equal to the sum of the squares of the two numbers diametrically opposite to them. 
The four numbers placed, as examples, must stand as they are. 
The square of 16 is 256, and the square of 2 is 4. 
Add these together, and the result is 260. 
Also—the square of 14 is 196, and the square of 8 is 64. 
These together also make 260. 
Now, in precisely the same way, B and C should be equal to G and H (the sum will not necessarily be 260), A and K to F and E, H and I to C and D, and so on, with any two adjoining squares in the circle.

All you have to do is to fill in the remaining six numbers. 
Fractions are not allowed, and I shall show that no number need contain more than two figures.

See answer