Transform 2016 to base 2 (I will write just number and remainder then read from bottom to top) 2016 0 1008 0 504. 0 252. 0 126. 0 63. 1 31. 1 15. 1 7. 1 3. 1 1. 1 So2016= 11111100000b= 2^11-2^5=2048-32 Now if you are looking for real solutions this is trivial problem. Let a=n then pairs (log2(2016+2^a),a) always satisfy condition
