A more formal writeup of Riley's method: Consider an odd divisor o of the target number N. Divide the target number up in that many pieces (n=N/o). Subtract (n-1)/2 from the first one, (n-1)/2-1 from the second etc (one less for each next number). So for example 12 = 3*4 = 4+4+4 = 3+4+5 0r: 14 = 7*2 = 2+2+2+2+2+2+2 = -1+0+1+2+3+4+5
As you can see two things can happen. Case one is if the first number is positive. Then you have a sum of consecutive positive integers consisting of an odd number of members (o). On the other hand, if you have a odd length sequence of positive consecutive number adding to the target value, you can average the values in the symmetric positions (transforming for example 1+2+3 to 2+2+2) which for going in the other direction (back to 1+2+3) mmust obviously be a case one again since it starts with the same positive number you started out with. Case two is if the first number is negative or zero. There will be more positive than neghative values (since the sequence is symmetric around a positive value), so you can cancel out the negative numbers against the positive ones and drop 0. So in 14 = -1+0+1+2+3+4+5 cancel -1 versus +1 and drop 0, getting: 14 = 2+3+4+5 And you end up with an even sized sequence. On the other hand, if you have a even length sequence of positive consecutive number adding to the target value starting with a+1, you can add -a,-a+1.....a-1,a in front, giving a sequence of odd length. So again average out the symmetric positions. This will give a positive average since you started out with a sequence of positive values and added as many new negative as positive cases. For example 2+3 becomes -1+0+1+2+3 = 1+1+1+1+1 Going forward from 1+1+1+1+1 to -1+0+1+2+3 again gives you a case two sequence since they uniquely determine the -1+0+1+2+3 sequence which was actually constructed to start with a negative number or 0. So both cases are completely reversable and it must therefore be a bijection.
