Solution of the Week #426 - House Numbers

The sum of the first n odd numbers is simply n^2. If I live at the kth house (ie house number 2k-1) then the houses to the left of it add to (k-1)^2. For example if I lived at number 7 (the fourth odd number) 1+3+5 is equal to 3^2 = 9.

For the houses to the right, if there are n houses altogether then we will have n^2 – k^2. This needs to be equal to (k-1)^2.

Effectively then, we are looking for two consecutive square numbers (k-1)^2 and k^2, which sum to another square number n^2.

Solutions exist when k is 1, 4, 21, 120, 697 etc. Only one of these corresponds to a three digit house number. The 120th odd number is 239. Therefore I live at number 239 in a street with houses up to number 337. House numbers either side of me add up to 14161.