Problem 7 · Pell-equation descent · 5 points
Made by AI and not reviewed by a real person.
Split the first N positive odd integers into two nonempty consecutive blocks. If the second block sums to exactly one more than the first, what is the smallest possible N greater than 100?
- 577
- 99
- 239
- 408
- 578
Show a hint
- Express the two block sums using the total of the first several odd integers. Then ask whether a solution could lead to a smaller one.
- Once you have N² − 2k² = 1, try replacing (N, k) with (3N − 4k, 3k − 2N).
Show the answer and solution
Answer: A — 577
If the first block contains k numbers, its sum is k² and the total is N². Thus N² − 2k² = 1. Every solution with N > 3 yields a smaller nonnegative solution (3N − 4k, 3k − 2N) to the same equation; reversing this step gives the successive solutions (N, k) = (1, 0), (3, 2), (17, 12), (99, 70), (577, 408). So 577 is the first possible N above 100. The tempting 239 reverses the required difference: 239² − 2·169² = −1.