Number theoryDifficulty 4.8AIMEProve itSoviet Union
Problem: Can both x2+y and x+y2 be squares for x and y natural numbers?
Solution
Solution: No. The smallest square greater than x2 is (x+1)2, so we must have y>2x. Similarly x>2y. Contradiction.
Want a route through all this instead of an archive? The track
puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.
Source: MathNet,
licensed CC-BY-4.0.
Statement reproduced verbatim; metadata (topic, difficulty) added by this project.