TITLE: Korea Math Olympiad 1993
QUESTION [5 upvotes]: An integer which is the area of a right-angled triangle with integer sides is called Pythagorean. Prove that for every positive integer n > 12 there exists a Pythagorean number between n and 2n.

REPLY [4 votes]: Here we exploit the Pythagorean triple $3k,4k,5k$.
Notice because of this Pythagorean triple $6k^2$ is Pythagorean for every $k$.
The case where $n \ge 54$:
Choose $m\ge 3$ so that 
$$6m^2 \le n < 6(m+1)^2 < 12m^2$$ 
This gives us the result for $n \ge 54$. 
Various smaller $n$: 
Using again our $3k,4k,5k$ triangles we can get: 
For $13 \le n \le 23$ the number $24$ works (taking $k = 2$).
And for $30 \le n \le 53$ the number $54$ works (taking $k = 3$). 
The only problem then comes for numbers $24 \le n \le 29$, and this is the one time we stray from our $3k,4k,5k$  method and use the triangle $5,12,13$. 
For $24 \le n \le 29$ the number $30$ works.
Combining all these we have the result.