Problem:
A triangle is called a parabolic triangle if its vertices lie on a parabola y=x2. Prove that for every nonnegative integer n, there is an odd number m and a parabolic triangle with vertices at three distinct points with integer coordinates with area (2nm)2.
Solution:
Let A=(a,a2),B=(b,b2), and C=(c,c2), with a<b<c. We have AB=[b−a,b2−a2] and AC=[c−a,c2−a2]. Hence the area of triangle ABC is equal to
Setting b−a=x and c−b=y (where both x and y are positive integers), the above equation becomes
(2nm)2=2xy(x+y)​(3)
If n=0, then (m,x,y)=(1,1,1) is clearly a solution to (3). If n≥1, it is easy to check that,
(m,x,y)=((24n−2−1,22n+1,(22n−1−1)2))
satisfies (3).
Alternate Solution from Jacek Fabrykowski:
The beginning is the same up to (2nm)2=2xy(x+y)​. If n=0, we take m=x=y=1. If n=1, we take m=3,x=1,y=8. Assume that n≥2. Let a,b,c be a primitive Pythagorean triple with b even.
Let b=2rd where d is odd and r≥2. Let x=22k,y=22kb and z=22kc where k≥0. We let m=adc and r=2 if n=3k+2,r=3 if n=3k+3 and r=4 if n=3k+4.
Assuming that x=aâ‹…2s,y=bâ‹…22, other triples are possible:
(a) If n=3k, then let m=1 and x=y=22k.
(b) If n=3k+1, then take m=3,x=22k,y=22k+3.
(c) If n=3k+2, then take m=63,x=49â‹…22k, and y=22k+5.