A123673 Smaller side of right triangles with integer sides and integer side inscribed squares with two vertices on the hypotenuse.

111, 222, 333, 444, 555, 666, 777, 888, 999, 1110, 1145, 1221, 1332, 1443, 1554, 1665, 1776, 1887, 1998, 2109, 2220, 2290, 2331, 2442, 2553, 2664, 2775, 2886, 2997, 3108, 3219, 3272, 3330, 3435, 3441, 3552, 3663, 3774, 3885, 3996, 4107, 4218, 4329, 4440

Offset

1,1

Comments

The side of the inscribed square having two vertices on the hypotenuse of a right triangle, sides x<y<z is 1. s = xyz/(z^2+xy). So if z and s are known we can solve for x and y. solving 1. for xy, squaring and substituting y^2 = z^2 - x^2 we get a quartic in x, z^2*x^2 - x^4 = z^4*s^2/(z-s)^2 which a quadratic in x^2. Let q=x^2, a=z^2 and b = z^4*s^2/(z-s)^2 then solving q^2 - aq + b = 0 we get q = (a +/- sqrt(a^2-4b))/2 and x = sqrt(q) = sqrt((a +/- sqrt(a^2-4b))/2) This implies z^4 >= 4*z^4*s^2/(z-s)^2. So it follows for a solution to exist, z >= 3s. For s=60 and z = 185 we have a = 185^2, b = 185^4*60^2/125^2=269879184 then x1 = sqrt((185^2 - sqrt(185^4-4*269879184))/2) = 111 x2 = sqrt((185^2 - sqrt(185^4+4*269879184))/2) = 148 So x= 111 and y = 148. It is interesting to note that 37 almost always divides these numbers. Some exceptions are 1145, 2290, 3272, and 3435.

Links

Table of n, a(n) for n=1..44.

Prog

(PARI) g(n)= { for(x=1, n, for(y=x, n, z=sqrt(x^2+y^2); s=x*y*z/(z^2+x*y); if(s==floor(s), print1(floor(x)", ") ) ) ) }

Crossrefs

Sequence in context: A291266 A033284 A333568 * A177769 A074842 A180115

Adjacent sequences: A123670 A123671 A123672 * A123674 A123675 A123676

Keyword

nonn

Author

Cino Hilliard, Nov 17 2006

Status

approved