Euler Builds a House

Does there exist a rectangular cuboid whose edges and face diagonals are all integer lengths?

Yes, finitely many (for relatively prime side lengths) Yes, infinitely many (for relatively prime side lengths) No

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

David Vreken
Dec 26, 2018

Cuboids whose edges and face diagonals are all integer lengths are called Euler bricks .

In 1740, Saunderson found that for any Pythagorean triple ( u , v , w ) (u, v, w) , the sides a = u 4 v 2 w 2 a = u|4v^2 - w^2| , b = v 4 u 2 w 2 b = v|4u^2 - w^2| , and c = 4 u v w c = 4uvw produce diagonals d = w 3 d = w^3 , e = u ( 4 v 2 + w 2 ) e = u(4v^2 + w^2) , and f = v ( 4 u 2 + w 2 ) f = v(4u^2 + w^2) , which are all integers.

With these formulas, if gcd ( u , v , w ) = 1 \gcd(u, v, w) = 1 , then gcd ( a , b , c ) = 1 \gcd(a, b, c) = 1 . Since there are infinitely many relatively prime Pythagorean triples, there are infinitely many rectangular cuboids with edges and face diagonals such that gcd ( a , b , c ) = 1 \gcd(a, b, c) = 1 .

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...