Other correct solutions were sent in by Will Putt. Further solutions were submitted by alumnus Brian Laughlin ('81), and from David Charles, Ivan Lisac, Lou Cairoli, Philippe Fondanaiche, Dane Brooke, Lenny Gibson and his AEGIS Math Squad, Burkart Venzke, Sudipta Das, Scott Powell, Mike Young, Jan Siwanowicz, Al Zimmermann, Jeff Anderson, Skip Kuzel, Denis Borris, Bob Margulies.

There are infinitely many integer-sided solutions, all of whose sides are distinct.  The dimensions of the rectangle are 104 x 105, almost a square.

Label the squares as below; I'll let the labels stand for the length of the square's side, too.  

We have the following equations:

Note that F + B = G + A, so the third and fifth equations give 2B + C = 4A + B and C = 4A - B.  The last three equations now become F = 4A,  I = 8A - B, J = 12A - 2B.  Further note that B + D + H = C + J which becomes B + (A + B) + (3A + 2B) = (4A - B) + (12A - 2B) or 7B = 12A.  The smallest solution in integers comes from A = 7, B = 12, then C = 16, D = 19, E = 26, F = 28, G = 33, H = 45, I = 44, J = 60, which are indeed distinct integers.   The dimensions of this smallest rectangle are 104 by 105, almost a square.  

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