Oops. I hit the reply, not the reply all, but I see that went to the list as well as Dan. Sorry about that. On Wed, Feb 8, 2012 at 10:57 PM, James Buddenhagen <jbuddenh@gmail.com>wrote:
1943??
How about a=-119, b=120, c=13^2 ?
On Wed, Feb 8, 2012 at 4:51 PM, Dan Asimov <dasimov@earthlink.net> wrote:
According to what I've read, Fermat posed this problem to Mersenne in 1943. This puzzle is intended for people who haven't seen the solution already:
Find a primitive Pythagorean triple a^2 + b^2 = c^2 such that both a+b and c are each square numbers.
(Please e-mail your answers straight to me so they don't keep others from solving this.)
--Dan
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