Re: TI-M: How's about something different?

To: ti-math@lists.ticalc.org
Subject: Re: TI-M: How's about something different?
From: Acidic113@aol.com
Date: Tue, 16 May 2000 00:05:27 EDT
Reply-To: ti-math@lists.ticalc.org


Just out of curiosity... that math problem that JayEll posted earlier.. I 
used triangles and I don't think there is a positive integral answer.  Maybe 
I am doing something wrong but that problem violates the Pythagorean Theorem. 
 Ok, first off, you want four consecutive integers...  one of those 
consecutive integers must be part of a 3-4-5 triple because every three 
integers you have a multiple of three (Duh!). If you say that A^2 + B^2 = the 
hypotenuse^2 = N then B^2 + C^2 = N can't be true because C^2 would be the 
hypotenuse of a special triangle and therefore bigger than the squares of A 
and B put together so the other equation could nver be equal.. it would 
always be less or greater depending on the location of the triple in the set 
of four consecutive integers... I think the proof is adequate if I am 
correct.. I don't think it's possible.. but I could always be wrong.. 
Mathematics are famous for "stupid" mistakes... right everybody?  The one 
problem you always miss is ALWAYS the easiest one...