TI-M: N = ...

To: ti-math@lists.ticalc.org
Subject: TI-M: N = ...
From: JayEll64@aol.com
Date: Tue, 16 May 2000 17:04:44 EDT
Reply-To: ti-math@lists.ticalc.org

In a message dated 5/15/00 10:05:55 PM Mountain Daylight Time, 
Acidic113@aol.com writes:

> 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...

It never said N had to be a perfect square ;)  Someone else I showed this 
problem to also assumed N was a perfect square...

Actually, it's easier if you find the smallest positive integer that can be 
expressed as the sum of two different squares and involving consecutive 
squares; ie, N = A^2 + B^2 = (A+1)^2 + C^2.  Then it might be easier to find 
the N = A^2 + B^2 = (A+1)^2 + C^2 = (A+2)^2 + D^2.  From what I've read, 
there's no solution to the version of this problem that involves four 
consecutive squares, but I haven't gotten around to investigating why this 
is...

JayEll