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Set of triplets (Posted on 2008-05-08) Difficulty: 2 of 5

A2 + B2 = C2

where A=n*X, B=n*(X+1) and C=n*Y.

Determine triples (A,B,C) which satisfy the constraints of n,X,Y noting that A, B and C are all 3 digit numbers when n is 1 through 5.

See The Solution Submitted by brianjn    
Rating: 2.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(3): Start the ball rolling...? | Comment 7 of 8 |
(In reply to re(2): Start the ball rolling...? by ed bottemiller)

History of problem:

Brian Smith put up a proble "Rectangle Around A Rhombus".  While examining this with a spreadsheet I became aware that 119 and 120, as two legs of a Pythagorean triangle were one digit apart.  I thenbegan to look for multiples of that, much like one would with a 3,4,5 triple (6,8,10; 9,12,15...).

K Sengupta pointed me in the direction of the link which is in my solution.  It was only then that I realised that what I saw as an oddity was actually a little more common, eg (3,4,5), (20,21,29).  It was from there  "The defining equation is X^2 + (X+1)^2 = Z^2, which when doubled gives 2Z^2 = (2X+1)^2 + 1."
that the parameters arose thereby giving the structure for each term, with X being the base upon which a triple and subsequent multiples are based.  We would lose base if we substituted as you have suggested, yes?

As for negative values, from where I was coming from they did not enter my thoughts; note a few comments prior to this one Dej Mar actually demonstrated this structure with some of those.





  Posted by brianjn on 2008-05-09 22:45:01

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