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.

(In reply to re: Start the ball rolling...? by brianjn)

Thanks for the clarification, though I am still not sure the exact meaning of this problem.

You apparently seek a DIFFERENT triplet for each of the values of n=1..5 (so A, B, and C do not need to be divisible by 60).  However, in that case, B-n=A still seems to follow from "A=n*X" and "B=n*(X+1)" = (n*X) + (n*1), by substituing "A" for "(n*X)" giving B=A + (n*1) == B=A+n == B-n=A.

I presume the same value of n must be used for all three constraints, for any one triplet solution.  Also, I was also asking if we should assume that X and Y are positive integers in all cases. If this is so, I think it should be stated; otherwise we should assume that they may not be.  I am not sure if you preclude one or more of the "3 digit numbers" starting with one or more zeroes, since you do not so state.  Do you intend that X and Y have their same values in all three constraint equations for a given n?  Also, should we assume that you are asserting that there is only one 5-tuple of triplets which will satisfy your conditions?

 

Posted by ed bottemiller on 2008-05-09 10:19:29