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.

1) a^2+b^2=c^2

substituting a=nx b=n(x+1) and c=ny into 1 we get

n^2x^2+n^2(x+1)^2=n^2y^2  we can divide the n^2 out and get

x^2+(x+1)^2=y^2

now since a and c are 3 digits for n=1 then x,y must also both be 3 digits thus we are looking for a pythagorean triple of the form (x,x+1,y) with all 3 values having 3 digits.  Using mathematica I found only 2 such triples which gives us either x=119 y=169 or x=696 y=985

we can throw out the second triple because it would cause A to be 4 digits for n=2 thus we have x=119 y=169 which gives us the 5 triples

(119,120,169)

(238,240,338)

(357,360,507)

(476,480,676)

(595,600,845)

Posted by Daniel on 2008-05-08 22:11:51