C cutemouse Account Closed Joined Apr 23, 2007 Messages 2,250 Gender Undisclosed HSC N/A Jul 18, 2009 #1 Find 6 integer solutions for x and y such that: x2+2009=y2
Aquawhite Retiring Joined Jul 14, 2008 Messages 4,946 Location Gold Coast Gender Male HSC 2010 Uni Grad 2013 Jul 18, 2009 #2 That's a really good problem solving question. I personally don't know how to solve it, but I'd like to see some others give it a go.
That's a really good problem solving question. I personally don't know how to solve it, but I'd like to see some others give it a go.
I Iruka Member Joined Jan 25, 2006 Messages 544 Gender Undisclosed HSC N/A Jul 18, 2009 #3 Re-write as y^2-x^2=2009 Factorise using difference of 2 squares: (y-x)(y+x) =2009 If x and y are both integers, then so are y-x and y+x. Find the prime factorisation of 2009 and then check what all the possibilities for x and y are.
Re-write as y^2-x^2=2009 Factorise using difference of 2 squares: (y-x)(y+x) =2009 If x and y are both integers, then so are y-x and y+x. Find the prime factorisation of 2009 and then check what all the possibilities for x and y are.
I Iruka Member Joined Jan 25, 2006 Messages 544 Gender Undisclosed HSC N/A Jul 18, 2009 #4 Doing that I got: y= +/- 1005, x=+/-1004 y= +/- 147, x=+/-140 y= +/- 45, x=+/-4