Problem 3
Prove that there exists a triangle which can be cut into 2005 congruent triangles.
Step 1 of 5: Represent 2005 as two squares
Detailed analysis
Multiply out (4+1)(400+1): it equals 2005, and rearranging gives 40 squared+20 squared+2 squared+1=39 squared+22 squared. Thus 2005 is a sum of two integer squares.