None. Here's why: x and y cannot be equal since x²+x²=2007 2x²=2007 x²=2007/2 Thus x cannot be an integer. Thus x² and y² are not equal. 0²=0, 1²=1, 2²=4, 3²=9, 4²=16, 5²=25, 6²=36, 7²=49, 8²=64, 9²=81 Therefore squares only end with 0,1,4,5,6, or 9. Therefore the only way the sum of two different squares could end with 7 is for one of them to end with 1 and the other to end with 6. By symmetry, there is no loss in generality by supposing x² ends in 1 and y² ends in 6. There are 4 cases to consider. x ends in 1 or 9 and y ends in 4 or 6 Case 1: x ends in 1 and y ends in 4 x = 10u+1 and y = 10v+4, for some positive integers u and v (10u+1)² + (10v+4)² = 2007 100u²+20u+1 + 100v²+80v+16 = 2007 100u² + 100v² + 20u + 80v = 1990 The left side is divisible by 20 but the right side isn't. Thus Case 1 is out. Case 2: x ends in 1 and y ends in 6 x = 10u+1 and y = 10v+6, for some positive integers u and v (10u+1)² + (10v+6)² = 2007 100u²+20u+1 + 100v²+120v+36 = 2007 100u² + 100v² + 20u + 120v = 1970 The left side is divisible by 20 but the right side isn't. Thus Case 2 is out. Case 3: x ends in 9 and y ends in 4 x = 10u+9 and y = 10v+4 (10u+9)² + (10v+4)² = 2007 100u²+180u+81 + 100v²+80v+16 = 2007 100u² + 100v² + 180u + 80v = 1910 The left side is divisible by 20 but the right side isn't. Thus Case 3 is out. Case 4: x ends in 9 and y ends in 6 x = 10u+9 and y = 10v+6 (10u+9)² + (10v+6)² = 2007 100u²+180u+81 + 100v²+120v+36 = 2007 100u² + 100v² + 180u + 120v = 1890 The left side is divisible by 20 but the right side isn't. Thus Case 4 is out. -------------------- Thus there are no solutions. Edwin