Question 1199579
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The solutions in integers of the equation x^2+y^2 = 289 are the following:<br>
(0,17)
(0,-17)
(8,15)
(-8,15)
(-8,-15)
(8,-15)<br>
Since the problem specifies a greater than or equal to 0, a can have any of the values 0, 8, 15, or 17.<br>
[1] a=0, a-b=17 --> (a,b)=(0,-17)
[2] a=0, a-b=-17 --> (a,b)=(0,17)
[3] a=8, a-b=15 --> (a,b)=(8,-7)
[4] a=8, a-b=-15 --> (a,b)=(8,23)
[5] a=15, a-b=8 --> (a,b)=(15,7)
[6] a=15, a-b=-8 --> (a,b)=(15,-23)
[7] a=17, a-b=0 --> (a,b)=(17,17)<br>
ANSWER: 7 ordered pairs (a,b): (0,17), (0,-17), (8,-7), (8,23), (15,7), (15,-23), (17, 17)<br>