Find the number of ordered pairs (a, b) of integers that satisfy \(a^2 + ab + b^2 = 79\).
+33615 Express this as \((a+b)^2 -ab=79\)
Rewrite as \((a+b)^2 = 79 - ab\)
Look at all the perfect squares, say ps, less than 79. Must have 79 - ab = ps or ab = 79 - ps
Look for two integers that satisfy ab = 79 - ps and (a + b)2 < 79.
I think there are only four pairs:
(-5, -3)
(-3, -5)
(3, 5)
(5, 3)
