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Find the number of ordered pairs (a, b) of integers that satisfy \(a^2 + ab + b^2 = 79\).
 

 Apr 26, 2020
 #1
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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)

 Apr 26, 2020

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