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Find the number of triples  (a, b, c) of positive integers, such that \( 1 \le a, b,c \le 100,\)   and \(a^2 + b^2 + c^2 = ab + ac + bc.\)
 

 Apr 14, 2021
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notice that\(a^2 + b^2 + c^2 - ab - ac - bc = \frac{(a-b)^2 + (b-c)^2 + (a-c)^2}{2}\) and this equals to 0, so a - b = b - c = a - c = 0, therefore a = b = c, so there are 100 ordered triples. 

 Apr 14, 2021

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