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# Algebra

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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

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.