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Find the number of triples  (a, b, c) of positive integers, such that 1 <= a, b, c <= 100   and a^2 + b^2 + c^2 = + ab + ac + bc = 81.

 Jun 14, 2021
 #1
avatar+2106 
+1

a^2 + b^2 + c^2 = 81

Let's list the squares under 81. 

1, 4, 9, 16, 25, 36, 49, 64

 

Now the ones that add up to 81. 

1, 16, 64 (1, 4, 8)

16, 16, 49 (4, 4, 7)

9, 36, 36 (3, 6, 6)

 

Now let's check for the second criteria.

ab + ac + bc = 81

1*4 + 4*4 + 4*1 != 81

4*4 + 4*7 + 4*7 != 81

3*6 + 6*6 + 3*6 != 81

 

I found 0 triples, but there are probably some more. 

Sorry I couldn't help more.

 

=^._.^=

 Jun 14, 2021
 #2
avatar+287 
+2

Perhaps you meant to say $a^2+b^2+c^2+ab+ac+bc=81$ instead.  If that's the case, then there are three solutions:
    (3, 4, 4)
    (4, 3, 4)
    (4, 4, 3)
 

 Jun 14, 2021

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