2
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Given:

3*(c^3 - c^2*b - c*a^2 + a^2*b) == 
 a^3 + b^3 + c^3 + 3*(a + b + c)*(a b + a c + b c) - 3 a b c

Need to solve for all three variables as integers. Problem is to find one solution where all variables are positive.

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  • $\begingroup$ please review my edit. Some of your formatting of the formula made it unclear and I'm not sure that I got it completely right. $\endgroup$ – Carl Lange Jan 27 '19 at 9:56
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    $\begingroup$ @carl lange Gostei muito do seu canal no YouTube $\endgroup$ – LCarvalho Jan 27 '19 at 11:19
  • $\begingroup$ @LCarvalho Obrigado pelas suas palavras gentis!! $\endgroup$ – Carl Lange Jan 27 '19 at 15:14
7
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Use FindInstance.

f = 3*(c^3 - c^2 b - c a^2 + a^2 b) == 
  a^3 + b^3 + c^3 + 3*(a + b + c)*(a b + a c + b c) - (3 a b c)

FindInstance[f, {a, b, c}, Integers, 3]

{{a -> -130, b -> -130, c -> 65}, {a -> -1, b -> 0, c -> 1}, {a -> 1, b -> 0, c -> -1}}

We can test the solutions out:

f /. {a -> -130, b -> -130, c -> 65}

True

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  • $\begingroup$ Nice work! Do you think it would be possible to have all variables positive? I meant to add this to the question/will edit. $\endgroup$ – Dale Jan 27 '19 at 14:08
  • $\begingroup$ Found this: mathematica.stackexchange.com/questions/74215/… $\endgroup$ – Dale Jan 27 '19 at 14:14
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    $\begingroup$ @Dale Yes, simply add the extra constraints to the first argument in FindInstance: FindInstance[{f, a > 0, b > 0, c > 0}, {a, b, c}, Integers, 1] However, in this case it does not appear that a solution exists. $\endgroup$ – Carl Lange Jan 27 '19 at 15:13

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