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As there are infinite solutions you must limit the scope of your search. You can find your desired a == b == c ==1 with f.e.: FindInstance[ a/(b + c) == b/(c + a) == c/(a + b) && 0 < a < 10 && 0 < b < 10 && 0 < c < 10, {a, b, c}, Integers, 9] {{a -> 1, b -> 1, c -> 1}, {a -> 2, b ...


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Mathematica can solve this problem completely. It provides three types of non-trivial solutions. Here we go The two defining eqations are eq1 = a/(b + c) == b/(c + a); eq2 = b/(c + a) == c/(a + b); Solving these, one by one, using Solve without specifying the variables to solve for Solve [eq1] {{a -> b}, {a -> -b - c}} Solve [eq2] {{a ...


3

Perhaps what you looking for is Reduce. It will give the whole solution space of your equations. Reduce[a/(b + c) == b/(a + c) == c/(a + b), {a, b, c}, Reals] This should make it clear that a == b == c == 1 is not the answer. At the request of the OP in a comment below, here is a non-trivial solution where the three variables are not equal. ratios = ...



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