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The following is based on this question which shows mathematically that if

$$a = \sqrt[3]{1 + \sqrt{\frac{152}{27}}}- \sqrt[3]{-1 + \sqrt{\frac{152}{27}}}$$

then

$$a^3 + 5 a$$

is an integer.

In Mathematica, though, the result does not hold:

a = Power[1 + Sqrt[152/27], (3)^-1] - Power[-1 + Sqrt[152/27], (3)^-1];

IntegerQ[a^3 + 5 a]

(* False *)

How can we "prove" or compute the proper result?

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  • 7
    $\begingroup$ Using IntegerQ[FullSimplify[a^3+5*a]] gives True. Relevant quote from the documentation: "IntegerQ[expr] returns False unless expr is manifestly an integer (i.e. has head Integer)". In the "Possible Issues" section of the documentation, this kind of problem is discussed for GoldenRatio - 1/GoldenRatio. $\endgroup$
    – user293787
    Aug 3 at 23:00
  • $\begingroup$ Thanks. That does it! $\endgroup$ Aug 3 at 23:05
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    $\begingroup$ I vastly prefer using AlgebraicNumber[] for these sorts of things: IntegerQ[AlgebraicNumber[Power[1 + Sqrt[152/27], 1/3] - Power[-1 + Sqrt[152/27], 1/3], CoefficientList[a^3 + 5 a, a]]]. $\endgroup$ Aug 3 at 23:06
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    $\begingroup$ A bit disappointing that ResourceFunction["RadicalDenest"][a^3 + 5*a] splats. $\endgroup$ Aug 3 at 23:27

2 Answers 2

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A couple more ways:

ToNumberField[a^3 + 5a] ∈ Integers
True
AlgebraicIntegerQ[a^3 + 5a] && Exponent[MinimalPolynomial[a^3 + 5a, x], x] == 1
True
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1
  • $\begingroup$ Thanks. Your first solution is particularly valuable. ($+1$) $\endgroup$ Aug 4 at 20:57
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Clear["Global`*"]

a = (1 + Sqrt[152/27])^(1/3) - (-1 + Sqrt[152/27])^(1/3);

Use RootReduce

a^3 + 5 a // RootReduce

(* 2 *)

Or FullSimplify

a^3 + 5 a // FullSimplify

(* 2 *)
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    $\begingroup$ Yep... and then perform IntegerQ... $\endgroup$ Aug 3 at 23:48

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