# Some Non-Trivial Algebraic(Rational) Number

Every problem about algebraic-ness (rational-ness) of numbers that I have seen is in one of the below types:

• The number is algebraic(rational) and proving that it is algebraic(rational) is trivial, like proving that $\sqrt{7+\sqrt{2}}$ is algebraic.
• The number is not algebraic, and $\cdots$.

My Question:

• Is there any example of an algebraic(rational) number which is not trivial to prove that the number is algebraic(rational)?
• What is "simple"? I don't understand the question. Apr 5 '15 at 17:06
• I think you'll be interested in mathoverflow.net/questions/32967/…. In particular, I think Legendre's constant (en.wikipedia.org/wiki/Legendre%27s_constant, mentioned in the first answer) is a pretty good example. Apr 5 '15 at 17:08
• Conway's Constant might be an example: It is algebraic, the root of a polynomial of degee $71$. Apr 5 '15 at 17:13
• There certainly are examples that are hard if the student knows just the basic definition of algebraic and no actual related theory behind that property. For example, show $\sqrt[3]{2} + \sqrt[4]{2} + \sqrt[5]{2}$ is algebraic, or if $\alpha$ is a root of $x^3 + x + 1$ and $\beta$ is a root of $x^8 + 6x^3 + 4x - 10$ then prove $\alpha + \beta$ is algebraic. Do you think those are trivial? Apr 6 '15 at 15:01
• How about $${1\over\pi^2}\sum_1^{\infty}{1\over n^2}$$ Apr 7 '15 at 0:59

There are plenty of examples. See, for example this page from google books, which talks about non-trivial proofs of algebraicity of functions. In particular, if a function is algebraic, so are its special values at rational arguments (one example: function is algebraic if its power series has integer coefficients and is meromorphic in a domain of conformal radius $> 1$ - this result is due to Borel and Polya).
Let $E/\mathbb{Q}$ be an elliptic curve, let $\Omega$ be its real period (which is the value of an elliptic integral that's easy to write down using the equation of $E$), and let $L(E,s)$ be the $L$-series of $E$, which we know by Wiles' theorem has an analytic continuation to all of $\mathbb{C}$. Then $L(E,1)/\Omega$ is a rational number, but I'd say that that's not so easy to prove. Even harder, suppose that $E(\mathbb{Q})$ has rank 1, let $P$ be a non-torsion point, and let $\hat h(P)$ be the canonical height of $P$. Then $$\frac{L'(E,1)}{\Omega\cdot\hat h(P)}$$ is a rational number. In this latter problem, we don't actually know that $\hat h(P)$ is not itself rational (this is unknown for even a single example), but the consensus seems to be that $\hat h(P)$ should be transcendental if it's not zero.