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Assume that $\pi$ and $e$ are both transcendental over $\mathbb{Q}$.

I have proved $e$ and $\pi$ are both algebraic over the field $\mathbb{Q}(e+\pi, e\pi)$. The polynomial is $p(x)=x^2-(e+\pi)x+e\pi$

Now I need to deduce that at least one of the numbers $e+\pi$ and $e\pi$ is transcendental over $\mathbb{Q}$.

Any help?

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    If they are both algebraic, what can you say about $\mathbb Q(e, \pi)$? – 2017-01-15
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    A simpler way to look at it: consider the quadratic equation whose roots are $e$ and $\pi$. What can you say about its coefficients? – 2017-01-15
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    @MichaelWeiss They are algebraic? So the roots must be algebraic . Contradiction? – 2017-01-16
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    That's it, Bob! – 2017-01-16
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    Unfortunately, we neither know whether $e\pi$ is transcendental, nor whether $e+\pi$ is transcendental. But this is a consequence of the widely believed Schanuel's conjecture. It is not even known whether $e\pi$ is irrational or whether $e+\pi$ is irrational. – 2017-01-16

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We can generalize the result as follows :

Suppose, $a$ and $b$ are complex numbers, at least one of which is transcendental. Then, at least one of $a+b$ and $ab$ is transcendental:

Otherwise the polynomial $(x-a)(x-b)=x^2-(a+b)x+ab$ would have algebraic coefficients.

Since it is well known that the field of algebraic numbers is algebraically closed, we could conclude that the roots (which are $a$ and $b$) are both algebraic, contradicting our assumption.