Let $p(x_1,x_2, \dots , x_n) =a_0+ a_1x_1+a_2x_2+\dots+a_n{x_n}$ be a polynomial where coefficients $a_i$ are integer and variables $x_i$ are exclusively square roots of distinct primes $p_i, \ \ \ $ i.e. $x_i=\sqrt{p_i}$.

I would like to prove that such polynomial never can be equal to $0$.

Example of such polynomial for $n=3 \ \ \ \ $ $p(\sqrt{2},\sqrt{3},\sqrt{5})=1 + 4\sqrt{2}-\sqrt{3} +3\sqrt{5}$.

It's relatively easy to prove that for small $n$.

For $n=1$ expresion $a_0+a_1\sqrt{p_1}$ can't be equal to $0$ because it would mean that from $-a_0/a_1=\sqrt{p_i}$ right side expression would be rational what is not the case.

For $n=2$ expression $a_0+a_1\sqrt{p_1}+a_2\sqrt{p_2} =0$ also can't be equal to $0$.
Squarring both sides of $a_0=-a_1\sqrt{p_1} -a_2\sqrt{p_2} $ would lead to $a_0^2= a_1^2 p_1 +a_2^2 p_2 +2a_1 a_2\sqrt{p_1p_2} $ which also leads to the contradiction as above.

The same technique can be used also for $n=3$ and $n=4$, squarring appropriately sides of equation leads to reduction of the number of square roots in the equation and the contradiction rational vs. irrational can be detected.

For bigger $n$ however this method doesn't act because expressions like $(\sqrt{p_1}+\sqrt{p_2}+\sqrt{p_3})^2$ have the same number of square roots as before squarring, for bigger $n$ the number of square roots in the final expression even increases.

So my question is

  • how to prove that $a_0+ a_1x_1+a_2x_2+\dots+a_n x_n \neq 0 $ for arbitrarily big $n$ with constraints imposed on coefficients (integer numbers) and variables (square roots of primes) as above ?
  • $\begingroup$ Your claim is equivalent to showing that $\{1,\sqrt2,\sqrt3,\sqrt5,\sqrt7,\ldots\}$ is linearly independent over $\Bbb{Q}$. This has been explained on our site earlier. Give me a minute. $\endgroup$ – Jyrki Lahtonen Jul 30 '17 at 16:12
  • $\begingroup$ @JyrkiLahtonen Yes, but primes may be chosen randomly only they should be distinct.. $\endgroup$ – Widawensen Jul 30 '17 at 16:16
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    $\begingroup$ See here or here for somewhat more general results. The conclusion is that for any polynomial $f(x_1,x_2,\ldots,x_n)$ with integer coefficients we have $$f(\sqrt{p_1},\sqrt{p_2},\cdots,\sqrt{p_n})\neq A\sqrt{p_{n+1}},$$ where $p_i$ are distinct primes and $A\in\Bbb{Z}\neq0$. This is just easier to formulate using the language of field extensions. What it means is that you cannot write $\sqrt p_1$ as an integer linear combination of $\sqrt{p_2},\sqrt{p_3},\sqrt{p_2p_3}$ et cetera. $\endgroup$ – Jyrki Lahtonen Jul 30 '17 at 16:16
  • $\begingroup$ Linear independence of an infinite set means that no finite subset will satisfy a linear equation, so the claim in my first comment covers all finite combinations of square roots of arbitrarily chosen primes. $\endgroup$ – Jyrki Lahtonen Jul 30 '17 at 16:21
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    $\begingroup$ I might not be understanding the question, but can I ask how it's not simply possible to select an extremely large prime, then square root it, and then just subtract it from the convex opening polynomial to force it to encounter the x-axis at some point? Otherwise, isn't any polynomial with an odd degree guaranteed to equal zero at least one time? $\endgroup$ – user468211 Jul 30 '17 at 16:59

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