4

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 ?
Michael Hardy
  • 1
Widawensen
  • 8,172
  • 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. – Jyrki Lahtonen Jul 30 '17 at 16:12
  • @JyrkiLahtonen Yes, but primes may be chosen randomly only they should be distinct.. – Widawensen Jul 30 '17 at 16:16
  • 2
    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. – Jyrki Lahtonen Jul 30 '17 at 16:16
  • 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. – Jyrki Lahtonen Jul 30 '17 at 16:21
  • Anyway, because I answered one of the linked questions I won't vote to close this as a duplicate. It just wouldn't be kosher to use a power vote. – Jyrki Lahtonen Jul 30 '17 at 16:23
  • @JyrkiLahtonen very deep answers requiring a deep thinking over.. – Widawensen Jul 30 '17 at 16:23
  • Widawensen, not very deep, but familiarity with field theory is required. Familiarity with Galois theory will open other avenues of attacking this. My answer reduces the general case essentially to the way you handled the case $n=2$, but uses Galois theory. Bill Dubuque's answer only uses basic field theory, and may be better for you. – Jyrki Lahtonen Jul 30 '17 at 16:27
  • @JyrkiLahtonen I'm afraid that Galois theory is too difficult for me ( maybe it is somewhere easy introduction into it?) , yes, I was hoping naively that explanation is possible without very much advanced theory, anyway, I guess it is the example when things are becoming much more complicated when $n$ increases.. – Widawensen Jul 30 '17 at 16:34
  • @JyrkiLahtonen To say truth positive confirmation of the claim means also that it is the positive answer also for this question https://math.stackexchange.com/questions/2372391/can-we-always-find-such-a-vector#comment4889915_2372391 which I had in mind having asked this question.. do you agree Jyrki? – Widawensen Jul 30 '17 at 16:45
  • 1
    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? –  Jul 30 '17 at 16:59
  • @JyrkiLahtonen To add one more additional remark comparing my question to the quoted by you it seems that my question is asked with the use of much simpler words, not everything from both other questions is clear for me, they are formulated on the much more advanced level.. – Widawensen Jul 30 '17 at 16:59
  • @DaneJoe "to select an"? It's a question for me ? – Widawensen Jul 30 '17 at 17:02
  • @Widawensen It's a question for anyone who can answer it :) –  Jul 30 '17 at 17:02
  • @DaneJoe I see :) – Widawensen Jul 30 '17 at 17:03
  • @Widawensen But no your comment is concerning to me, because you're talking about every possible polynomial for these coefficients never equaling zero, so it would completely shatter my understanding of everything I know in mathematics if that was true, because odd degree polynomials are guaranteed to equal zero at least once. This means you would have to be wrong to assume every polynomial of these coefficients never equals zero. So that's why I'm hoping I am just misunderstanding the problem. –  Jul 30 '17 at 17:04
  • So I've meditated on it a little bit more and I have found another possible interpretation for your conjecture that may resolve my issue with it. So for every distinct instance, you're not actually talking about different powers in the general sense of a polynomial like x^2 or x^3, rather every single instance of "x" in your conjecture always has a degree of exactly one, and you are simply assigning coefficients to these linear combinations of only $x^1$. If that's the case, then your presumption makes sense. –  Jul 30 '17 at 17:12
  • @DaneJoe Yes, "always has a degree of exactly one" is correct meaning.. – Widawensen Jul 30 '17 at 17:14
  • @Widawensen Okay great, phew (out of breath) –  Jul 30 '17 at 17:16

0 Answers0