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The original question is to prove that $\sqrt{2} + \sqrt{3} + \sqrt{5} + \sqrt{7}$ is irrational, which already has many answers. But one of the answers to that question used this method:

They found out a monic polynomial(coefficient of leading term=1) with integer coefficients with the given number as one of its roots. Now the roots of this polynomial must either be irrational or integers(due to the rational root theorem). We can easily prove that the given number is not an integer and thus it must be irrational.

Now such a polynomial could be found out by the following method(or so I was told):

We multiply all the linear factors with the roots ($\sqrt{2} \pm \sqrt{3} \pm \sqrt{5} \pm \sqrt{7}$) (which are 16 roots in total) so we will get a $16$ degree polynomial. And it turns out that such a polynomial actually has integer coefficients and is also monic. But my question is, is there any way we prove that each of the coefficients are integers without actually expanding the whole thing?

My attempt: I have tried to use the Vieta's formulas but it is easy only to prove that the coefficient of $x^{15}$ is $0$. Everyhing else seems to be possible only through expanding the terms out.

Is there any way to do this without expanding it?

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  • $\begingroup$ Why would someone need such polynomial? $\endgroup$
    – nonuser
    Apr 4, 2020 at 12:49
  • $\begingroup$ Not only the coefficient of $x^{15}$ is $0$: this is so for all the coefficients of odd exponent. The exercise (hard enough!) is very good for a student of algebraic numbers. $\endgroup$
    – Piquito
    Apr 4, 2020 at 13:06
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    $\begingroup$ You should post a link to the original question. $\endgroup$
    – TonyK
    Apr 4, 2020 at 13:07
  • $\begingroup$ quora.com/… $\endgroup$
    – Aditya
    Apr 4, 2020 at 13:33
  • $\begingroup$ @Aditya We can get this polynomial by hand (maybe a bit of by using calculator). It's not so hard. I am ready to show it with a full explanation. $\endgroup$ Apr 4, 2020 at 13:45

3 Answers 3

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Are you familiar with the notion of Galois conjugates?

For instance, the product of all 16 conjugates will equal the norm of $\sqrt 2+\sqrt 3+ \sqrt 5 + \sqrt 7$, which is an integer, etc.

That's the easiest way to see why they are all integers.

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    $\begingroup$ Best not to answer a question with a question. Or to post an answer relying significantly on a link. $\endgroup$
    – amWhy
    Apr 4, 2020 at 13:26
  • $\begingroup$ And the trace also is an integer (in this case the zero found by the O.P.). $\endgroup$
    – Piquito
    Apr 4, 2020 at 13:45
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Let $x=\sqrt2+\sqrt3+\sqrt5+\sqrt7$ thus, $$(x-\sqrt2-\sqrt3)^2=12+2\sqrt{35},$$ which gives $$((x-\sqrt2-\sqrt3)^2-12)^2=140$$ or $$x^4+A(\sqrt2,x)\sqrt3+B(\sqrt2,x)=0,$$ where $A$ and $B$ are a polynomials of $\sqrt2$ and of $x$ with integer coefficients, with degrees by $x$ less than $4$.

Now, $$(x^4-B(\sqrt2,x))^2=3A(\sqrt2,x)^2$$ gives $$x^8+C(x)\sqrt2+D(x)=0,$$ where a degree of the polynomials $C$ and $D$ with integer coefficients are less than $8$.

Thus, $$(x^8-D(x))^2-2C(x)^2$$ gives a polynomial of degree $16$ with integer coefficients and a coefficient before $x^{16}$ is equal to $1$ and $\sqrt2+\sqrt3+\sqrt5+\sqrt7$ is a root of this polynomial.

Id est, it's enough to check that $\sqrt2+\sqrt3+\sqrt5+\sqrt7$ is not an integer number.

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    $\begingroup$ @Aditya: Regarding Michael Rozenberg's observation that in this case we only need to check that $\sqrt{2} + \sqrt{3} + \sqrt{5} + \sqrt{7}$ is not an integer, for another such example see my answer to Is number rational. For rationalizing more generally, see this and this. $\endgroup$ Apr 4, 2020 at 15:34
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    $\begingroup$ $\sqrt2+\sqrt3+\sqrt5+\sqrt7$ is suspiciously close to an integer! Proving that it is not, in fact, equal to $8$ will involve some complicated fractions. $\endgroup$
    – TonyK
    Apr 4, 2020 at 22:41
  • $\begingroup$ @TonyK: Perhaps by actually carrying out the process, at least sufficiently to obtain the constant term of the polynomial, we'll find that $8$ is not among the rational root possibilities (what I actually did in "my answer" I cited). As for actually doing this, I might possibly give it a try sometime (not now; I'm busy with something else), but for others interested, I'll mention that I actually did carry this out for $\sqrt 2 + \sqrt{10} + \sqrt{12} + \sqrt{56}$ (continued) $\endgroup$ Apr 5, 2020 at 9:02
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    $\begingroup$ Actually it's easier than I thought. We have: $8.01=1.41+1.73+2.23+2.64<\sqrt 2+\sqrt 3+\sqrt 5+\sqrt 7<1.42+1.74+2.24+2.65=8.05$. $\endgroup$
    – TonyK
    Apr 5, 2020 at 9:46
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    $\begingroup$ @TonyK: Nice! But to show each of the $4$ inequalities $1.41 < \sqrt 2 < 1.42$ and $1.73 < \sqrt 3 < 1.74$ and $2.23 < \sqrt 5 < 2.24$ and $2.64 < \sqrt 7 < 2.65$ (all from scratch) is a bit tedious. However, now that I think about it, this can be done by squaring, ${1.41}^{2} < 2 < {1.42}^{2}$ etc., which I imagine will be less work than determining the constant value of an appropriate polynomial needed here (I still haven't tried working this out, however). $\endgroup$ Apr 5, 2020 at 11:36
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Here's another method:

Let $p=\sqrt2$, $q=\sqrt3$, $r=\sqrt5$ and $s=\sqrt7$. Then here's a set of polynomials that when solved, give $x$:

$$x-p-q-r-s=0\\p^2-2=0\\q^2-3=0\\r^2-5=0\\s^2-7=0$$

Then, you can use resultants to eliminate the unwanted variables, for example, to eliminate $s$ using Bezout's resultant to $s^2+0.s+(-7)$ and $-1.s+(x-p-q-r)$, you find the determinant of $$\left[\begin{matrix}1&0&-7\\-1&x-p-q-r&0\\0&-1&x-p-q-r \end{matrix}\right]$$ which yields $$(x-r-p-q)^2-7=0$$

The algebra gets messy, but repeated applications of resultants will eliminate the unwanted variables one by one, leaving a polynomial in $x$.

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