Finding a monic polynomial with integer coefficients having $\sqrt{2} + \sqrt{3} + \sqrt{5} + \sqrt{7}$ as one of its roots.

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $x$.