The original problem asked to "Show that there exists a polynomial $f(x)$ with rational coefficients such that $f(\sqrt{2}+\sqrt{3})=\sqrt{2}$."
I was wondering if there exists an integer coefficient polynomial. Intuitively, I think the answer should be no.
We know that $\sqrt2 + \sqrt3$ is a root of the polynomial $$(x-\sqrt2-\sqrt3)(x-\sqrt2+\sqrt3)(x+\sqrt2-\sqrt3)(x+\sqrt2-\sqrt3) = x^4 - 10x^2+1.$$ So in any polynomial $f$ of degree at least $4$, we can replace the leading term $a_n x^n$ by $a_n(10x^{n-2} - x^{n-4})$, reducing the degree while not changing the value of $f(\sqrt2 + \sqrt3)$. This means that it's enough to check all cubic polynomials.
If $x=\sqrt2+\sqrt3$, then out of $x^0 = 1$, $x^1 = \sqrt2 + \sqrt3$, $x^2 = 5 + 2\sqrt6$, and $x^3 = 11\sqrt2 + 9 \sqrt3$, only $x$ and $x^3$ contain any square roots of $2$. There is only one linear combination of $\sqrt2 + \sqrt3$ and $11\sqrt2 + 9\sqrt3$ that gives $\sqrt2$, and it does not have integer coefficients.
