THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS

There is only one such $n$, namely, $n=2$, in which case we may take $a=1/2-\sqrt{2}$. Verifications are routine and hence omitted.

To rule out the case $n\ge 3$, let $a$ be a real number such that $a+\sqrt{2}$ is rational. For $a^n+\sqrt{2}$ to be rational, it is necessary and sufficient that $a+\sqrt{2}$ be a rational root of the degree $n-1$ polynomial $f_n={\sum }_{k=1}^{\lfloor (n+1)/2\rfloor }{2}^{k-1}\left(\genfrac{}{}{0ex}{}{n}{2k-1}\right)X^{n-2k+1}-1$. It is therefore sufficient to show that $f_n$ has no rational roots.

If $n$ is odd, then $f_n$ is a polynomial in $X^2$ whose coefficients are all positive, so it has no real roots.

If $n$ is even, then the constant term of $f_n$ is $-1$, and since the leading coefficient of $f_n$ is $n$, a rational root of $f_n$, if any, must be of the form $1/d$ for some divisor $d$ of $n$. If $d$ is one such, then $n+{\sum }_{k=3}^{n/2}{2}^{k-1}\left(\genfrac{}{}{0ex}{}{n}{2k-1}\right)d^{2(k-1)}-d^{n-1}=0$, showing that $d$ must be even.

Finally, use Legendre's standard valuation formula to infer that the highest power of $2$ dividing the greatest common divisor of the last $n/2-1$ summands in the left-hand member above exceeds the highest power of $2$ dividing $n$, and derive thereby a contradiction.