Let's prove the polynomial $P(x):=1+x^2+x^{n_1}+\dots +x^{n_k}$ has no complex roots in the disk $\left\{z:|z|\le \frac{\sqrt{5}-1}{2}\right\}$ where $2<n_1<n_2<\dots<n_k$ are integers. Assume $|x|\le \frac{\sqrt{5}-1}{2}.$ We have \begin{align*} |1+x^2+x^{n_1}+\dots +x^{n_k}|&\ge 1-|x|^2-|x|^{n_1}-\dots-|x|^{n_k}\\ &>1-|x|^2-|x|^3-|x|^4-\dots\\ &=1-|x|^2\frac{1}{1-|x|}\ge 0. \end{align*}Thus, when $|x|\le \frac{\sqrt{5}-1}{2}$ it follows $|P(x)|>0$, hence the claim holds. In our case, the real root of $P(x)$ cannot be positive, so it's negative and less than $\frac{1-\sqrt{5}}{2}.$ Comment. I've recently seen something like that, given at a certain competition. Can't remember exactly, but it was some top competition, I think. The fact that $P(z):=1+z^{n_1}+z^{n_2}+\dots+z^{n_k}$ does not have zeroes in $\left\{z:|z|\le \frac{\sqrt{5}-1}{2}\right\}$ where $1\le n_1<n_2<\dots<n_k$ are integers. The case $n_1\ge 2$ goes like above. In the case $n_1=1$ we consider $P_1(z):=(1-z)P(z)$ and apply again the above argument. EDIT: Here it is: Putnam 1972, B6