# Sum of radicals is rational only for perfect squares [duplicate]

How can we prove that if $a_1\sqrt{n_1}+a_2\sqrt{n_2}+\dots+a_k\sqrt{n_k}\in\mathbb{Q}$ with $a_1,a_2\dots, a_k\in\mathbb{Q}^{*}_{+}$, then the natural numbers $n_1,n_2,\dots, n_k$ are perfect squares?

I have proved it for $k=1,2,3$ but for $k\geq 4$ I don't find anything. I couldn't show even that $\sqrt{2}+\sqrt{3}+\sqrt{5}+\sqrt{7}\notin\mathbb{Q}$.

## marked as duplicate by Dietrich Burde, Namaste, Leucippus, user91500, Alex M.Nov 25 '16 at 9:44

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• You need some reformulation, because $(-1)\sqrt{12}+ 2\sqrt{3}$ is rational. – G Tony Jacobs Nov 24 '16 at 15:17
• What about $\sqrt{2}-\sqrt{2}$? – Oiler Nov 24 '16 at 15:17
• Coefficients must be positive... – Bogdan Nov 24 '16 at 15:18
• Ah. How did you prove it for $k=3$? – G Tony Jacobs Nov 24 '16 at 15:19
• @Bogdan check this answer: math.stackexchange.com/a/437374/133781 – Xam Nov 24 '16 at 15:31