<s>[INST] Why is $n_1 \sqrt{2} +n_2 \sqrt{3} + n_3 \sqrt{5} + n_4 \sqrt{7} $ never zero? Here $n_i$ are integral numbers, and not all of them are zero.
It is natural to conjecture that similar statement holds for even more prime numbers. Namely, 
$$ n_1 \sqrt{2} +n_2 \sqrt{3} + n_3 \sqrt{5} + n_4 \sqrt{7} + n_5 \sqrt{11} +n_6 \sqrt{13} $$ is never zero too.
I am asking because this is used in some numerical algorithm in physics
 [/INST] Assume that:
$$ n_1\sqrt{2}+n_2\sqrt{3}+n_3\sqrt{5}+n_4\sqrt{7}=0.\tag{1} $$
That implies:
$$ (2n_1^2+3n_2^2)+2n_1 n_2 \sqrt{6} = (5n_3^2+7n_4^2)+2n_3 n_4 \sqrt{35}\tag{2} $$
as well as (just keep rearranging and squaring):
$$ (2n_1^2+3n_2^2-5n_3^2-7n_4^2)^2 = 24 n_1^2 n_2^2 + 140 n_3^2 n_4^2 - 8n_1 n_2 n_3 n_4 \sqrt{2\cdot 3\cdot 5\cdot 7} \tag{3}$$
but the last line implies $\sqrt{2\cdot 3\cdot 5\cdot 7}\in\mathbb{Q}$, which we know to be false. 
The same approach works also in the second case.
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