Product of numbers $\pm\sqrt{1}\pm\sqrt{2}\pm\cdots\pm\sqrt{n}$ is integer

Question

Prove that the product of the $2^n$ numbers $\pm\sqrt{1}\pm\sqrt{2}\pm\cdots\pm\sqrt{n}$ is an integer.

I want to consider the polynomial $P(x)=(x-a_1)(x-a_2)\cdots(x-a_{2^n})$, where the $a_i$'s are the $2^n$ numbers. The desired product is the constant term of the polynomial. Can we show that this polynomial have some simple form?

Try finding an inductive proof. Suppose that $P_n$ is the given product; can you find a formula for $P_{n+1}$ in terms of $P_n$? — Steven Stadnicki, Dec 17 '14 at 16:36

Thomas Andrews · Accepted Answer · 2014-12-17T17:16:30.420

10

Hint: Let $P_n(x)$ be your polynomial. Then show $P_{n+1}(x)=P_n(x-\sqrt{n+1})P_n(x+\sqrt{n+1})$, and show inductively that $P_n(x)$ always has only integer coefficients.

edited Dec 17 '14 at 17:16

answered Dec 17 '14 at 16:36

Thomas Andrews

177,126

Product of numbers $\pm\sqrt{1}\pm\sqrt{2}\pm\cdots\pm\sqrt{n}$ is integer

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