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Let $A$ be the product of $2^{100}$ numbers of the form $$\pm \sqrt{1} \pm \sqrt{2} \pm \dots \pm \sqrt{100}$$ Show that $A$ is an integer, and moreover, a perfect square.

I found a similar problem here, but the induction doesn't seem to show that $A$ is a perfect square. And I think we can generalize the following problem: if $n$ is a perfect square then $A$ is a perfect square, too.

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The same argument can be used to show that the product of the $2^{n-1}$ numbers $\sqrt{1} \pm \sqrt{2} \pm \cdots \pm \sqrt{n}$ is an integer.

Then your number is exactly the square of this product.

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  • $\begingroup$ Those $2^{100}$ numbers are totally different (from those $2^{99}$ numbers). Their amount is indeed a square of the previous amount, but that's about the only obvious fact here... isn't it? $\endgroup$ Nov 15 '15 at 20:43
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    $\begingroup$ @barakmanos "Totally different"? Every term of primitiveroot's product is either equal to a term of my product, or $-1$ times a term of my product. Since the number of signs, $2^{n-1}$, is even (for $n>1$), my conclusion follows. And no, $2^{100} \neq (2^{99})^2$. $\endgroup$
    – Slade
    Nov 15 '15 at 20:48

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