If \(P(n)\) denotes the product of all the numbers of the form \(\pm \sqrt{1} \pm \sqrt{2} \pm \sqrt{3}\pm \cdots \pm \sqrt{n}\) where \(\pm\) are chosen arbitrarily.

Then prove that \(P(n)\) is a positive integer \(\forall n \geq 2\).

(Can we find that positive integer?)

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TopNewestMy try is as follows.

Let \(P_{n-1}(x) = a_{2^{n-1}}x^{2^{n-1}} + a_{2^{n-1}-1}x^{2^{n-1}-1} + \cdots + a_1 x^1 + a_0\)

be the polynomial having roots as numbers of the form \(\pm \sqrt{1} \pm \sqrt{2} \pm \sqrt{3} \pm \cdots \pm \sqrt{n-1}\)

Then, \(P_n(x)\) has roots as \(\text{Root of} P_{n-1}(x) \pm \sqrt{n}\). Thus we can transform it as-

\(P_n(x) = (a_{2^{n-1}}{(x-\sqrt{n})}^{2^{n-1}} + a_{2^{n-1}-1}{(x-\sqrt{n})}^{2^{n-1}-1} + \cdots + a_1 {(x-\sqrt{n})}^1 + a_0)(a_{2^{n-1}}{(x+\sqrt{n})}^{2^{n-1}} + a_{2^{n-1}-1}{(x+\sqrt{n})}^{2^{n-1}-1} + \cdots + a_1 {(x+\sqrt{n})}^1 + a_0)\)

Proposition:\(P_n(x) \text{is a polynomial having integer coefficients}\)We prove this using induction.

Base Case - \(n = 1 : x^2 -1 = 0\)

Assuming for \(n-1\), let's try for \(n\),

\(\displaystyle [x^k]{P_n(x)} = \sum_{b + c = k}{\left(\sum_{m=0}^{2^{n-1}}{\binom{m}{b} a_m {(\sqrt{n})}^{m-b}}\right)\left(\sum_{t=0}^{2^{n-1}}{\binom{t}{c} a_t {(-\sqrt{n})}^{t-c}}\right)}\)

(Oh, that is scary, but it is not required).

We can get a \(\sqrt{n}\) term left if \(m-b = \text{Odd}, t-c = \text{Even}\) or \(t-c = \text{Odd}, m-b = \text{Even}\)

Now, an irrational term or a non-integer term(in this case), will be left if either

\(m-b = \text{Odd}, t -c = \text{Even}\)

or\(m-b = \text{Even}, t -c = \text{Odd}\).

And now if there is a case like \((m-b, t-c) = (1,2)\), then there will also be a case like \((m-b,t-c) = (2,1)\). Note that this counter-case will not be there if \(m-b = t-c\) but then their parity is same too, so they will not be considered.

Since by interchanging \((m,b) \rightarrow (t,c)\), we change the coefficient of \(n^{(m+t-b-c)/2}\)from

\(\binom{m}{b}\binom{t}{c}a_m a_t \rightarrow -\binom{m}{b}\binom{t}{c}a_m a_t\)

all the numbers with fractional powers cancel out and only integers remain.

By observing the general polynomial recurrence we can see that leading coefficient is given by

\(a_n = a_{n-1}^2; a_1 = 1 \implies a_n = 1\)

And we have just proven that all the other coefficients are integers.

Therefore, all symmetric products of numbers of the form \(\pm \sqrt{1} \pm \sqrt{2} \pm \cdots \pm \sqrt{n}\) is an integer. – Kartik Sharma · 7 months ago

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In the final line, I believe you mean "product" instead of "sum". – Calvin Lin Staff · 6 months, 2 weeks ago

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– Kartik Sharma · 6 months, 2 weeks ago

Oops, yeah. Fixed it! What part exactly is not clear? Maybe the formatting isn't good.Log in to reply

Note that the induction should start with \( n = 2 \).

Note that you want to show the product is positive, which has been ignored. In particular, looking at \(n = 1 \), you should realize that the product is \( -1 \) (which is why we have to start with \( n = 2 \)).

Justifying the formula for \( [ x^k ] P_n(x) \). At least give a verbal description of how that is arrived at.

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Here's my attempt: Call this set \(S\). The set \(S\) has \(2^{n}\) members. No such member can be zero. There is a one-one correspondence between a positive member \(x \in S\) and \(-x \in S\) (note that both \(x\) and \(-x\) will belong to this set and this is seen by switching all the signs). Their product is \(-x^2 < 0\). And the number of such pairs is \(2^n/2 = 2^{n-1}\). For \(n \ge 2\), this is an even natural number. And the product of an even number of pairs, where the product of each pair is negative, is of course, positive. – Parth Kohli · 3 months, 3 weeks ago

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Hey,great solution.BTW you are in which class? – A E · 6 months, 4 weeks ago

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– Kartik Sharma · 6 months, 3 weeks ago

Class XII.Log in to reply

– A E · 6 months, 3 weeks ago

You have immense amount of knowledge as compared to a 12th grader!!!!Log in to reply

– Kartik Sharma · 6 months, 3 weeks ago

I just have a good amount of time to devote to science and mathematics.Log in to reply