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Show that the product of the $2^{2019}$ numbers of the form $\pm 1\pm \sqrt{2}\pm\cdots \pm \sqrt{2019}$ is the square of an integer. I'm aware very similar problems were asked before (e.g. here and here), but I'm not sure how to justify the answers provided on the two linked pages.

In particular, I'm not sure how to show that if $P$ is a polynomial with integer coefficients, then so is $P(x+\sqrt{n})P(x-\sqrt{n})$ for any positive integer n, which seems to be one of the claims in one of the posted answers.

If $P(x)$ is an even polynomial, then so is $Q(x) := P(x+\sqrt{k})P(x-\sqrt{k}).$ Indeed, $P(x)=P(-x)$ for all x, so $Q(-x) =P(-x+\sqrt{k}) P(-x-\sqrt{k}) = P(x-\sqrt{k})P(x+\sqrt{k}) = Q(x).$ Also note that if $P(x)$ is even, then since $P(x)-P(-x)$ is identically zero, only the coefficients of even degree terms in P may be nonzero.

Inductively, define $P_1(x) := x+1, P_n(x) := P_{n-1}(x+\sqrt{n}) P_{n-1}(x-\sqrt{n})$ for $n\ge 2.$ We will prove by induction that $P_n(x)$ has integer coefficients, $P_n(x)$ has degree $2^n,$ and that $P_n(x)$ is even for every positive integer n. In particular, the constant term of $P_n(x)$ is even. Assume the result holds for all $1\leq k < n,$ some $n\ge 2.$ Since $P_n(x) = P_{n-1}(x+\sqrt{n}) P_{n-1}(x-\sqrt{n})$ and $P_{n-1}(x\pm \sqrt{n})$ (where we choose either plus or minus) has degree $2^{n-1}$ by the inductive hypothesis, $P_n(x)$ has degree $2^n.$ Then writing $P_{n-1}(x) = \sum_{i=0}^{2^{n-1}} a_i x^i,$ we have that for $0\leq k\leq 2^n,$ the coefficient of $x^k$ in $P_n(x) $ is given by $\sum_{i=0}^{2^n}[x^i] P_{n-1}(x+\sqrt{n}) [x^{k-i}] P_{n-1}(x-\sqrt{n}) = \sum_{i=0}^{2^n} (\sum_{j=i}^{2^{n-1}} {j\choose i} (\sqrt{n})^{j-i}) (\sum_{j=k-i}^{2^{n-1}} {j\choose k-i} (-\sqrt{n})^{j-(k-i)})$. The latter expression seems too difficult to work with.

user33096
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  • I was confused by the way you described the linked answer, "if $P$ is a polynomial with integer coefficients, then so is $P(x+\sqrt{n})P(x-\sqrt{n})$ for any positive integer $n$". The linked answer writes $P_n$ rather than $P$ and makes it clear that this is a very specific polynomial $P_n,$ not just any polynomial $P.$ Later in the question, however, it seems you understand this, because you write out the very specific inductive step for that particular polynomial. It would be less confusing if you were more specific in how you quote previous answers. – David K Nov 23 '22 at 00:50

1 Answers1

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We may group these $2^{2019}$ sums $\sum_{i=1}^{2019}\pm\sqrt i$ into $2^{2018}$ pairs, each of which is composed of a sum where there is a $+$ sign in front of $\sqrt1,$ and of the opposite sum. Therefore, the product $N$ of the $2^{2019}$ sums is the square of the product $n$ of the $2^{2018}$ sums of the form $1+\sum_{i=2}^{2019}\pm\sqrt i$, $$N=n^2,$$ and there remains to prove that $n$ is an integer. Observe that $$n=P(\sqrt2,\sqrt3,\dots,\sqrt{2019}),$$ where the polynomial $P(X_2,X_3,\dots,X_{2019})$ is the product of the $2^{2018}$ sums of the form $1+\sum_{i=2}^{2019}\pm X_i.$ By construction, $P$ has integer coefficients and is even in each $X_i,$ hence $$P(X_2,X_3,\dots,X_{2019})=Q(X_2^2,X_3^2,\dots,X_{2019}^2)$$ for some polynomial $Q$ with integer coefficients. So $$n=Q(2,3,\dots,2019)$$ is an integer.

Anne Bauval
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