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Find all positive integers $n$ for which$$N=\left({1^4+\frac14}\right)\left({2^4+\frac14}\right)...\left({n^4+\frac14}\right)$$ is the square of a rational number.

$$N=\prod_{k=1}^n\left(k^4+\frac14\right)$$ we have : $$k^4+\frac{1}{4}=k^4+\frac{1}{4}+k^2-k^2=\left({k^2+\frac{1}{2}}\right)^2-k^2=\left({k^2+k+\frac{1}{2}}\right)\left({k^2-k+\frac{1}{2}}\right)$$

therefore: $$\begin{align}N&=\prod_{k=1}^n\left({k^4+\frac{1}{4}}\right)\\&=\prod_{k=1}^n\left({k^2+k+\frac{1}{2}}\right)\left({k^2-k+\frac{1}{2}}\right)\\&=\prod_{k=1}^n\left({k^2+k+\frac{1}{2}}\right)\prod_{k=1}^n\left({k^2-k+\frac{1}{2}}\right)\\&=\frac{1}{2}\left({n^2+n+\frac{1}{2}}\right)\left({\prod_{k=1}^{n-1}\left({k^2+k+\frac{1}{2}}\right)}\right)^2 \end{align}$$

Now we can see that the equality : $$\frac{1}{2}\left({n^2+n+\frac{1}{2}}\right)=m^2$$ where $m\in \mathbb{R}$

Delta
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    $$8m^2=\cdots=(2n+1)^2+1$$ See https://mathworld.wolfram.com/PellEquation.html OR https://math.stackexchange.com/questions/1045127/how-to-find-a-fundamental-solution-to-pells-equation – lab bhattacharjee Mar 27 '24 at 13:53
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    Extending the comment of @labbhattacharjee, set $~X = (2n+1), Y = 2m.~$ Then, you are looking for all integer solutions to $~X^2 - 2Y^2 = -1.~$ In the theory of continued fractions, with $~X,Y~$ further constrained to each be a positive integer, the infinite set of all satisyfing ordered pairs $~(X = P_k, Y = Q_k) ~$ is defined by $$P_k + Q_k\sqrt{2} = \left[ ~1 + \sqrt{2} ~\right]^k ~: k,P_k,Q_k \in \Bbb{Z^+}, ~k ~\text{odd}.$$ – user2661923 Mar 27 '24 at 14:02
  • Re last comment, see also Khinchin pdf, chapter 1 (specifically) and see also Old's pdf. See also this answer. – user2661923 Mar 27 '24 at 14:11
  • Please answer in tge answer, not in comments, or not at all. Note my attempt to obey this rule resulted in a drive-by downvote, which then led to deletion. – Oscar Lanzi Mar 27 '24 at 14:48
  • Based on the first few successful cases (found via Mathematica) I'd suggest that the relevant OEIS equence is https://oeis.org/A001652. This is supported by the following entry on said OEIS entry: "Numbers m such that Product_{k=1..m} (4*k^4+1) is a square (see A274307)" . – Semiclassical Mar 27 '24 at 14:57
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    @labbhattacharjee for integers, congruences say $(2n+1)^2 + 1 $ is never a multiple of $8$ as $(2n+1)^2 \equiv 1 \pmod 8$ – Will Jagy Mar 27 '24 at 18:15
  • I guess we should have $\left({\prod_{k=1}^n\left({k^2{\color{red}-}k+\frac{1}{2}}\right)}\right)^2$ instead of $\left({\prod_{k=1}^n\left({k^2+k+\frac{1}{2}}\right)}\right)^2$. – Alex Ravsky Apr 08 '24 at 11:27

1 Answers1

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Equivalently, we wish to find all values of $n$ for which $2n^2+2n+1$ is a square of a positive integer. Therefore, we should find integer solutions to $2n^2+2n+1=m^2$ or $(2n+1)^2-2(m)^2=-1$. Through a renaming of variables, the following equation should be solved: $$ x^2-2y^2=-1\quad,\quad x\in\Bbb O\quad $$ where $\Bbb O$ is the set of positive odd integers, resp. The latter equation is a negative Pell's equation with $n=2$ with the fundamental solution of $(x_1,y_1)=(1,1)$. Other solutions are given by: $$ { x_{k+1}=3x_k+4y_k \\ y_{k+1}=2x_k+3y_k, } $$ for $k\in\Bbb N$. It is observed that $(x_k,y_k)\in\Bbb O\times \Bbb O$. Also, $$ x_k=\frac{(1-\sqrt2)^{2k+1}+(1+\sqrt2)^{2k+1}}{2}\quad,\quad k\in\Bbb N $$ therefore, $n=\frac{x-1}{2}=\frac{(1-\sqrt2)^{2k+1}+(1+\sqrt2)^{2k+1}-2}{4}$ for $k\in\Bbb N$ $\blacksquare$

Mostafa Ayaz
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