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For any positive integer $n$, prove that $\displaystyle \prod_{j=1}^{2n}\left(j^j(2n+j)^{j-1}\right)$ is a perfect square.

Let $f(n) = \prod_{j=1}^{2n}\left(j^j(2n+j)^{j-1}\right)$. Then $$f(1) = 2^4, \quad f(2) = 2^{20} \cdot 3^4 \cdot 7^2, \quad f(3) = 2^{32} \cdot 3^{18} \cdot 5^8 \cdot 11^4.$$ I didn't see a way to prove the expression is a perfect square for all $n$.

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  • $\begingroup$ We can write explicitly $f(n) = \prod_{j=1}^{2n}\left(j^j(2n+j)^{j-1}\right)$ as $$ 1^1 (2n + 1)^0 \cdot 2^2 (2n+2)^1 \cdot 3^3 (2n+3)^2 \cdots (2n)^{2n} (2n+2n)^{2n-1}.$$ Then throw away all the terms that are already squares. It suffices to show that $(2n-1)!\,2^n(n+1)(n+2)\cdots(n+n) = 2^n (2n-1)!\frac{(2n)!}{n!}$ is square. By induction then I end up with needing to show $(2n)$ is a square. I am confused then. $\endgroup$
    – swoopin_swallow
    Mar 26, 2017 at 14:23

1 Answer 1

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Extracting even powers, we see that, up to a perfect square, this expression is equal to $$ \prod_{\text{odd }j\le 2n} j \cdot \prod_{\text{even }j\le 2n}(2n+j) = (2n-1)!! \cdot \prod_{k=1}^n(2n+2k) = (2n-1)!! \cdot 2^n \frac{(2n)!}{n!} \\ = (2n-1)!! \cdot 2^{2n} \frac{(2n)!}{(2n)!!} = \bigl(2^n (2n-1)!!\bigr)^2. $$

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  • $\begingroup$ How did you get the last equality? $\endgroup$
    – user19405892
    Mar 26, 2017 at 14:29
  • $\begingroup$ @user19405892, $(2n)!/(2n)!!= (2n-1)!!$. $\endgroup$
    – zhoraster
    Mar 26, 2017 at 14:30
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    $\begingroup$ I get $\prod_{odd j\leq 2n}j=(2n)!/(2^n n!)$ and $\prod_{(even j\leq 2n...j>0)}(2n+j)=2^n(2n)!/n!$ and the product of these is $((2n)!/n!)^2.$ $\endgroup$
    – DanielWainfleet
    Mar 27, 2017 at 6:17
  • $\begingroup$ @user254665, yes, this is the same. $\endgroup$
    – zhoraster
    Mar 27, 2017 at 7:56
  • $\begingroup$ Yes. But I'm not familiar with the $!!$ notation. $\endgroup$
    – DanielWainfleet
    Mar 27, 2017 at 7:58

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