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It's a question from the Bangladesh Mathematical Olympiad. It still haunts me a lot. I want to find an answer to this question.

Find, with proof, all the perfect squares each of which is the product of four consecutive odd natural numbers.

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    $\begingroup$ Baltimore & Delaware Motor Organization? (The point being, you shouldn't assume the everyone or even the majority of us know what this means; I think you posted with the same acronym and haunted rhetoric yesterday. Bangladesh Mathematical Olympiad) $\endgroup$ – Simon S Nov 9 '14 at 1:03
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Consider that any such number must be in the following form, for integer $m$: $$(2m-3)(2m-1)(2m+1)(2m+3)=(4m^2-9)(4m^2-1).$$ Notice that both factors on the left equal $3\pmod 4$ and so cannot individually be square. Moreover, from the Euclidean algorithm $\gcd(4m^2-9,4m^2-1)=\gcd(8,4m^2-1)=1$, so the numbers are coprime, implying that their product is square if and only if they are both squares. This never happens. ($m=0$ is a solution, since both terms are the negation of a square, but it represents the product $-3\cdot -1\cdot 1\cdot 3$, which doesn't remain in the naturals)

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let n be an odd number:[n-2, n, n+2, n+4]# all numbers in this set are odd if n is odd

$m^2 = (n)(n+2)(n+2)(n-4)$

$m^2 = (n)(n-4)(n^2-4)$#difference of squares

$m^2 =(n^2-4n)(n^2-4)$

$m*m$ $ $ $!= (n^2-4n)(n^2-4)$

m cannot be both $(n^2-4n)$ and $(n^2-4)$ and $(n^2-4n)=(n^2-4)$ only has one solution "1" however that would make our original set [-1, 1, 3, 5] which is not in the set of natural numbers therefore there is no solution.

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We observe that at most two of the four numbers are divisible by 3, and that any larger odd prime can only divide one of them, hence must occur to an even power in the number it divides. But this means that two of the numbers which are not divisible by three are odd squares having a difference of 2, 4 or 6, which is impossible.

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