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Find all such positive integers n that the set $(n,n+1,n+2,n+3,n+4,n+5)$ can be partitioned into two subsets such that product of the two subsets is equal.

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closed as off-topic by S.C.B., JKnecht, Pragabhava, Daniel W. Farlow, Ben Sheller Apr 12 '16 at 18:43

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  • $\begingroup$ What did you try? $\endgroup$ – S.C.B. Apr 12 '16 at 7:12
  • $\begingroup$ i tried to do some calculations such as looking at possibilities as n*(n+1)=(n+2)*(n+3)*(n+4)*(n+5) and all such possibilities but could not arrive at solution.i got this problem in an imo book $\endgroup$ – satyajeet jha Apr 12 '16 at 7:14
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    $\begingroup$ Rigge, Uber ein diophantisches Problem, 9th Congress Math. Scand. 155-160, proved that a product of two or more consecutive positive integers cannot be a square. $\endgroup$ – Gerry Myerson Apr 12 '16 at 7:17
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If $p>=5$ is a prime number, then $p$ cannot divide $n+1$, $n+2$, $n+3$ or $n+4$ (if it did, it would divide EXACTLY ONE of the six numbers, and you couldn't break them into two subsets with equal products). So the four numbers in the middle are divisible, at most, by 2 and 3. Two of them are odd, so they must be powers of 3, and the difference between them is 2. So either $n+1$, $n+3$ or $n+2$, $n+4$ must be 1 and 3. Which is not possible. So there are no solutions.

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