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Prove that the product of five consecutive positive integers cannot be the square of an integer.

I don't understand the book's argument below for why $24r-1$ and $24r+5$ can't be one of the five consecutive numbers. Are they saying that since $24-1$ and $24+5$ aren't perfect squares it can't be so? Also, the argument after that about how $24r+4$ is divisible by $6r+1$ and thus is a perfect square is unclear.

Book's solution:

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  • $\begingroup$ No perfect square is equal to $5$ modulo $6$. (Can you check this?) So no perfect square is equal to $-1$ or $5$ modulo $24$ either. $\endgroup$ – mjqxxxx Jul 12 '16 at 0:27
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$24r-1$ and $24r+5$ are also divisible neither by $2$ nor by $3$. So they must also be coprime to the remaining four numbers, and thus must be squares.

But this is impossible, because we already know that $24r+1$ is a square, and two non-zero squares can't differ by $2$ or $4$.

For the second part: $6r+1$ is coprime to $24r,24r+1,24r+2$, and $24r+3$. So it must be a square. Hence $24r+4=4(6r+1)$ is a square. But then the two perfect squares $24r+1$ and $24r+4$ differ by $3$, and the only two squares differing by $3$ are $1$ and $4$. This forces $r=0$, which contradicts $r=k(3k\pm 1)/2$.

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  • $\begingroup$ How are $24r-1$ and $24r+5$ not divisible by $2$ and $3$? $\endgroup$ – John Ryan Jul 12 '16 at 0:30
  • $\begingroup$ @JohnRyan: Because $24r$ is divisible by both $2$ and $3$, $\endgroup$ – TonyK Jul 12 '16 at 0:31
  • $\begingroup$ What about the last part about the question about $24r+4$ being a perfect square? $\endgroup$ – John Ryan Jul 12 '16 at 0:33
  • $\begingroup$ @JohnRyan: See update. $\endgroup$ – TonyK Jul 12 '16 at 0:43
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The numbers $24r-1$ and $24r+5$ are also divisible neither by $2$ nor by $3$, so by the previous argument, if they are part of the list they must be perfect squares. However, they are odd and respectively congruent to $-1$ and $5$ modulo $8$. But any odd perfect square must be congruent to $1$ modulo $8$.

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  • $\begingroup$ What about the last part of the question about $24r+4$ being a perfect square? $\endgroup$ – John Ryan Jul 12 '16 at 0:39
  • $\begingroup$ Sorry, I was away. Your question has been answered. $\endgroup$ – André Nicolas Jul 12 '16 at 0:57

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