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Let $k\ge 2$ be an integer. Can a product of $k$ consecutive integers be a perfect square?

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    $\begingroup$ @THELONEWOLF.: You have vandalised this question! Changing "can" to "Can" is perhaps laudable, but surely unnecessary. Changing "doesn't" to "does not" is officious and annoying. Changing "perfect" to "peqrfect" is beyond comprehension. $\endgroup$ – TonyK Dec 11 '16 at 20:04
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    $\begingroup$ @123 why not prime numbers?? $\endgroup$ – Vidyanshu Mishra Dec 11 '16 at 20:09
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    $\begingroup$ Any good theorem should have a proof without using primes, we have to learn to live without primes. —123 $\endgroup$ – user372272 Dec 11 '16 at 20:12
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    $\begingroup$ See also product of six consecutive integers being a perfect square for references and an idea of the difficulty of this question. $\endgroup$ – punctured dusk Dec 11 '16 at 20:23
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    $\begingroup$ @THELONEWOLF. Yea, using primes is nothing bad, I would say quite opposite. Primes are building blocks in number theory, seems quite unnatural to avoid them. I would understand avoiding using advanced theorems, but primes... $\endgroup$ – Sil Dec 11 '16 at 20:31
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What you want is HERE $1$, and HERE $2$. I think copying/describing this text will take an entire hour which is not good for my fingers. So, just look at the paper I have given.

Note that the theorem in the paper $1$ is a generalised one. It states that The product of two or more consecutive positive integers is never a power. And your squares also come under this section.

Hope it will help you.

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  • $\begingroup$ Firstly, thank you for your helpful comments. I have already read the article that you've linked, it is using primes. I want a proof that doesn't using primes, but thank you for your answer. $\endgroup$ – user372272 Dec 11 '16 at 20:53
  • $\begingroup$ Thanks @123. I have linked another paper. $\endgroup$ – Vidyanshu Mishra Dec 11 '16 at 20:56
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    $\begingroup$ @123 Why don't you accept primes in the proof? This is almost as saying "I want a proof that doesn't use integers, but thank you for your efforts". $\endgroup$ – Dietrich Burde Dec 11 '16 at 22:25
  • $\begingroup$ I would agree with you @Dietrich Burde in this case. I m trying to convince OP in this matter for hours but no result. $\endgroup$ – Vidyanshu Mishra Dec 11 '16 at 22:55
  • $\begingroup$ The article that you've linked after the first one is helped me. This answer is what I'm looking for! $\endgroup$ – user372272 Dec 12 '16 at 13:22