# Can the product of $k$ conseuctive integers be a perfect square? [duplicate]

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

• @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. – TonyK Dec 11 '16 at 20:04
• @123 why not prime numbers?? – Vidyanshu Mishra Dec 11 '16 at 20:09
• Any good theorem should have a proof without using primes, we have to learn to live without primes. —123 – user372272 Dec 11 '16 at 20:12
• See also product of six consecutive integers being a perfect square for references and an idea of the difficulty of this question. – punctured dusk Dec 11 '16 at 20:23
• @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... – Sil Dec 11 '16 at 20:31

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.