factorials to perfect squares

[Mohaimin]

Tahmid, check if this google search result help.
Google is our friend!

[Masum]

Theorem (ERDOS):
The product of more than two consecutive numbers is not a perfect power.

could you please give the proof or a link to the proof.

See this now.

[Labib]

Rewrite as $(n-2)!((n+2)(n+1)n(n-1)+1)$
But $(n+2)(n+1)n(n-1)+1=(n^2-n+1)^2$
So we need $(n-2)!$ a square.

How do you factorise and defactorise so well??
Some tips please...

[Masum]

That's simple. In fact, it is an exercise of our secondary level. You should face this in factorizing chapters.
Re-write as \[n(n+1)(n-1)(n+2)+1=(n^2+n)(n^2+n-2)+1=(n^2+n-1)^2\]

[Labib]

Masum vai, I got the factorization here.
But how do you know when to add that additional $1$ to get such a nice figure. [$(n^2+n-1)^2$]
Factorizing is really helpful to solve NT problems....
But how did you learn to use it so perfectly??

[Tahmid Hasan]

thank you very much Masum vai.

[Tahmid Hasan]

Masum vai, I got the factorization here.
But how do you know when to add that additional $1$ to get such a nice figure. [$(n^2+n-1)^2$]
Factorizing is really helpful to solve NT problems....
But how did you learn to use it so perfectly??

well this is a very well known identity:the product of 4 consecutive integers is $1$ less from a perfect square.for proof you can check the first chapter(approximately page 4-5) of "ART AND CRAFT OF PROBLEM SOLVING"
but i do admit that factorization can be very helpful.

[Labib]

Acknowledge your advice, Tahmid.

[Masum]

Theorem (ERDOS):
The product of more than two consecutive natural numbers is not a perfect power.
Corollary:
$a!$ is never perfect square for $a>1$
So we must have $n-2=0,1\Rightarrow n=2,3$

We can prove this corollary for $n!$ without this theorem very easily for $n>1$. Let $p(n)$ denotes the greatest prime which is less than $n$. Then, it is obvious that $p(n)$ comes only once in the prime factorization of $n!$, hence $n!$ can't be a square.

[Tahmid Hasan]

i remember an IMO problem(longlist maybe) asking this,i solved it with the same idea too.