Product of 5 consecutive integers cannot be a perfect square Problem: Show that the product of any five consecutive positive integers cannot be a perfect square.
Proof: Let $N=n \times (n+1) \times (n+2) \times (n+3) \times (n+4)$, where $n \in \mathbb{N}$. We know the that N must have at least: 


*

*one factor is divisible by 5, 

*two or three factors are even, 

*one factor of the even factors, is divisible by $4$, 

*and one factor is divisible by $3$.


To proof the statement above we need to assume that the product of any five consecutive positive integers is a perfect square to lead us to a contradiction. Hence, I know there will be cases for us to test for. However, I am unsure where to start after mentioning everything above. Can anyone help me from here?
 A: Consider the product $n(n+1)(n+2)(n+3)(n+4)$. A perfect square $k$ will have a factorization $k=a_1^{2m_1}a_2^{2m_2}...a_n^{2m_n}$ , i.e., every factor will appear an even number of times. But this will not happen for $n(n+1)(n+2)(n+3)(n+4)$; there will always be a factor that appears an odd number of times. There are two main cases here:
1) The middle term $(n+2)$ is not a perfect square ,  and,
2)The middle term is a perfect square .
Case 1): If $(n+2)$ is not a perfect square, the product, then, cannot be a perfect square, because the factors of the middle term $(n+2)$ will appear an odd number of times, unless $(n+2)$ is itself a perfect square, since $$ 2(n+2)>(n+4); g(\frac{(n+2)}{2}) < n$$, where $g(\frac{n+2}{2})$ is the greatest integer less than $(n+2/2)$. 
Basically, any factor $a_i$ of $(n+2)$ will be between  $ka_i$ and $(k+1)a_i$ for some positive integer $k$. This is because of the fact that $$ka_i < n(n+1)(n+2)(n+3)(n+4)< (k+1)a_i$$  when $n+2$ is not a perfect square. 
If $(n+2)$ is itself a perfect square, then neither $(n+1)$ , nor $(n+3)$ are perfect squares, and you can repeat a similar argument.
A: We need to think about prime factorisations.
Assume the product is a perfect square.
It is intuitively clear that for any prime p, greater than (or equal to) 5, p is a factor of at most one of the integers. We also know that for p greater than (or equal to) 5, p must have even power, because we're square rooting and getting a rational.
This leaves us to consider primes 2 and 3. These are the only primes which can appear in the prime decomposition of our integers with an odd power. Now we need to split into cases.
Integers can be of 4 forms:
power(2)=even, power(3)=even implies m^(2)
power(2)=odd, power(3)=even implies 2m^(2)
power(2)=even, power(3)=odd implies 3m^(2)
power(2)=odd, power(3)=odd implies 6m^(2)
We have 5 integers total, so two must be of the same form.
Supposing we have 2 of the form m^(2) means we must have 1 and 4 in our list. Checking this, we get (1)(2)(3)(4)(5)=120 which isn't a square.
Supposing we have 2 of the form 2m^(2) means we have two squares with a difference of 1 or 2, which clearly doesn't happen.
Supposing we have 2 of the form 3m^(2) gives another contradiction, as for this to happen, two squares need to have a difference of 1, which is false.
Supposing we have 2 of the form 6m^(2), we get two multiples of 6 but with a difference of only 4.
In all cases, we find contradictions.
