What is the significance of using prime numbers in proving: $x$ is a multiply of $y$? I came to a problem where it asks me to prove, for example, $n^4-n^2$ is a multiple of $12$. 
Now, factorize the multiple: $n\times n\times (n-1)\times (n+1)$. Here we have $3$ consecutive integers. Any three consecutive integers say something that one of them must be a multiple of $3$. Likewise for $2$ in any $2$ consecutive integers . Here I cannot say any thing about $6$ or $4$ because we do not have $4$ consecutive integers nor $6$ although $12$ can be written as $6\times 2$ or $3\times 4$. 
But, assume that we have $4$ consecutive  integers in some how, I "heard" that you cannot prove that the product of these $4$ is divisible by $4$ but you can use $2$ since it is a prime. 
Why $2$ and not $4$? Why to use a prime and not a non prime? 
 A: If $n$ is even then it's divisible by $4$ because it's divisible by $n^2$.  Now if $n$ is odd then you have two even terms $n-1$ and $n+1$ so it's still divisible by $4$
A: You can still rewrite it to make use of what you want: the product of $4$ consecutive integers is divisible by $4! = 24$, and the product of $3$ consecutive integers is divisible by $3! = 6$. We have: $n^4 - n^2 = n\times n\times (n-1)\times (n+1) = (n-1)\times n\times (n+1)\times ((n+2) - 2)= (n-1)\times n\times (n+1)\times (n+2) - 2\times (n-1)\times n\times (n+1)$. This expression is divisible by $12$ since its the difference of two integers that is each divisible by $12$.
A: Why will you take $4$?
Look at $n\cdot n\cdot(n-1)(n+1)$.
Assume $n$ to be $2$.Then the factors become $2,2,1,3$.You don't have $4$ anywhere.But,you can break $4$ into $2\cdot 2$.So you see the two $2$'s will always cancel.So you may now realize that we are doing all works using the smallest constituent prime factors. 
In general all composite numbers can be expressed as prime factors (by the Fundamental Theorem of Arithmetic).So,its same to show that a number $p$ has a factor which is a composite number $z$ by showing that all prime factors $z$ are factors of $p$
