Clarification on Proof by Contradiction

[Delta2]

@Jonathanlikesmath now you are very rigorous but you did it the very hard way, I had in mind something like what @PeroK suggests at post #29.

More specifically, if we write the product as $n(n+1)(n+2)(n+3)(n+4)$ then it will be either $n\pmod 5=0$ in which case we are finished, or one of the cases of $n\pmod 5=1,n\pmod 5=2,n\pmod 5=3,n\pmod 5=4$. If for example $n\pmod 5=1$ then $(n+4)\pmod 5=n\pmod 5+4\pmod 5=1+4\pmod 5=5\pmod 5 =0$, hence in this case 5 divides n+4. Similarly we can handle the other cases.

 

[Jonathanlikesmath]

@Jonathanlikesmath now you are very rigorous but you did it the very hard way, I had in mind something like what @PeroK suggests at post #29.

More specifically, if we write the product as $n(n+1)(n+2)(n+3)(n+4)$ then it will be either $n\pmod 5=0$ in which case we are finished, or one of the cases of $n\pmod 5=1,n\pmod 5=2,n\pmod 5=3,n\pmod 5=4$. If for example $n\pmod 5=1$ then $(n+4)\pmod 5=n\pmod 5+4\pmod 5=1+4\pmod 5=5\pmod 5 =0$, hence in this case 5 divides n+4. Similarly we can handle the other cases.

Well, I soon won't forget this method. Thank you all for the help!

 

[PeroK]

A better approach if you want to use modular arithmetic is to take:
$$P = (n - 2)(n-1)n(n+1)(n+2) = n(n^2 -1)(n^2 - 4)$$If $n$ is not divisible by $3$, then $n = 1$ or $2 \ (mod \ 3)$; $n^2 = 1 \ (mod \ 3)$, so $n^2 -1$ is divisible by $3$. Etc.

Just to say that I thought reducing the product to three terms looked like a good idea (and it is always something to consider), but in this case it doesn't really help. @Delta2 's method is much simpler!