# Thread: Revision related number theory question(s)

1. ## Revision related number theory question(s)

I don't have a clue to even start the following question so any pointers would be welcome!

Let n be an integer with n greater than or equal to 2. Show that n! + 2 is even.
Now show that m divides n! + m, for all m such that 2<m<n (here it is a less than or equal to sign) and show why n! +m is composite.
Use this to find a sequence of n-1 consecutive integers (that is a sequence a2, a3,...,an, with ai+1=ai +1 and all ai composite).

2. Originally Posted by schteve
I don't have a clue to even start the following question so any pointers would be welcome!

Let n be an integer with n greater than or equal to 2. Show that n! + 2 is even.
Hint: $n! = n(n - 1)(n - 2) \cdots 3 \cdot {\color{red}2} \cdot 1$

Now show that m divides n! + m, for all m such that 2<m<n (here it is a less than or equal to sign) and
Again, $m$ will be one of the factors of $n!$ (i hope by now you see why. if not, think about it some more)

show why n! +m is composite.
by the above, it would be divisible by $m > 2$. hence, it cannot be prime, it must be composite.

Use this to find a sequence of n-1 consecutive integers (that is a sequence a2, a3,...,an, with ai+1=ai +1 and all ai composite).
take the sequence $\{ a_i \}_{i = 2}^{n} = n! + i$

do you see why this works?

3. $n!$ with $n$ greater than $1$ is even, because it has $2$ as a factor. I think you should be ok for the first question.

Since $2 < m < n$, $m$ divides $n!$ (because m is a factor of n! as $m < n$).
Thus you can say that $n! + m = m \times \frac{n!}{m} + m = m(\frac{n!}{m} + 1)$ with $\frac{n!}{m} \in \mathbb{N}$. Thus, $n! + m$ is composite because it has $m$ as a factor, and we know that $m > 2$.

4. Thanks for the help. Yeah its all clicked into place after just understanding the 1st stage.