Largest divisor of $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$ 
Let $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$. What is the largest integer that is a divisor of $P(n)$ for all positive even integers $n$?

I can see that obviously it is divisible by $3$ and $5$ and hence answer is $15$ but I can't prove it.
 A: As  $P(0)=1\cdot3\cdot5\cdot7\cdot9,$  and $9\nmid P(10),7\nmid P(8)$
The highest divisors of $P(n)$ must divide $3\cdot5$
$P(n)\equiv(n+1)n^2(n+2)^2\pmod3$
Clearly, $3\mid n(n+1)(n+2)$
$P(n)\equiv(n-2)(n-1)n(n+1)(n+2)\pmod5$
Clearly, $5\mid(n-2)(n-1)n(n+1)(n+2)$
A: We have that $\gcd(P(1),P(2))=15$, hence a number $N$ with the property that $N\mid P(n)$ for every $n$ has to be a divisor of $15$. We may notice that $3\mid P(n)$ for every $n$ since exactly one number among $n+3,n+1,n+5$ is a multiple of three. In a similar way, exactly one number among $n+5,n+1,n+7,n+3,n+9$ is a multiple of $5$, hence $\color{red}{15}$ is the answer.
A: In order to prove that $5$ is a divisor of $P(n)$ for even $n$, consider the following cases:


*

*$n\equiv\color\red0\pmod{10} \implies n+5\equiv\color\red0+5\equiv5\pmod{10} \implies 5|P(n)$

*$n\equiv\color\red2\pmod{10} \implies n+3\equiv\color\red2+3\equiv5\pmod{10} \implies 5|P(n)$

*$n\equiv\color\red4\pmod{10} \implies n+1\equiv\color\red4+1\equiv5\pmod{10} \implies 5|P(n)$

*$n\equiv\color\red6\pmod{10} \implies n+9\equiv\color\red6+9\equiv15\equiv5\pmod{10} \implies 5|P(n)$

*$n\equiv\color\red8\pmod{10} \implies n+7\equiv\color\red8+7\equiv15\equiv5\pmod{10} \implies 5|P(n)$


That said, I'm not sure how to prove that $5$ is indeed the largest divisor of $P(n)$...
A: Note that if a prime $p$ divides all the values of $P(n)$, then $-1,-3,-5,-7,-9$ cover all residues modulo $p$. Hence, $p\leq 5$. So for primes you need to check only $2,3,5$ (and you did) and $3,5$ are these primes.
Now we need to go to prime powers; by inspection we see that $9$ does not divide $P(2)$ and that $25$ does not divide $P(1)$. So the maximal prime powers that divide all values are $3,5$.
By the Chinese Remainder Theorem one may conclude that the maximal number is $15$, as you suggested.
This means that the largest number that divides all the values is of the form $3^k 5^l$ with $k,l\geq 1$. By inspection, one may note that $3^2$ and $5^2$ do not satisfy
A: $$P(10)=(11)(13)(15)(17)(19)=3.5.11.13.17.19$$
$$P(20)=(21)(23)(25)(27)(29)=3^4.5^2.7.23.29$$
we can see that
 $$gcd(P(10),P(20))=15$$
$$P(n)=(n+1)(n+3)(n+5)(n+7)(n+9) \equiv (n+1)n(n+2)(n+1)n \equiv 0 \mod 3$$
since $n(n+1)(n+2) \equiv 0 \mod 3.$, that is product of 3 consecutive numbers must be divisible by $3$.
Also, 
$$P(n)=(n+1)(n+3)(n+5)(n+7)(n+9) \equiv (n+1)(n+3)n(n+2)(n+4)\equiv 0 \mod 5$$
since product of 5 consecutive numbers is divisible by $5$.
Hence $15$ is the largest number that divides $P(n)$ for all even $n$.
