0

I'm looking to prove that $$ n^2(n - 4)(n - 3)(n - 2)(n - 1)(n + 1)^2(3n^2 - n - 6) $$ is divisible by $11520$ for all integers $n > 4$. I honestly have no clue where to start, I've never seen a problem like this before and I probably chose a really daunting one to start learning. Any advice?

FleebJuice
  • 11
  • 1
    How to start: factor $11520=2^83^25$. So all you need is to show that this beast is divisible by $5$, by $9$, and by $2^8=256$. Also, if you've never proved that a polynomial is always divisible by certain integers, you should definitely start with a simpler one. For example, prove that $n(n-1)$ is always even. – vadim123 Nov 22 '15 at 16:27
  • Related : http://math.stackexchange.com/questions/40429/what-is-the-greatest-integer-that-divides-p4-1-for-every-prime-number-p-gre – lab bhattacharjee Nov 22 '15 at 16:31

2 Answers2

3

$$n^2(n - 4)(n - 3)(n - 2)(n - 1)(n + 1)^2(3n^2 - n - 6)\\= (n - 4)(n - 3)(n - 2)(n - 1)n(n + 1) \bigl[n (n+1)(3n^2 - n - 6) \bigr] $$

Now, the product of $6$ consecutive integers is always divisible by $5$.

Also note that $$ 3n^2 - n - 6=4n(n+1)-(n^2+5n+6)=4n(n+1)-(n+2)(n+3) $$

Thus $$ n^2(n - 4)(n - 3)(n - 2)(n - 1)(n + 1)^2(3n^2 - n - 6)\\= 4(n - 4)(n - 3)(n - 2)(n - 1)n(n + 1) n (n+1) n(n+1) - \\ (n - 4)(n - 3)(n - 2)(n - 1)n(n + 1) n (n+1) n(n+1)(n+2)(n+3) $$

Now, use the fact that among any $6$ consecutive integers you can find a multiple of $4$ and two even numbers. Also among any $8$ consecutive integers you can find a multiple of $8$, another multiple of $4$ and two even numbers.

Finally among any 6 consecutive integers you have 2 multiples of 3....

N. S.
  • 132,525
0

The polynomial has six consecutive integers in its factorization. In any such sequence, one integer is divisible by $5$ and two are divisible by $3$. Thus, we've proved its divisibility of the powers of $3$ and $5$ from the prime factorization of $11520$.

The following is for the divisibility of $2^8$. $n$ or $(n+1)$ is divisible by 2, so $n^2$ or $(n+1)^2$ is divisible by $2^2$. Besides that, there are $(n-4)$, $(n-3)$, $(n-2)$, $(n-1)$, a sequence of four consecutive integers, in the factorization of the polynomial. In this sequence, one integer is divisible by $2^2$ and two are divisible by $2$. Since divisibility by $2^2$ implies divisibility by $2$, the integer divisible by $2^2$ is one of the two that are divisible by 2. So the product of these integers is divisible by $2^3$. The quadratic factor $(3n^2 - n - 6)$ can be rewritten as $((n-1)(n-2) - 2^3 + 2n(n+1))$, and since $(n-1)$ or $(n-2)$ is divisible by $2$, so is $(3n^2 - n - 6)$. Overall, we have so far proved the divisibility of $2^6$. If $n$ or $(n+1)$ is divisible by $2^2$, we have an additional $2^2$ for the divisibility of $2$, giving the desired divisibility by $2^8$. Otherwise, either $(n-1)$ or $(n-2)$ is divisible by $2^2$. In both cases, looking at the rewrited version of the quadratic expression, $(n-1)(n-2)$ is divisible by $2^2$, and since $n$ or $(n+1)$ is divisible by $2$, we have $2n(n+1)$ divisible by $2^2$, so the entire quadratic expression is divisible by $2^2$, meaning we have an additional $2^1$ in the factorization. This brings the overall divisibility of $2$ to $2^7$. If $(n-1)$ or $(n-2)$ is divisible by $2^3$, we have an additional $2^1$ and thus a total divisibility of $2^8$. Finally, otherwise, looking at the rewrited quadratic expression, both $(n-1)(n-2)$ and $2n(n+1)$ are odd multiples of $2^2$, so $((n-1)(n-2) + 2n(n+1))$ is an even multiple of $2^2$ and thus divisible by $2^3$, so the whole quadratic expression is divisible by $2^3$, giving us an additional $2^1$ for a total of $2^8$.

Since the above deduces the divisibility by the powers of primes of $11520$ in all possible integer configurations of $n$, it follows that the original polynomial expression is divisible by $11520$ for all integers $n$. The proof is finished.

$Q.E.D.$

Måns Nilsson
  • 323