The Biggest Divisor

We can see the expression above can be factored as $n(n-1)(n+1)(n-4)(n+4)$ . Now, if $n$ is even then $n, n-4, n+4$ are even, so $8$ divides the expression. If $n$ is odd, we see $n-1, n+1$ are even, but one of them is divisible by $4$ , so $8$ divides the expression. Then we see $8$ always divides the expression. Now, $n-1, n, n+1$ are distinct modulo $3$ , so 3 divides the expression. Hence, 24 always divides the expression.

Plugging $n=5$ , we obtain $2^{3}\cdot 3^{3}\cdot 5$ . Plugging $n=12$ , we get $2^{9}\cdot 3\cdot 11\cdot 13$ . With this, we see the greatest common divisor of the expression divides the greatest common divisor of $2^{3}\cdot 3^{3}\cdot 5$ and $2^{9}\cdot 3\cdot 11\cdot 13$ , which is $2^{3}\cdot 3=24$ ,

Therefore, $24$ is the number we are looking for.

[LaTex edits - Calvin]

$n^5-17n^3+16n = n (n^4-17n^2+16)$ [common factoring out n] $=n(n^2-16)(n^2-1)$ [by difference of squares] $=n(n+4)(n-4)(n+1)(n-1)$ [by difference of squares again] Rearranging, we have $(n-4)(n-1)(n)(n+1)(n+4)$ .

Since S is a divisor of the above expression, the factors of S must all be included in the factors of $n-4, n-1, n, n+1$ and $n+4$ . We will consider the products of these five factors.

Note the case $n=5$ gives the five factors 1, 4, 5, 6, and 9. The prime factorization of the product of these is $2^3 * 3^3 * 5$ . From the question, S must be a factor of this, so we will consider each of these prime factors separately.

Firstly, we consider if $2^3$ is a factor of S. If it is, then 2 must appear at least 3 times in the prime factorizations of our 5 divisors.
Case 1: if $n$ is even, then $n-4$ and $n+4$ are also even, and we have our 3 factors of 2.
Case 2: if $n$ is odd, then both $n-1$ and $n+1$ are even. However, we require 3 factors of 2. This requirement would be fulfilled if either of the factors were also divisible by 4. We will demonstrate this to be true. Note that since $n-1$ is even, $(n-1)\pmod{4} \equiv 0$ or $2$ . If $0 \equiv (n-1)\pmod{4}$ then we have our factor of 4. If not, then from $(n-1)\pmod{4} \equiv 2$ it follows that $(n-1 +2)\pmod{4} \equiv 2 +2$ and $(n+1)\pmod{4} \equiv 0$ , and we have our factor of 4. Therefore, $2^3$ is a factor of S.

Next, consider if $3^3$ is a factor of S. Setting $n=6$ gives a clear counter-example to both, since its five factors are 2, 5, 6, 7, and 10, of which the only multiple of 3 is 6, which is a multiple of neither $3^2$ nor $3^3$ . Therefore, S is not a multiple of $3^2$ or $3^3$ . However, S may still be a multiple of 3. Consider three of our five factors: $n-1, n,$ and $n+1$ . These three are consecutive integers. It may be easily seen or demonstrated that for every n consecutive integers, exactly one of them is divisible by n (through the pigeonhole principle, proof by induction or otherwise). Therefore, one of these three factors is divisible by 3, and thus 3 is a multiple of S.

Finally, we consider if S is a multiple of 5. A counter-example is found when $n=7$ , with the five factors of 3, 6, 7, 8, and 11. None of these are divisible by 5, and thus 5 is not a factor of S.

We have shown that S has a prime factorization of $2^3 * 3^2$ and that S may have no more factors than these (since S must be a factor of $2^3 * 3^3 * 5$ ). Therefore, S = 24.

The polynomial $n^5 - 17n^3 + 16n$ factors as $n(n^2 - 1)(n^2 - 16) = (n-4)(n-1)(n)(n+1)(n+4).$ First note that 3 must divide one of $n-1$ , $n$ , and $n+1$ . If $n$ is even, then so are $n-4$ and $n+4$ , so 8 divides the product. If $n$ is odd, then $n-1$ and $n+1$ are even, and 4 divides one of $n-1$ and $n+1$ and 2 divides the other; hence, 8 divides the product. Therefore, $3 \cdot 8 = 24$ must divide the product.

Plugging $n = 5$ and $n = 6$ into the polynomial $n^5 - 17n^3 + 16n$ yields the values $1 \times 4 \times 5 \times 6 \times 9 = 1080$ and $2 \times 5 \times 6 \times 7 \times 10 =4200$ , respectively. By considering the divisors of these 2 numbers ( Worked Examples 1 in blog post), we see that their GCD is $8 \times 3 \times 5$ . Considering $n=7$ , we have $3 \times 6 \times 7 \times 8 \times 11$ , and 5 does not divide this number. Hence, the GCD of the sequence is 24.

To answer this problem, we need to find out the guaranteed multiples of the expression.

The first step is to factor: n^5 - 17n^3 + 16n = n(n+4)(n-4)(n+1)(n-1)

Next, we note that the factors (n-1), n, and (n+1) are consecutive. This guarantees that one of the factors is a multiple of three. Thus, the expression is at least divisible by 3.

We then consider two cases: when n is even and when n is odd.

When n is even, (n-4) and (n+4) are also even. This ensures that there are three factors that are each a multiple of two. This ensures that the expression is divisible by 8 when n is even.

When n is odd, (n-1) and (n+1) are even factors. Since they are consecutive, one of them is a multiple of four while the other would still be a multiple of 2. This guarantees that the expression is also divisible by 8 when n is odd.

For both odd and even cases, the expression is divisible by 8. Since the expression is also at least divisible by 3, we multiply 3 by 8 -- giving us 24. Thus, the largest integer which is a divisor of n^5 - 17n^3 + 16n for all n >= 4 is 24.

n(n^2 -1)(n^2 - 4 - 12) = n(n-1)(n+1)(n-2)(n+2) - 12n(n-1)(n+1).

We know that, product of consecutive r numbers is divisible by r ! . So 1st part of the expression is divisible by 5 ! = 120 and 2nd part is divisible by 12.( 3 !) = 72. Again (120, 72) = 24. So largest divisor is 24.

The main concept behind the problem is the use of the Euclidean Algorithm. Originally I thought to un-FOIL the equation, obtaining:

$n(n^2 - 16)(n^2 - 1)$

However the easiest way is to plug in the integers $n \geq 4$ and find the GCD using the Euclidean Algorithm. If you plug in 5 and 6 and find the GCD, you'll obtain 120. The GCD of 5 and 7 is 72, and the GCD of 6 and 7 is 168. You now have 3 different numbers to reapply the Euclidean Algorithm. The GCD of 72 and 120 is 24, of 72 and 168 is 24, therefore the GCD of 120 and 168 must also be 24. You can keep going on plugging in 8, 9, 10, and comparing GCD's, however there are repeated numbers, i.e. 7 and 8 have a GCD of 168 as well, 6 and 8, so on and so forth.

Put n=5, the answer is 216.So S divides 216. Checking mod 8 the answer is always 0. So S is 2^3 3^a. When n is a multiple of 3 not multiple of 9 we get a number exactly divisible by 3. so a is 1 and answer is 8 3. (It is easy to see the the function is always a multiple of 3)