How to determine the smallest value of $N=n^4+6n^3+11n^2+6n$ if 13 and 19 both divide N?

Question

I tried to solve for an integer solution by making N equal to multiples of 247 but this is not leading me anywhere. I then tried using the tests for divisibility which did not seem to lead me anywhere either.

Answer 1

It means that $n$ is a natural number and $N=n^4+6n^3+11n^2+6n.$

I think the fact $$N=n(n+1)(n+2)(n+3)$$ can help.

I see that $n=36$ is valid and check smaller values of $n$.

Answer 2

$N=n^4 + 6n^3 + 11n^2 + 6n = n(n+1)(n+2)(n+3)$ so $n, n+1, n+2$ and $n+3$ all divide $N$.

$19$ and $13$ are prime so we need $19|K$ and $13|M$ where $K$ and $M$ are each one of $n,n+1,n+2,$ or $n+3$

So basically we need $|13a + 19b| \le 3$ and the smallest value that is so.

Do a hobbled Euclid's Algorithm.

$19 - 13 = 6$ that's too big.

But $13 = 2*6 + 1$ so $13 = 2(19 - 13) + 1$ so $3*13-2*19= 1$. So for $38,39 \in \{n,n+1,n+2, n+3\}$ will be a solution. Obviously the smallest product will but $n+3 = 39$ and $n+2 =38$ and $n = 36$.

Is there any smaller?

Well, the next smaller multiple of $13$ is $26$ and $26 - 1*19 = 7> 3$ and $2*19-26=12 > 3$ so none of $26= n,n+1,n+2,n+3$ will involve $26$. ($19$ divides none of $23... 29$.)

And the next smaller multiple of $13$ is $13$ and $19-13 =6 > 3$ so none of $13=n, n+1,n+2,n+3$ will have solution.

So $n = 36$ and $N= 36*37*38*39$ is the smallest (positive) such value.

If $N \le 0$ is allowed then $N=0$ is a smaller solution. As is $-38,-39\in \{n,n+1,n+2,n+3\}$. For $n<-3$, $N > 0$ anyway, the value of $N$ when $n\le -39$ is the same as $N$ when $n \ge 36$.

SO $N= 0$ is the smallest integer value.