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.

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    $\begingroup$ Presumably you mean $11 n^2$, not $11 n$. $\endgroup$ – Robert Israel Mar 29 at 17:29
  • $\begingroup$ I should point out I assume you mean smallest positive value of $N$. $N= 0$ is a solution. Oh, wait. As $N\to \infty$ as $n\to -\infty$ maybe you DO mean least negative value of $N$? $\endgroup$ – fleablood Mar 29 at 19:37

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 $n$.

  • $\begingroup$ Yes, I have explored this route as well. How can I find the smallest $n$ and thus the smallest $N$ that is divisible by 247? $\endgroup$ – user603256 Mar 29 at 17:31
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    $\begingroup$ I see $n=36$ gives something and check the less values. $\endgroup$ – Michael Rozenberg Mar 29 at 17:32
  • $\begingroup$ How would you prove $n=36$ is the smallest product? $\endgroup$ – user603256 Mar 29 at 17:34
  • $\begingroup$ There is not much to prove. Obviously $n=13$ is too small, since then 19 doesn't divide $N$. The next candidate can only be $n=16$, so 19 divides $N$, but then 13 doesn't divide $N$. The same can be checked for $n=17,18,19$ since then still each factor is $<2\cdot 13$ and 13 is prime. So now you need to check the smallest composite numbers. The smallest one can only be $n+3=38$, since otherwise 19 doesn't divide $N$. But divisible by 13 is only 13,26,39. The first ones are too low, and 39 arises first at $n=36$. $\endgroup$ – Diger Mar 29 at 21:12

$N=n^4 + 6n^3 + 11n^2 + 6n = n(n+1)(n+2)(n+3)$ 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.


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