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

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.

• Presumably you mean $11 n^2$, not $11 n$. – Robert Israel Mar 29 at 17:29
• 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 $N$? – fleablood Mar 29 at 19:37

## 2 Answers

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 $$n$$.

• 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? – user603256 Mar 29 at 17:31
• I see $n=36$ gives something and check the less values. – Michael Rozenberg Mar 29 at 17:32
• How would you prove $n=36$ is the smallest product? – user603256 Mar 29 at 17:34
• 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 $n=36$. – 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.