Question

Is $13(13)^n-11n-13$ divisible by 144 for infinitely many $n \in N$?

Answer 1

Yes

Answer 2

We need to show that

$$E(n)=13\cdot 13^n-11n-13=13^{n+1}-11n-13$$

is divisible by 144 for infinitely many $n\in\mathbb{N}$. Since

$$144=16\cdot9,$$

we check divisibility modulo 16 and modulo 9 separately.

1. Modulo 16

Since $13\equiv 13\pmod{16}$, compute the order of 13 modulo 16:

$$13^2\equiv169\equiv9,\quad 13^3\equiv13\cdot9=117\equiv5,\quad 13^4\equiv5\cdot13=65\equiv1\pmod{16}.$$

Thus, the cycle length is 4, and

$$13^{n+1}\pmod{16}$$

depends on $n+1\pmod{4}$.

Let $n=4m+r$ where $r=0,1,2,3$:

• Case $r=0$ (i.e. $n=4m$): Then $n+1\equiv1\pmod{4}$ so $13^{n+1}\equiv 13$. Thus,

  $$E(n)\equiv 13-11(4m)-13 = -44m\equiv -12m\pmod{16}.$$

  For $-12m\equiv0\pmod{16}$, note that $\gcd(12,16)=4$; hence, $m$ must be a multiple of $4$. That is, $m=4k$ and so $n=16k$.

• For other residues $r=1,2,3$, one does not get a congruence of 0 modulo 16.

Thus, the necessary condition modulo 16 is:

$$n\equiv0\pmod{16}.$$

2. Modulo 9

Rewrite modulo 9: Since $13\equiv4\pmod{9}$ and $11\equiv2\pmod{9}$, we have

$$E(n)\equiv 4^{n+1}-2n-4\pmod{9}.$$

The powers of 4 modulo 9 are:

$$4^1\equiv4,\quad 4^2\equiv16\equiv7,\quad 4^3\equiv4\cdot7=28\equiv1\pmod{9}.$$

Thus, the order is 3 and we consider cases modulo 3 by letting $n=3k + r$ for $r=0,1,2$:

• Case $r=0$ ($n=3k$): Then

  $$4^{n+1}=4^{3k+1}=4\cdot (4^3)^k\equiv4\pmod{9}.$$

  So,

  $$E(n)\equiv 4-2(3k)-4 = -6k\pmod{9}.$$

  For $-6k\equiv0\pmod{9}$, since $\gcd(6,9)=3$, it is required that $k$ be a multiple of 3. That is, $k=3m$ and hence $n=9m$.

• The other cases ($r=1,2$) do not yield a solution.

Thus, the necessary condition modulo 9 is:

$$n\equiv0\pmod{9}.$$

Combining Conditions

We require:

$$n\equiv0\pmod{16}\quad\text{and}\quad n\equiv0\pmod{9}.$$

Thus, $n$ must be a multiple of the least common multiple of 16 and 9:

$$\text{lcm}(16,9)=144.$$

So, for every $n=144k$ (with $k\in \mathbb{N}$), $E(n)$ is divisible by 144.

Conclusion

There are infinitely many natural numbers $n$ for which $13\cdot 13^n-11n-13$ is divisible by 144.