Why is it that $(n+(n+1)^2+(n+2)^3) \bmod 144 =0$ starting with $n=3$ and incrementing $n=n+6$?

Question

The series involves: $(n+(n+1)^2+(n+2)^3) \bmod 144 =0$ starting with $n=3$ and incrementing $n=n+6$

Examples:

$3+4^2+5^3 = 3+16+125 =144$ and $144 / 144 =1$

$9+10^2+11^3 = 9+100+1331 =1440$ and $1440 / 144 =10$

$15+16^2+17^3 = 15+256+4913 =5184$ and $5184 / 144 =36$

$21+22^2+23^3 = 21+484+12167 =12672$ and $12672 / 144 =88$

$27+28^2+29^3 = 27+784+24389 =25200$ and $25200 / 144 =175$

...

I have tried using OEIS for the sequence $144,1440,5184,12672,25200...$ and couldn't find anything and then I have tried using the sequence $1, 10,36,88,175...$ and one result came about. OEIS link

It mentions something about Pentagonal prism and so I checked wikipedia's page in regards to Pentagonal prism Wikipedia link, and then i tried searching the subject here on stackexchange. I have found many questions in regards to the subject, but either they are too complicated for my current level of understanding or I simply can't find the answer to the question as I am presenting:

Why is it that $(n+(n+1)^2+(n+2)^3) \bmod 144 =0$ starting with $n=3$ and incrementing $n=n+6$?

Answer 1

You are talking about values of $n$ of the form $3 + 6k$. If we plug that in we get $$(3+6k) + (6k+4)^2 + (6k+5)^3 = 216k^3 + 576k^2 + 504k + 144$$ $$ = 72\cdot3 k^3 + 4\cdot144k^2 + 72\cdot 7k + 144$$ The $k^2$ term and the constant term are both automatically divisible by $144$. Factoring out $72$ from the remaining terms, we get $3k^3 + 7k$, which is either the sum of two even numbers or the sum of two odd numbers, which either way is even, giving us the remaining factor of two needed to make the sum evenly divisible by $144.$ QED.

Answer 2

$$(6n+3)+(6n+4)^2+(6n+5)^3=6n+3+36n^2+48n+16+216n^3+540n^2+450n+125\\=216n^3+576n^2+504n+144=144(\frac32n^3+4n^2+\frac72n+1)=144P(n)$$

It remains to prove $P(n)\in \Bbb N$ for each $n\in \Bbb N$. $P(2k)$ is trivial, and $$P(2k+1)=\frac32(8k^3+12k^2+6k+1)+4(4k^2+4k+1)+\frac72(2k+1)+1\\=12k^3+34k^2+23k+10\in\Bbb N$$

Answer 3

hint

It is the same to prove that for any integer $ k\ge 0$, we have

$$(3+6k)+(3+6k+1)^2+(3+6k+2)^3\equiv 0 \pmod {144}$$ or equivalently $$(3+6k)+(16+48k+36k^2)+(125+450k+540k^2+216k^3)\equiv 0\pmod {144}$$

and $$k(504+576k+216k^2)\equiv 0 \pmod {144}$$

or $$72k(k+1)\equiv 0 \pmod{144}$$

Which is true since $ k(k+1)$ is always even.

Answer 4

If you expand, you find $$ n^3+7n^2+15n+9 $$ and you want to show when this is divisible by $144$.

In particular it ought to be divisible by $3$, so $n+n^2$ should be divisible by $3$, which only happens if $n$ is already divisible by $3$. In this case the expression is also divisible by $9$. (I used that $n^3\equiv n\pmod{3}$.)

Let $n=3m$, so we get that $27m^3+63m^2+45m+9$ should be divisible by $144$, which is the same as requiring that $3m^3+7m^2+5m+1$ is divisible by $16$. In particular $m=2p+1$ should be odd and this becomes $$ 24p^3+64p^2+56p+16\equiv0\pmod{16} $$ that's the same as $$ 8p^3-8p\equiv0\pmod{16} $$ which is true because $p^3-p$ is even for every $p$. Thus we have proved that $$ n+(n+1)^2+(n+2)^3\equiv0\pmod{144} \text{ if and only if } n=6p+3\text{ for some integer }p $$ which is even more general than your statement.