# $n + (n+1)^2 + (n+2)^3$ is a multiple of $16$ where $n$ is odd number

$n + (n+1)^2 + (n+2)^3$

is a multiple of $16$

where $n$ is odd number

And how do i prove it. And i tried to do with expanding but could not figured it out. I have attached an image

• Do you know the congruence modulo? – Jlamprong Feb 3 '14 at 10:58
• $n$ is odd, so now write $n=2r+1$ for some $r$. – Gerry Myerson Feb 3 '14 at 11:00
• no, I dont know congruence modulo – Rafee Feb 3 '14 at 11:00
• You did a great job expanding the number. Now, simply write $n$ as $2r+1$ since $n$ is odd. – Traklon Feb 3 '14 at 11:18
• even my expanding the formula is wrong, well i got the solution thanks again. – Rafee Feb 3 '14 at 11:34

Let $n=2k+1$, since it is an odd integer.

Now expand to get: $$2k+1+(2k+2)^2+(2k+3)^3=8(k+1)(k+2)^2$$ but $(k+1)(k+2)$ is divisible by 2, since one of $k+1,k+2$ is even. Therefore the whole expression is divisible by 16.

• how did we got 8(k+1)(k+2)^2 and i got the final answers as 8k^3 + 40k^2 + 64k + 32 – Rafee Feb 3 '14 at 11:53

HINT:

Setting $n=2m+1,$

We get $$8m^3+40m^2+64m+32=8m^2(m-1)+48m^2+64m+32$$

Now, $\displaystyle m^2(m-1)$ is divisible by $\displaystyle m(m-1)$ which being product of two consecutive integers is even