1. ## Divisible by 16

Prove that if m and n are odd integers, then $m^3n^2+m^2n^3-m-n$ is divisible by 16.

2. Originally Posted by Sigyn3
Prove that if m and n are odd integers, then $m^3n^2+m^2n^3-m-n$ is divisible by 16.
this seems like a straight forward computation will work

let m = 2a + 1 and let n = 2b + 1

where a and b are integers

expand the expression and simplify. your goal is to show that the expression can be written as 16k where k is an integer

3. Hello, Sigyn3!

Prove that, if $m$ and $n$ are odd integers, then $m^3n^2+m^2n^3-m-n$ is divisible by 16.
I used your approach, Jhevon . . . and hit a wall.

We have: . $P \;=\;m^2n^2(m + n) - (m + n) \;=\;(m+n)(m^2n^2-1) \;=\;(m+n)(mn-1)(mn+1)$

Let . $\begin{array}{ccc}m &=&2a+1 \\ n &=&2b+1 \end{array}$

Then:. . $\begin{array}{ccccccc}m + n &=& (2a+1) + (2b+1) &=& \quad 2a+2b + 2 &\;\;=& \quad2(a+b+1)\\
mn-1 &=& (2a+1)(2b+1) - 1 &=& \quad 4ab + 2a +2b &\;\;=& \;\;2(2ab + a + b) \end{array}$

. . . . . . $\begin{array}{ccccccc}mn+1 &=& (2a+1)(2b+1)+1 &=& 4ab + 2a + 2b + 2 &=& 2(2ab + a + b + 1)\end{array}$

And we have: . $P \;=\;2(a + b + 1)\cdot2(2ab + a + b)\cdot2(2ab + a + b + 1)$

. . . . . . . . . . $P \;=\;8(a+b+1)(2ab + a + b)(2ab + a + b + 1)$

So we have a multiple of 8.
. . And here's where I hit the wall . . .

I tried various ways to find another factor of 2,
. . and settled on an exhaustive listing.

. . $\begin{array}{c|c|c|c}
(a,b) & (a+b+1) & (2ab + a + b) & (2ab + a + b + 1) \\ \hline
\text{even, even} & \text{odd} & \text{even} & \text{odd} \\
\text{even, odd} & \text{even} & \text{odd} & \text{even} \\
\text{odd, odd} & \text{odd} & \text{even} & \text{odd} \end{array}$

So for any combination of $a$ and $b$, at least one of the factors is even.

Therefore, $P$ is a multiple of 16.

Surely there must a more elegant method . . .
.

4. Soroban,
Whenever you said:
$
P \;=\;m^2n^2(m + n) - (m + n) \;=\;(m+n)(m^2n^2-1) \;=\;(m+n)(mn-1)(mn+1)$

Because:
m+n is even.
mn-1 is even
mn+1 is even (AND next even number after mn-1)
which means either mn-1 or mn+1 is divisible by 4.
$
(m+n)(mn-1)(mn+1) = 2k_1 2k_2 4k_3 = 16k_4$

5. Originally Posted by Sigyn3
Prove that if m and n are odd integers, then $m^3n^2+m^2n^3-m-n$ is divisible by 16.
First note that the square of any odd integer is congruent to 1 modulo 8. This is because $(2k+1)^2=4k(k+1)+1$ and one of $k$ and $k+1$ is even.

Since $m$ snd $n$ are odd, so is $mn$. $\therefore\ m^2n^2\equiv1\pmod8$

$\Rightarrow\ m^2n^2-1$ is divisible by 8.

And since $m+n$ is even, $m^3n^2+n^3m^2-m-n=(m+n)(m^2n^2-1)$ is divisible by 16.