Prove that if m and n are odd integers, then $\displaystyle m^3n^2+m^2n^3-m-n$ is divisible by 16.
Hello, Sigyn3!
I used your approach, Jhevon . . . and hit a wall.Prove that, if $\displaystyle m$ and $\displaystyle n$ are odd integers, then $\displaystyle m^3n^2+m^2n^3-m-n$ is divisible by 16.
We have: .$\displaystyle P \;=\;m^2n^2(m + n) - (m + n) \;=\;(m+n)(m^2n^2-1) \;=\;(m+n)(mn-1)(mn+1)$
Let .$\displaystyle \begin{array}{ccc}m &=&2a+1 \\ n &=&2b+1 \end{array}$
Then:. . $\displaystyle \begin{array}{ccccccc}m + n &=& (2a+1) + (2b+1) &=& \quad 2a+2b + 2 &\;\;=& \quad2(a+b+1)\\
mn-1 &=& (2a+1)(2b+1) - 1 &=& \quad 4ab + 2a +2b &\;\;=& \;\;2(2ab + a + b) \end{array}$
. . . . . . $\displaystyle \begin{array}{ccccccc}mn+1 &=& (2a+1)(2b+1)+1 &=& 4ab + 2a + 2b + 2 &=& 2(2ab + a + b + 1)\end{array}$
And we have: .$\displaystyle P \;=\;2(a + b + 1)\cdot2(2ab + a + b)\cdot2(2ab + a + b + 1)$
. . . . . . . . . . $\displaystyle P \;=\;8(a+b+1)(2ab + a + b)(2ab + a + b + 1)$
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.
. . $\displaystyle \begin{array}{c|c|c|c}
(a,b) & (a+b+1) & (2ab + a + b) & (2ab + a + b + 1) \\ \hline
\text{even, even} & \text{odd} & \text{even} & \text{odd} \\
\text{even, odd} & \text{even} & \text{odd} & \text{even} \\
\text{odd, odd} & \text{odd} & \text{even} & \text{odd} \end{array}$
So for any combination of $\displaystyle a$ and $\displaystyle b$, at least one of the factors is even.
Therefore, $\displaystyle P$ is a multiple of 16.
Surely there must a more elegant method . . .
.
Soroban,
Whenever you said:
$\displaystyle
P \;=\;m^2n^2(m + n) - (m + n) \;=\;(m+n)(m^2n^2-1) \;=\;(m+n)(mn-1)(mn+1)$
you've already solved it.
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.
$\displaystyle
(m+n)(mn-1)(mn+1) = 2k_1 2k_2 4k_3 = 16k_4$
First note that the square of any odd integer is congruent to 1 modulo 8. This is because $\displaystyle (2k+1)^2=4k(k+1)+1$ and one of $\displaystyle k$ and $\displaystyle k+1$ is even.
Since $\displaystyle m$ snd $\displaystyle n$ are odd, so is $\displaystyle mn$. $\displaystyle \therefore\ m^2n^2\equiv1\pmod8$
$\displaystyle \Rightarrow\ m^2n^2-1$ is divisible by 8.
And since $\displaystyle m+n$ is even, $\displaystyle m^3n^2+n^3m^2-m-n=(m+n)(m^2n^2-1)$ is divisible by 16.