# Math Help - How to do this algebra "trick"

1. ## How to do this algebra "trick"

I'm trying to prove that $7|(3^{2n+1}+2^{n+2})$

Anyway, I've simplified the RHS to $3*9^n+4*9^n$

How can I "pull out" $(3+4)$ here? ie, make it so that

$3*9^n+4*9^n=(3+4)*(???)$

Thanks!

2. Originally Posted by paupsers
I'm trying to prove that $7|(3^{2n+1}+2^{n+2})$

Anyway, I've simplified the RHS to $3*9^n+4*9^n$

How can I "pull out" $(3+4)$ here? ie, make it so that

$3*9^n+4*9^n=(3+4)*(???)$

Thanks!
$\displaystyle 3^{2n + 1} + 2^{n + 2} = 3 \cdot 9^n + 4 \cdot 2^n$

Note that 9 = 2 mod 7.

-Dan

3. (3+4).9^n no ? ( but i think what you made a mistake in simplifying RHS is wrong )
You can actually get the result in one/two lines using congruence.
$3^2\equiv 2[7]$ so $3^{2n}\equiv 2^n[7]$ and since $3\equiv -4[7] \equiv -2^2[7]$ we get (by multupliying) the result.

Topsquark is faster

4. Originally Posted by paupsers
I'm trying to prove that $7|(3^{2n+1}+2^{n+2})$

Anyway, I've simplified the RHS to $3*9^n+4*9^n$

How can I "pull out" $(3+4)$ here? ie, make it so that

$3*9^n+4*9^n=(3+4)*(???)$

Thanks!
That's the "distributive law": (ac+ bc)= (a+ b)c