# A prime number problem

• September 28th 2009, 12:07 AM
Normis
A prime number problem
It's part of a larger problem but this is what I got left;
if 3^a - 2^a is a prime number, then a is a prime number.
If anyone could point me to the right direction to prove/disprove this, it would be nice. Thanks!
• September 28th 2009, 06:02 AM
gdmath
I do not know if i am going to help you, but:

Why dont you try to expand the identity as follow:
$
3^a-2^a=(2+1)^a-2^a
$

From binomial expansion we have:
$
(2+1)^a-2^a=2^a+\binom{a}{1}2^{a-1}+...+\binom{a}{a-1}2^{1}+1-2^a=...
$

And then we have the equality:
$
\binom{a}{1}2^{a-1}+...+\binom{a}{a-1}2^{1}=p-1
$

From there i have nothing more...

Some ideas:

If we rise both sides to a expand the right side and the again rise to a,
mayby we could compare the addition terms.

or

if we try to simplify left side by divisions of 2.

again i do not know if this helps to prove (or disprove) your identity.
• September 28th 2009, 06:41 AM
Prove It
Quote:

Originally Posted by Normis
It's part of a larger problem but this is what I got left;
if 3^a - 2^a is a prime number, then a is a prime number.
If anyone could point me to the right direction to prove/disprove this, it would be nice. Thanks!

The contrapositive is easier to prove.

So you need to show that:

If $a$ is NOT a prime number, then $3^a - 2^a$ is NOT a prime number.

Recall that a prime number is one that only has 1 and itself as factors.

So to show that $3^a - 2^a$ is not prime, we would have to show that it has a common factor.

First, if $a$ is not prime, then it can be expressed as $b\cdot c$, where $b \neq 1, b \neq a, c \neq 1, c\neq a$.

So $a = b \cdot c$

Therefore $3^a - 2^a = 3^{b \cdot c} - 2^{b \cdot c}$

$= \left(3^b\right)^c - \left(2^b\right)^c$.

Now we make use of the factorisation rule

$x^n - y^n = \left(x - y\right)\left(x^{n - 1} + x^{n - 2}y + x^{n - 3}y^2 + \dots + xy^{n - 2} + y^{n - 1}\right)$.

So $\left(3^b\right)^c - \left(2^b\right)^c$

$= \left(3^b - 2^b\right)\left(3^{b(c - 1)} + 3^{b(c - 2)}\cdot 2^b + 3^{b(c - 3)} \cdot 2^{2b} + \dots + 3^b\cdot 2^{b(c - 2)} + 2^{b(c - 1)}\right)$.

Thus $3^a - 2^a$ is not prime.

So, if $a$ is not prime, then $3^a - 2^a$ is not prime.

Therefore, if $3^a - 2^a$ IS prime, then $a$ is prime.
• September 28th 2009, 01:03 PM
Normis
Quote:

Originally Posted by Prove It
First, if $a$ is not prime, then it can be expressed as $b\cdot c$, where $b \neq 1, b \neq a, c \neq 1, c\neq a$.

I got to that and the factors when I tried the problem a few hours later but I wasn't sure if I was right. Can you just say that $b\neq 1, b \neq a, c \neq 1, c\neq a$? I mean, $1\cdot4$ (or doesnt it count because it CAN be written $2\cdot2$ and then **** it up?) would'nt be a prime and it wouldn't screw up the $(3^b-2^b)$ factor later on. I dont need to show that the factor with the sum is $\neq$ prime if for example $b=1$ and $c=4$ or any other number $n$ where $b=1$ and $c\neq$ prime?

Sorry for the stupid questions, I'm kinda new to this proving thing. Really appreciate the help!
• September 28th 2009, 02:13 PM
artvandalay11
in 1*4 you let b=1 which is a no no