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!

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- Sep 28th 2009, 12:07 AMNormisA 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! - Sep 28th 2009, 06:02 AMgdmath
I do not know if i am going to help you, but:

Why dont you try to expand the identity as follow:

$\displaystyle

3^a-2^a=(2+1)^a-2^a

$

From binomial expansion we have:

$\displaystyle

(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:

$\displaystyle

\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. - Sep 28th 2009, 06:41 AMProve It
The contrapositive is easier to prove.

So you need to show that:

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

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

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

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

So $\displaystyle a = b \cdot c$

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

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

Now we make use of the factorisation rule

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

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

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

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

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

Therefore, if $\displaystyle 3^a - 2^a$ IS prime, then $\displaystyle a$ is prime. - Sep 28th 2009, 01:03 PMNormis
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 $\displaystyle b\neq 1, b \neq a, c \neq 1, c\neq a$? I mean, $\displaystyle 1\cdot4$ (or doesnt it count because it CAN be written $\displaystyle 2\cdot2$ and then **** it up?) would'nt be a prime and it wouldn't screw up the $\displaystyle (3^b-2^b)$ factor later on. I dont need to show that the factor with the sum is $\displaystyle \neq$ prime if for example $\displaystyle b=1$ and $\displaystyle c=4$ or any other number $\displaystyle n$ where $\displaystyle b=1$ and $\displaystyle c\neq$ prime?

Sorry for the stupid questions, I'm kinda new to this proving thing. Really appreciate the help! - Sep 28th 2009, 02:13 PMartvandalay11
in 1*4 you let b=1 which is a no no