
Induction!
I'm not sure how to go about setting up the steps for this problem.
a1 = 1, a2 = 1 and an = 2an1 + an2 for all n ≥ 3. prove an is odd for all n≥1.
first six are 1,1,3,7,17,41
I could also use help on another problem...
proving (3,5,7) is the only prime triple  consecutive odd integers which are all prime 
to do this i prove that for any odd integer k, one of the integers 2k+1, 2k+3, and 2k+5 must be divisible by 3.
I know i need to use cases for this.
how does this then show that (3,5,7) is the only prime triple?
thanks for the input guys.
kyle

The characteristic equation is $\displaystyle a^{2}2a1=0$
It has solutions $\displaystyle a=\sqrt{2}+1, \;\ a=1\sqrt{2}$
$\displaystyle b(\sqrt{2}+1)^{n}+c(1\sqrt{2})^{n}$
Use the initial conditions to solve:
$\displaystyle 1=b(\sqrt{2}+1)+c(1\sqrt{2})$
$\displaystyle 1=b(\sqrt{2}+1)^{2}+c(1\sqrt{2})$
Solve the system and get $\displaystyle b=\frac{\sqrt{2}1}{2}; \;\ c=\frac{(\sqrt{2}+1)}{2}$
Therefore, we have:
$\displaystyle a_{n}=(\frac{\sqrt{2}1}{2})(\sqrt{2}+1)^{n}+(\frac{(\sqrt{2}+1)}{2})(1\sqrt{2})^{n}$
Now, perhaps you can use this to prove the solutions are always odd.

$\displaystyle a_1=1$; $\displaystyle a_2=1$ and $\displaystyle a_n=2\cdot{a_{n1}}+a_{n2}$
We have the base cases, so we'll go to the inductive step
Suppose $\displaystyle a_{n1}$ and $\displaystyle a_{n2}$ are odd, we shall prove that $\displaystyle a_n$ is also odd
Indeed, $\displaystyle a_n=2\cdot{a_{n1}}+a_{n2}\equiv{2\cdot{1}+1}\equiv{1}(\bmod.{2})$
Or if you prefer: $\displaystyle a_{n1}=2a+1$ and $\displaystyle a_{n2}=2b+1$ ( $\displaystyle a$ and $\displaystyle b$ are natural numbers)
So $\displaystyle a_n=2\cdot{(2a+1)}+2b+1=2(2\cdot{a}+b+1)+1 $ an odd number