Prove that if (2 to the power n) - 15 = x square, then n = 4 or n = 6.
Thanks very much guys
We need,
$\displaystyle 2^n-15=x^2$
If,
$\displaystyle n=2k$
Then,
$\displaystyle (2^k)^2-15=x^2$
Thus,
$\displaystyle (2^k)^2-x^2=15$
Thus,
$\displaystyle (2^k-x)(2^k+x)=15$
The factorization of 15 is,
$\displaystyle 1\cdot 15,3\cdot 5$
Examine each of the cases to get the result.
Note,
$\displaystyle 2^k-x<2^k+x$
Thus, we have only two possibilities,
$\displaystyle \left\{ \begin{array}{c}2^k-x=1\\2^k+x=15\end{array} \right\}$
$\displaystyle \left\{ \begin{array}{c}2^k-x=3\\2^k+x=5\end{array} \right\}$
The equations respectively,
$\displaystyle 2\cdot 2^k=2^{k+1}=16\to k+1=4\to k=3$
$\displaystyle 2\cdot 2^k=2^{k+1}=8\to k+1=3\to k=2$
In each case we have,
$\displaystyle n=2(3)=6$
$\displaystyle n=2(2)=4$
----
If,
$\displaystyle n=1$ then,
$\displaystyle 2-15=x^2$--->Impossible.
If $\displaystyle n>1$ and
$\displaystyle n=2k+1$
Then,
$\displaystyle 2(2^k)^2-15=x^2$
Thus,
$\displaystyle x^2-2(2^k)^2=-15$
This is a Pellian look-a-like equation.
(Now I am hoping that it has no solution to it).
I have to think about this diophantine equation.
Basically I need to show,
$\displaystyle a^2-2b^2=-15$
Has no solutions.
Aha! Solved it.
The diophantine equation,
$\displaystyle a^2-2b^2=-15$
Has no solution!
The left hand side needs to be divisible by 15
Thus,
$\displaystyle a^2-2b^2\equiv 0(\mbox{ mod }3)$
$\displaystyle a^2-2b^2\equiv 0(\mbox{ mod }5)$
This means the Legendre symbols have the value,
$\displaystyle (2b^2/3)=(2/3)=1$
$\displaystyle (2b^2/5)=(2/5)=1$
But that is not true!
Because by Euler's criterion,
$\displaystyle 2^{\frac{3-1}{2}}\equiv -1 (\mbox{ mod } 3)$
$\displaystyle 2^{\frac{5-1}{2}}\equiv -1 (\mbox{ mod }5)$