Prove that if (2 to the power n) - 15 = x square, then n = 4 or n = 6.

Thanks very much guys:)

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- Dec 3rd 2006, 06:36 PMsuedenationnumber theory, solve diophantine equation
Prove that if (2 to the power n) - 15 = x square, then n = 4 or n = 6.

Thanks very much guys:) - Dec 3rd 2006, 06:47 PMThePerfectHacker
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. - Dec 3rd 2006, 06:58 PMThePerfectHacker
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,

$\displaystyle 2^{\frac{3-1}{2}}\equiv -1 (\mbox{ mod } 3)$

$\displaystyle 2^{\frac{5-1}{2}}\equiv -1 (\mbox{ mod }5)$ - Dec 3rd 2006, 07:39 PMsuedenation
Thanks sooooo much, you are a genius....hehe:D