# number theory, solve diophantine equation

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• December 3rd 2006, 06:36 PM
suedenation
number 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:)
• December 3rd 2006, 06:47 PM
ThePerfectHacker
Quote:

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

Thanks very much guys:)

We need,
$2^n-15=x^2$
If,
$n=2k$
Then,
$(2^k)^2-15=x^2$
Thus,
$(2^k)^2-x^2=15$
Thus,
$(2^k-x)(2^k+x)=15$
The factorization of 15 is,
$1\cdot 15,3\cdot 5$
Examine each of the cases to get the result.
Note,
$2^k-x<2^k+x$
Thus, we have only two possibilities,
$\left\{ \begin{array}{c}2^k-x=1\\2^k+x=15\end{array} \right\}$
$\left\{ \begin{array}{c}2^k-x=3\\2^k+x=5\end{array} \right\}$
The equations respectively,
$2\cdot 2^k=2^{k+1}=16\to k+1=4\to k=3$
$2\cdot 2^k=2^{k+1}=8\to k+1=3\to k=2$
In each case we have,
$n=2(3)=6$
$n=2(2)=4$
----
If,
$n=1$ then,
$2-15=x^2$--->Impossible.
If $n>1$ and
$n=2k+1$
Then,
$2(2^k)^2-15=x^2$
Thus,
$x^2-2(2^k)^2=-15$
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,
$a^2-2b^2=-15$
Has no solutions.
• December 3rd 2006, 06:58 PM
ThePerfectHacker
Aha! Solved it.

The diophantine equation,
$a^2-2b^2=-15$
Has no solution!

The left hand side needs to be divisible by 15
Thus,
$a^2-2b^2\equiv 0(\mbox{ mod }3)$
$a^2-2b^2\equiv 0(\mbox{ mod }5)$
This means the Legendre symbols have the value,
$(2b^2/3)=(2/3)=1$
$(2b^2/5)=(2/5)=1$
But that is not true!
Because by Euler's criterion,
$2^{\frac{3-1}{2}}\equiv -1 (\mbox{ mod } 3)$
$2^{\frac{5-1}{2}}\equiv -1 (\mbox{ mod }5)$
• December 3rd 2006, 07:39 PM
suedenation
Thanks sooooo much, you are a genius....hehe:D