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$2^{x}=3^{y}+5$

Posted: Sun Sep 01, 2013 12:31 pm
Find all positive integers x,y such that $2^{x}=3^{y}+5$

Re: $2^{x}=3^{y}+5$

Posted: Mon Sep 09, 2013 11:59 am
asif e elahi wrote:Find all positive integers x,y such that $2^{x}=3^{y}+5$
It is easy to see that $(x,y)=(3,1),(5,3)$.
If $x \ge 6$ then $64|2^x$. Therefore $3^y \equiv 59 \pmod{64}$. It follows that $y \equiv 11 \pmod{16}$. Hence $3^y \equiv 3^{11} \equiv 7 \pmod{17}$.
Since $3^y+5 \equiv 2 \pmod{3}$ then $2^x \equiv 2 \pmod{3}$. It follows that $x$ is odd. Hence $x \equiv 1,3,5,7 \pmod{8}$.
If $x \equiv 1 \pmod{8}$ then $2^x \equiv 2 \pmod{17}$, a contradiction.
If $x \equiv 3 \pmod{8}$ then $2^x \equiv 8 \pmod{17}$, a contradiction.
If $x \equiv 5 \pmod{8}$ then $2^x \equiv 15 \pmod{17}$, a contradiction.
If $x \equiv 7 \pmod{8}$ then $2^x \equiv 9 \pmod{17}$, a contradiction.
Thus, the solution is $\boxed{ (x,y)=(3,1),(5,3)}$.

Re: $2^{x}=3^{y}+5$

Posted: Mon Sep 09, 2013 12:01 pm
More stronger: $2^x=3^y+5^z$.

Re: $2^{x}=3^{y}+5$

Posted: Tue Sep 10, 2013 11:26 am
harrypham wrote: Therefore $3^y \equiv 59 \pmod{64}$. It follows that $y \equiv 11 \pmod{16}$. Hence $3^y \equiv 3^{11} \equiv 7 \pmod{17}$.
How $y \equiv 11 \pmod{16}$ ?

Re: $2^{x}=3^{y}+5$

Posted: Tue Sep 10, 2013 12:16 pm
asif e elahi wrote:
harrypham wrote: Therefore $3^y \equiv 59 \pmod{64}$. It follows that $y \equiv 11 \pmod{16}$. Hence $3^y \equiv 3^{11} \equiv 7 \pmod{17}$.
How $y \equiv 11 \pmod{16}$ ?
You note that $3^{16} \equiv 1 \pmod{64}$ and $3^{11} \equiv 59 \pmod{64}$.