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Nice Equation

Posted: Thu Apr 12, 2012 12:27 pm
by Phlembac Adib Hasan
If $x,y,z\in \mathbb {N}_0$ solve the equation \[2^x+3^y=z^2\]
(This problem is a simple joke if you use Zsigmondy's theorem.But there is another pretty solution using no advanced ideas.That's why I am posting it in secondary level.I also showed it to some persons in the main camp.)

Re: Nice Equation

Posted: Thu Apr 12, 2012 10:00 pm
by Nadim Ul Abrar
:|
Case 1 , $x\geq 2$ .
Case 2 , $x=0,1$ .

Case 1 :
$\bmod 4$ says that ,
$y$ is even ,
$\bmod 3$ says that ,
$x$ is even too .

let $y=2k$
then $(z-3^k)(z+3^k)=2^x$
So , $z+3^k=2^a,z-3^k=2^b$ .
or $2z=2^a+2^b$

this leads that $b=1$ .
Now $z-3^k=2$
or $2z=2.3^k+2=2^a$.
or $3^k+1=2^{a-1}$ .
$\bmod8$ says that $a-1\leq 2$
So $k=0,1$ ;$ a=2,3$

that leads $x=3,4$
for $x=3, k=0$ and $y=0$ .
for$ x=4, k=1$ and $ y=2$ .

So solutions for this case are $(x,y,z)=(3,0,3)(4,2,5)$

Case 2
For $x=1$ , $L.S\equiv 2 mod3$
For $x=0$ , $(z+1)(z-1)=3^y$ or $z=2$ and $y=1$ .
So solution for Case 2 is $(x,y,z)=(0,1,2)$

Re: Nice Equation

Posted: Sat Apr 14, 2012 9:44 am
by Phlembac Adib Hasan
Good solution, Nadim vaia.Here is mine:
Taking $mod\; 3$ shows $x$ is even.So $x=2k$.Therefore \[3^y=(z+2^k)(z-2^k)\]
So $z+2^k=3^a$ and $z-2^k=3^b$.But they differ by $2^{k+1}$.In particular \[3^b(3^{a-b}-1)=2^{k+1}\]Which easily gives the solution.

Re: Nice Equation

Posted: Thu Jan 30, 2014 10:45 pm
by asif e elahi
Phlembac Adib Hasan wrote:If $x,y,z\in \mathbb {N}_0$ solve the equation \[2^x+3^y=z^2\]
(This problem is a simple joke if you use Zsigmondy's theorem.But there is another pretty solution using no advanced ideas.That's why I am posting it in secondary level.I also showed it to some persons in the main camp.)
What is Zsigmondy's theorem?

Re: Nice Equation

Posted: Fri Jan 31, 2014 9:38 am
by Thanic Nur Samin
You should google it.