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July 28th, 2016, 01:30 AM  #1 
Senior Member Joined: Jul 2011 Posts: 405 Thanks: 16  non negative integer ordered pair
Total number of non negative integer ordered pair $\displaystyle (x,y,z)$ in $\displaystyle 2^x+2^y=z!$

July 28th, 2016, 04:05 AM  #2 
Senior Member Joined: Feb 2010 Posts: 707 Thanks: 142 
Well there are no ordered pairs ... but here are a few ordered triples: (0, 0, 2) (1, 2, 3) (2, 1, 3) (3, 4, 4) (4, 3, 4) There may be others. 
July 28th, 2016, 04:10 AM  #3 
Senior Member Joined: Jul 2011 Posts: 405 Thanks: 16 
Thanks mrtwhs, But How can we prove that there is no solution for $\displaystyle z\geq 5$ please explain me, Thanks 
July 28th, 2016, 04:27 AM  #4 
Senior Member Joined: Feb 2010 Posts: 707 Thanks: 142  
July 28th, 2016, 05:28 AM  #5 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
Factorials > 4 finishes by 0`s. How could the sum of 2 powers of 2 (2^x+2^y) give you a number finishing by 0? It will be easy to prove it using modular arithmetic. 
July 28th, 2016, 05:57 AM  #6  
Senior Member Joined: Feb 2010 Posts: 707 Thanks: 142  Quote:
$\displaystyle 2^1 + 2^3 = 10$, $\displaystyle 2^2 + 2^4 = 20$, $\displaystyle 2^{12} + 2^2 = 4100$  
July 28th, 2016, 07:06 AM  #7 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
That was just a hint. You need more than that to have a robust proof.

July 28th, 2016, 08:14 AM  #8 
Senior Member Joined: Feb 2010 Posts: 707 Thanks: 142  
July 28th, 2016, 08:29 AM  #9 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41  

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