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 July 28th, 2016, 02:30 AM #1 Senior Member   Joined: Jul 2011 Posts: 407 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, 05:05 AM #2 Senior Member     Joined: Feb 2010 Posts: 714 Thanks: 151 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. Thanks from panky
 July 28th, 2016, 05:10 AM #3 Senior Member   Joined: Jul 2011 Posts: 407 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, 05:27 AM   #4
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 Originally Posted by panky Thanks mrtwhs, But How can we prove that there is no solution for $\displaystyle z\geq 5$ please explain me, Thanks
If that was the question, you should have asked it at the start. In any event, I don't have a clue.

 July 28th, 2016, 06:28 AM #5 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 Factorials > 4 finishes by 0s. 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, 06:57 AM   #6
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 Originally Posted by mobel Factorials > 4 finishes by 0s. 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.
You are absolutely right! $\displaystyle 2^x + 2^y$ can never end in zeros ... well, except for these three plus an infinite number of others:

$\displaystyle 2^1 + 2^3 = 10$, $\displaystyle 2^2 + 2^4 = 20$, $\displaystyle 2^{12} + 2^2 = 4100$

 July 28th, 2016, 08:06 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, 09:14 AM   #8
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 Originally Posted by mobel How could the sum of 2 powers of 2 (2^x+2^y) give you a number finishing by 0?
This statement is factually incorrect and cannot be used in any valid proof.

July 28th, 2016, 09:29 AM   #9
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 Originally Posted by mrtwhs This statement is factually incorrect and cannot be used in any valid proof.
You are right.
I forget to write 0`s instead of 0
That was hint not proof.

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