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July 28th, 2016, 02:30 AM   #1
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non negative integer ordered pair

Total number of non negative integer ordered pair $\displaystyle (x,y,z)$ in $\displaystyle 2^x+2^y=z!$
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July 28th, 2016, 05:05 AM   #2
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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.
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July 28th, 2016, 05:10 AM   #3
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Thanks mrtwhs, But How can we prove that there is no solution for $\displaystyle z\geq 5$

please explain me, Thanks
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July 28th, 2016, 05:27 AM   #4
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Quote:
Originally Posted by panky View Post
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.
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July 28th, 2016, 06:28 AM   #5
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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.
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July 28th, 2016, 06:57 AM   #6
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Quote:
Originally Posted by mobel View Post
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.
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:

$\displaystyle 2^1 + 2^3 = 10$, $\displaystyle 2^2 + 2^4 = 20$, $\displaystyle 2^{12} + 2^2 = 4100$
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July 28th, 2016, 08:06 AM   #7
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That was just a hint. You need more than that to have a robust proof.
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July 28th, 2016, 09:14 AM   #8
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Quote:
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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.
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July 28th, 2016, 09:29 AM   #9
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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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