# A pretty hard problem .. please help

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• March 19th 2013, 07:43 AM
stephanx
A pretty hard problem .. please help
I have to solve that equation :

11n-2n=k2

where n,k are from N .
• March 19th 2013, 07:48 AM
Shakarri
Re: A pretty hard problem .. please help
Well by observation n=1, k=3 is one solution. I don't think there are other solutions but I have no way of proving their aren't
• March 19th 2013, 07:57 AM
stephanx
Re: A pretty hard problem .. please help
Yes , I have also find that . And n can be also =0 and k=0 .
So there are only 2 solutions,I think .. but how should I demonstrate ?
• March 19th 2013, 08:45 AM
Shakarri
Re: A pretty hard problem .. please help
0 is generally not considered a natural number
• March 19th 2013, 09:03 AM
Plato
Re: A pretty hard problem .. please help
Quote:

Originally Posted by Shakarri
0 is generally not considered a natural number

"There is no universal agreement about whether to include zero in the set of natural numbers: some define the natural numbers to be the positive integers {1, 2, 3, ...}, while for others the term designates the non-negative integers {0, 1, 2, 3, ...}. The former definition is the traditional one, with the latter definition having first appeared in the 19th century. Some authors use the term natural number to exclude 0 and whole number to include it; others use whole number in a way that includes both 0 and the negative integers, i.e., as an equivalent of the integer term."

See this web entry.

I have a great many different textbooks on set theory. In the vast majority of them use $0\in\mathbb{N}$.
• March 19th 2013, 09:05 AM
Shakarri
Re: A pretty hard problem .. please help
It's one of the things in mathematics that irks me :(
• March 19th 2013, 09:59 AM
stephanx
Re: A pretty hard problem .. please help
Quote:

Originally Posted by Plato
"There is no universal agreement about whether to include zero in the set of natural numbers: some define the natural numbers to be the positive integers {1, 2, 3, ...}, while for others the term designates the non-negative integers {0, 1, 2, 3, ...}. The former definition is the traditional one, with the latter definition having first appeared in the 19th century. Some authors use the term natural number to exclude 0 and whole number to include it; others use whole number in a way that includes both 0 and the negative integers, i.e., as an equivalent of the integer term."

See this web entry.

I have a great many different textbooks on set theory. In the vast majority of them use $0\in\mathbb{N}$.

Ok , firstly , thanks for the explanation . I've read the article about the natural numbers about one year ago :)
I haven't seen that the author had mentioned that the numbers are different from 0 . And also , in the some problems that i've seen if the numbers aren't null they are included in N* and in the problem the number are only from N .