Can a number of this form be a perfect square? (4 * 10^{2n-1} - 31) / 9

blamocur said:
Actually, I did not use induction.
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I never said what method you used as I have no idea how or even if you solved this problem. The OP mentioned induction so I wanted them to try it again.
 
blamocur said:
Actually, I did not use induction.
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It seems to me that the contradiction approach does seem a bit more straightforward. Did you go in a similar direction as I have or did you use a completely different approach?
 
anonym4= said:
It seems to me that the contradiction approach does seem a bit more straightforward. Did you go in a similar direction as I have or did you use a completely different approach?
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Your post #13 has a good start.
 
Thank you all for your help kind people. I think I'll take this from here. See ya
 
It would be a good idea to post your solution to close the thread. But since you haven't, and since it's been a week since the original post I'll post mine.
Here is my proof by contradiction: if the fraction is a square then:
[math]\frac{4\cdot 10^{2n}-31}{9} = k^2 \;\Longleftrightarrow\; {4\cdot 10^{2n}-31} = (3k)^2 \;\Longleftrightarrow\;[/math][math]\;\Longleftrightarrow\; 4\cdot 10^{2n}-(3k)^2 = 31 \;\Longleftrightarrow\; \left(2\cdot 10^{n}-3k\right) \left(2\cdot 10^{n}+3k\right) = 31[/math]In the latest equality 31, which is prime, is represented as a product of two numbers, which means that one of the terms is 1 and another 31. Since both numbers must be positive we know that [imath]n\leq 1[/imath]. A quick check shows that neither [imath]n=0[/imath] nor [imath]n=1[/imath] work.
 
blamocur said:
It would be a good idea to post your solution to close the thread. But since you haven't, and since it's been a week since the original post I'll post mine.
Here is my proof by contradiction: if the fraction is a square then:
[math]\frac{4\cdot 10^{2n}-31}{9} = k^2 \;\Longleftrightarrow\; {4\cdot 10^{2n}-31} = (3k)^2 \;\Longleftrightarrow\;[/math][math]\;\Longleftrightarrow\; 4\cdot 10^{2n}-(3k)^2 = 31 \;\Longleftrightarrow\; \left(2\cdot 10^{n}-3k\right) \left(2\cdot 10^{n}+3k\right) = 31[/math]In the latest equality 31, which is prime, is represented as a product of two numbers, which means that one of the terms is 1 and another 31. Since both numbers must be positive we know that [imath]n\leq 1[/imath]. A quick check shows that neither [imath]n=0[/imath] nor [imath]n=1[/imath] work.
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The exponent of 10 in OP was (2n - 1) ....... that might throw a monkey-wrench in the scheme.
 
blamocur said:
It would be a good idea to post your solution to close the thread. But since you haven't, and since it's been a week since the original post I'll post mine.
Here is my proof by contradiction: if the fraction is a square then:
[math]\frac{4\cdot 10^{2n}-31}{9} = k^2 \;\Longleftrightarrow\; {4\cdot 10^{2n}-31} = (3k)^2 \;\Longleftrightarrow\;[/math][math]\;\Longleftrightarrow\; 4\cdot 10^{2n}-(3k)^2 = 31 \;\Longleftrightarrow\; \left(2\cdot 10^{n}-3k\right) \left(2\cdot 10^{n}+3k\right) = 31[/math]In the latest equality 31, which is prime, is represented as a product of two numbers, which means that one of the terms is 1 and another 31. Since both numbers must be positive we know that [imath]n\leq 1[/imath]. A quick check shows that neither [imath]n=0[/imath] nor [imath]n=1[/imath] work.
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Hey, I know it's been a while but I'd be interested in as to how n must be less or equal than 1, in order for the factors to be positive. I don't seem to understand that. Thanks
 
anonym4= said:
Can any number of the form [math]\frac{4\cdot 10^{2n-1}-31}{9}[/math] ever be a perfect square? This statements needs to be mathematically proven (in detail!).
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I'm just wondering what the graph of [imath]f(n)[/imath] will look like. There should be, sensu amplissimo, holes at [imath]y \in \{0, 1, 4, 9, ... \}[/imath] i.e. these points - [imath](0, 0), (1, 1), (2, 4), (3, 9), (n, n^2), ... [/imath] - are off the curve.


[imath]4 \times 10^{2n - 1} - 31 = 31 \cdot 10^{2n - 1} - 27 \cdot 10^{2n - 1} - 31[/imath]

= [imath]31(10^{2n - 1} - 1) - 27(10^{2n - 1})[/imath]

[imath]f(n) = 31(1_11_2...1_{2n - 1}) - 3(10^{2n - 1})[/imath]

[imath]f(3)[/imath] is not a perfect square.
 
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