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

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anonym4=

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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!).
 
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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Okay. What has been covered recently in class? What are your thoughts? What have you tried? How far have you gotten?

Please be complete. Thank you!
 
stapel said:
Okay. What has been covered recently in class? What are your thoughts? What have you tried? How far have you gotten?

Please be complete. Thank you!
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My apologies. The initial question was if any number which consists an odd amount of consecutive digits of 4 and ends in a 1 (4441, 444441, and so on) can be a perfect square. So far I have found a general formula to construct such a number (as shown above) and thought about a proof using induction, which I am currently stuck at.
 
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!).
Click to expand...
I would probably start by plugging in some values for n, just to get a sense of how the expression works. I'd also check the definition of "perfect square" in your context.

Then I might think about what it would take to prove this "can it ever" (that is, existence) claim, one way or the other (existence or non-existence).
 
Dr.Peterson said:
I would probably start by plugging in some values for n, just to get a sense of how the expression works. I'd also check the definition of "perfect square" in your context.

Then I might think about what it would take to prove this "can it ever" (that is, existence) claim, one way or the other (existence or non-existence).
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I actually came up with that expression as the question was initially different (my mistake I should have said that in the question). It's for number like 4441, 444441, 44444441 (odd amount of 4 followed by 1). I have thought about using induction as a proof since I believe it's fine for existence or non-existence
 
I'm terribly sorry, It seems like I have made a typo in my original post.
Here is the updated term[math]\frac{4*10^{2n}-31}{9}[/math]
 
anonym4= said:
I actually came up with that expression as the question was initially different (my mistake I should have said that in the question). It's for number like 4441, 444441, 44444441 (odd amount of 4 followed by 1). I have thought about using induction as a proof since I believe it's fine for existence or non-existence
Click to expand...
anonym4= said:
I'm terribly sorry, It seems like I have made a typo in my original post.
Here is the updated term[math]\frac{4*10^{2n}-31}{9}[/math]
Click to expand...
Please (1) show us the actual original problem as given to you; and (2) show us your attempt at induction.

Proofs of existence and non-existence are very different. For existence, you just need to show an example; no induction is needed. So I suppose you must be trying to show non-existence?

(For your initial post, I found that n=1 gave an example.)
 
The answer is 'no', and the proof (by contradiction) is pretty straightforward.
 
Dr.Peterson said:
Please (1) show us the actual original problem as given to you; and (2) show us your attempt at induction.

Proofs of existence and non-existence are very different. For existence, you just need to show an example; no induction is needed. So I suppose you must be trying to show non-existence?

(For your initial post, I found that n=1 gave an example.)
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My mistake, certainly. The problem was this: "For any given number of the form 44...41, which consists of an odd number of digits of 4, followed by a digit of 1, could this number ever be a perfect square?". Perfect square meaning in this case, can the square root of such a number be a positive integer.
My first induction attempt looked something like this:
[math]\frac{4\cdot10^{2n}-31}{9} = k^2[/math], where k^2 is a positive integer
Starting condition: n =2 (not 1 in this case, since the smallest considerable number we can look at is 4441)
[math]\frac{4\cdot10^{4}-31}{9} = 4441[/math]Since [math]\sqrt{4441} \approx 66.641[/math], the condition is false, so this doesn't work. Now I'm stuck...
So, I need to show non-existence, which obviously doesn't work with induction and which I am rather unfamiliar with.
 
For any given number of the form 44...41 could this number ever be a perfect square?
They just want to know if there is at least one number of this form. So far you haven't found any. So either you find 1 or proof that there are none.

Can a perfect square even end in a 1? I would settle that before I look any further! How about 41?
 
Steven G said:
For any given number of the form 44...41 could this number ever be a perfect square?
They just want to know if there is at least one number of this form. So far you haven't found any. So either you find 1 or proof that there are none.

Can a perfect square even end in a 1? I would settle that before I look any further! How about 41?
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Sure, 81 for example or 121
I am certain there are none, I just have to prove it...
 
Can you factor the numerator into a perfect square? can you factor out a 9 from that perfect square (if it exists).
 
blamocur said:
The answer is 'no', and the proof (by contradiction) is pretty straightforward.
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Could you comment on this approach?
Suppose k is an integer, where
[math]9k^2=4\cdot10^{2n}-31[/math]With this, the right side is always odd, since 4*10^{2n} is even and subtracting 31 makes it odd. However the left side could be even (e.g 9*2^2 = 36). This is a contradiction, so the statement can not be true.
 
anonym4= said:
Could you comment on this approach?
Suppose k is an integer, where
[math]9k^2=4\cdot10^{2n}-31[/math]With this, the right side is always odd, since 4*10^{2n} is even and subtracting 31 makes it odd. However the left side could be even (e.g 9*2^2 = 36). This is a contradiction, so the statement can not be true.
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No, your logic is not correct. The problem is not asking you to show that every number of the form 44....41 is a perfect square. They are asking if ANY number of the form 44...41 is a perfect square.
You clearly showed in you example that if k is even that the lhs and the rhs can't be equal. Wonderful!! Now show that the two sides are never equal when k is odd or show an example of a number in the form 44...41 being a perfect square!

For the record, you already showed an example of a number in the form 44...41 that was not a perfect square. Then you go again and show us an infinite amount of numbers in the form 44...41 that are not perfect squares. You keep thinking that is enough. If it was enough then you solved this problem twice. Unfortunately, what you showed is not enough. Again, now show that when k is odd that ....
 
anonym4= said:
I actually came up with that expression as the question was initially different (my mistake I should have said that in the question). It's for number like 4441, 444441, 44444441 (odd amount of 4 followed by 1). I have thought about using induction as a proof since I believe it's fine for existence or non-existence
Click to expand...
If such a number exist, but NOT for all n, then you can't use induction. You can only use induction to show that such a number doesn't exist. That is all numbers of the form 44...41 are NOT perfect squares.
 
Steven G said:
No, your logic is not correct. The problem is not asking you to show that every number of the form 44....41 is a perfect square. They are asking if ANY number of the form 44...41 is a perfect square.
You clearly showed in you example that if k is even that the lhs and the rhs can't be equal. Wonderful!! Now show that the two sides are never equal when k is odd or show an example of a number in the form 44...41 being a perfect square!

For the record, you already showed an example of a number in the form 44...41 that was not a perfect square. Then you go again and show us an infinite amount of numbers in the form 44...41 that are not perfect squares. You keep thinking that is enough. If it was enough then you solved this problem twice. Unfortunately, what you showed is not enough. Again, now show that when k is odd that ....
Click to expand...
I see, thank you for pointing this out. Since any odd k does makes both sides odd, the parity argument doesn't work. I have though about other arguments like checking the modulo on both sides but they seem to be always the same when k is odd. Could you think of any other way of showing a contradiction when k is odd, because I can't seem to figure one out?
 
If both sides are both odds then the two sides can be equal and be equal to a perfect square. That is the type of number you are looking for.

You were told in a previous post by blamocur that there is no such number!

Therefore you should try to prove this by induction.

I'll start you you off.

n=1: [4*101-31]/9 = 1is not of the correct form.

n=2: [4*102-31]/9 = 369/9 = 41 which is not a perfect square.

for n=k assume that [4*10k-31]/9 is not a perfect square.

Possibly using the assumption show that [4*10k+1-31]/9 is not a perfect square
 
blamocur said:
81, 121, ...
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Yes, but the poster needs to 1st really that ending in 1 is even possible! Perfect squares can end in 0, 1, 4, 5, 6 or 9. I realized that on my own while in JHS.

If the number ended in 3, then the problem is done as no square ends in a 3. I'm just trying to get the OP to think in a logical order.
 
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