Help needed in proving a number to be perfect square

[murshid_islam]

the number 11...122...25 has 1997 1's and 1998 2's. now how do i show that the number is a perfect square? i don't even know where to start. any help would be appreciated.

 

[Gib Z]

I don't understand what you mean by that number, or with how many 1997/1998s are in it, you mean exactly, or that's how many whole 2's/1's are in it. eg there may be a remainder.

 

[murshid_islam]

the first 1997 digits of the number are 1s.
the next 1998 digits of the number are 2s.
the last digit of the number is 5.

i have to show that the number is a perfect square. i hope i have stated it clearly now. can anyone help me? thanks in advance.

 

[uart]

It's and interesting question. I don't know how to prove it but I've got a hunch that the square root of that number is the 1998 digit integer whos first (leftmost) 1997 digits are all 3 and whos 1998th digit (rightmost) is 5.

 

[murshid_islam]

It's and interesting question. I don't know how to prove it but the square root of that number is the 1998 digit integer whos first (leftmost) 1997 digits are all 3 and whos 1998th digit (rightmost) is 5.

i guessed that much by seeing that
$$\sqrt{1225} = 35$$
$$\sqrt{112225} = 335$$
$$\sqrt{11122225} = 3335$$

but what i want is a proof.

 

[AlphaNumeric]

Prove it by induction. Use that $$\underbrace{3...3}_{k}5 = 3\underbrace{0...0}_{k-2} + \underbrace{3...3}_{k-1}5$$ and then consider $$(\underbrace{3...3}_{k}5)^{2}$$ using that expansion.

 

[murshid_islam]

sorry, but i still can't prove that $$\underbrace{1...1}_{1997}\underbrace{2...2}_{1998}5$$ is a perfect square. this is probably an incredibly stupid question. but how does the expansion you suggested help?

 

[matt grime]

You have a number. You *know* its square root. So what is the issue? Just square the number you think is the square root. Ok, so its too big to do on your calculator, but so what? Do it by long hand if necessary (and that doesn't mean write out all the digits). Alphanumeric's hint would appear to be the perfect answer.

 

[murshid_islam]

Alphanumeric's hint would appear to be the perfect answer.

if i understood AlphaNumeric's hint, i wouldn't have asked for more help. anyway,

Use that $$\underbrace{3...3}_{k}5 = 3\underbrace{0...0}_{k-2} + \underbrace{3...3}_{k-1}5$$

shouldn't it be $$\underbrace{3...3}_{k}5 = 3\underbrace{0...0}_{k} + \underbrace{3...3}_{k-1}5$$?

 

[matt grime]

Whatever. Just square it and use the obvious induction argument that you'd already noticed in post 5.

 

[Gib Z]

To answer your question, just square what you know is the answer. To show how it would be derived on an occasion you don't know the answer, which isn't your question, I have no idea. If that's the question, your just going for overkill.

 

[murshid_islam]

ok i think i have got it. just let me know if i am wrong. this is what i have done:

i want to prove that
$$\left(\underbrace{3...3}_{n}5\right)^{2} = \underbrace{1...1}_{n}\underbrace{2...2}_{n+1}5$$ for any $$n \in \mathbb{N}$$

this is obviously true for n = 1, since $$(35)^2 = 1225$$

next i assume that it is true for n = k, i.e.,
$$\left(\underbrace{3...3}_{k}5\right)^{2} = \underbrace{1...1}_{k}\underbrace{2...2}_{k+1}5$$

now i have to prove it for n = k + 1. For n = k + 1,

$$\left(\underbrace{3...3}_{k+1}5\right)^{2}$$

$$= \left(3\underbrace{0...0}_{k+1} + \underbrace{3...3}_{k}5\right)^{2}$$

$$= \left(3\underbrace{0...0}_{k+1}\right)^{2} + 2\left(3\underbrace{0...0}_{k+1}\right)\left(\underbrace{3...3}_{k}5\right) + \left(\underbrace{3...3}_{k}5\right)^{2}$$

$$= 9\underbrace{0...0}_{2k+2} + 2\underbrace{0...0}_{k-1}1\underbrace{0...0}_{k+2} + \underbrace{1...1}_{k}\underbrace{2...2}_{k+1}5$$

$$= 9\underbrace{0...0}_{k}\underbrace{0...0}_{k+2} + 2\underbrace{0...0}_{k-1}1\underbrace{0...0}_{k+2} + \underbrace{1...1}_{k}\underbrace{2...2}_{k+1}5$$

$$= 11\underbrace{1...1}_{k-1}2\underbrace{2...2}_{k+1}5$$

$$= \underbrace{1...1}_{k+1}\underbrace{2...2}_{k+2}5$$

did i do anything wrong? or is it correct?

 

[slider142]

That's an excellent use of mathematical induction, and every step is correct.