Prove that an integer with digits '1' is not a perfect square.

[futb0l]

I am having trouble with this one...

Prove that any positive integer whose ALL digits are 1s (except 1) is not a perfect square.

 

[Muzza]

What base? 11 in base 8 is a perfect square.

 

[futb0l]

no, I mean not like 11^8.. I mean prove that 11, 111, 111, 1111... is a perfect square...

 

[Muzza]

I didn't mean 11^8 either, I meant 11 written in base 8. For example, 34 written in base 8 is equal to 3 * 8^1 + 4 * 8^0 = 28 in base 10.

 

[futb0l]

I think it should be:

<br /> N = \sum_{n=1}^\infty (1*10^{n-1})<br />

btw, it should be base 10 then.

 

[matt grime]

I don't think you quite mean that infinite (divergent) sum either, but we get the idea.

 

[futb0l]

ok then cool, i am only 15 - my maths skill is pretty crap.

 

[futb0l]

i think it should be:

<br /> N = \sum_{n=1}^k (1*10^{n-1})<br />

where k can be any positive integer.

 

[matt grime]

suppose x^2 = 1+10+100+...

then mod 10, x=+/-1, that is x = 10r+1 or 10r-1 for some r.

consider (10r+1)^2 = 100r^2+20r+1

this is equal to 1+10+100+...

assume r is not zero, 100r^2+20r=10+100+...
divide by 10,

10r^2+2r=1+10+..

but the rhs is even and the lhs is odd, so it must be that r=0. similarly we reason for x=10r-1 and decide r=0 too.

 

[futb0l]

how come the RHS is even and LHS is odd??

i don't quite understand this stuff, can you recommend me a book or an article that is simple to learn?

 

[matt grime]

my apologies, i meant to say the lhs is even (it is divisible by two) and the rhs is odd since it is 1 plus a multiple of 10

 

[Gokul43201]

Explanation of matt's proof :
x^2 = 1+10+100+...=1+ 10(1+10+...)=1+10k, say, where k=1+10+..., the number within ().

x^2 = 1+10k means that the units digit of x^2 is 1 (example, 81, 121, etc.)

Now for the units' digit of x^2 to be 1, the units digit of x must be 1 or 9. (example : 11^2 =121, 19^361, but 23^2 must end with 9 and 45^2 ends with 5 etc.)

If the units' digit of x is 1, this means that x can be written as 1 + 10a + 100b +... = 1 + 10r (example : 151 = 1 + 50 + 100 = 1 + 10*15 )

If the units' digit of x is 9, this means that x can be written as 9 + 10a + 100b + ... (10-1) + 10a + 100b +... = 10 + 10a + 100b + ... -1 = 10r -1
(example 69 = 70 - 1 = 7*10 - 1)

Having shown that x must be of the form 10r + 1 or 10r - 1, the rest is just algebra...and is easy to follow in matt's proof.

In the last line, you have 10r^2 + 2r = 1 + 10 + ...
LHS = 2*5r^2 + 2r = 2*(5r^2 + r) = 2n, an even number
RHS = 1+10+100+... = 1 + 2(5 + 50 + ...) = 2n + 1, an odd number

 

[futb0l]

So, if the LHS is an even number and RHS is odd, it should not be a perfect square?? if not please explain the very basics.

 

[futb0l]

i seem to have another proof, every odd perfect square when divided by 8 has a remainder of 1... and noting that if an odd integer is squared the answer is always odd.
and with the exception of 1,11, every positive integer whose all digits are 1 have a remainder of 7. But 11 is clearly not a perfect square.

I think this should be enough proof..
but please explain Matt Grime's proof, because I'd like to learn.

 

[Gokul43201]

matt's proof is known as reductio ad absurdum or proof by contradiction. You make an assumption (in this case, you assume that some perfect square can be a number of the form 11...1) and then arrive at a logical contradiction (such as the statement "even number = odd number"). This implies that the assumption that was made was incorrect (so it should not be possible to find any perfect square of the form 11...1), and that's what you wanted to prove.

 

[Gokul43201]

i seem to have another proof, every odd perfect square when divided by 8 has a remainder of 1... and noting that if an odd integer is squared the answer is always odd.
and with the exception of 1,11, every positive integer whose all digits are 1 have a remainder of 7. But 11 is clearly not a perfect square.

I think this should be enough proof..
but please explain Matt Grime's proof, because I'd like to learn.

This is a perfectly good proof, as long as you show how these statements are true. They are true, no doubt...but I would imagine that you would have to show, for instance that odd perfect squares leave residues of 1 when divided by 8. You might also have to prove that 11...1 always leaves a remainder of 7 when divided by 8.

The first is proved as follows :
N=(2k+1), say, some odd integer.
N^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1
Since k and k+1 are consecutive numbers, one of them must be even. So their product must also be an even number. So k(k+1) = 2m, say.
Then N^2 = 4*2m + 1 = 8m + 1.
This last equation tells you that N^2, when divided by 8, gives a remainder of 1.

To prove that 11...1 leaves a remainder of 7, do the following :
i. Write this as the sum of powers of 10, as in post#8
ii. What happens when you divide 1000 by 8 ? And higher powers of 10 ?
iii. This argument leaves you with having to find the remainder for only 111, which is 7 (all other numbers of the form 11...1 have the same remainder as 111).

Now you are really done.

For a 15 year old, your math skills are far removed from "pretty crap."

 

[futb0l]

just the answer that i wanted, thanks a lot Goku|43201 and also matt grime.

 

[supundika]

The answer to this question

I am having trouble with this one...

Prove that any positive integer whose ALL digits are 1s (except 1) is not a perfect square.

Tn=a+(n-1)d

a=11

...n
d=10
......1
T1=11+(1-1)10 = 11

......2
T2=11+(2-1)10 = 111

......3
T3=11+(3-1)10 = 111
.
.
.
.
.

Tn=a+(n-1)d
.......2
To be a perfect square = x = Tn

11 is not a squared number and it is an odd number so (a=11) is not a squared number and it is an odd number.
...............n
So (n-1)d is always an even number because ( d=10 ) is always an even number So (n-1)d is always an even number

So if you add an even number with an odd number the answer is always an
odd number. So Tn is always an odd number. So an odd number can never be a square.

Even though this could be obvious once you read the question this is the way to prove that all the positive integers where digits are 1s (except 1) is not a perfect square.

 

[Gokul43201]

Supundika,

This is NOT the way to prove this. You proof is fatally flawed.

So if you add an even number with an odd number the answer is always an
odd number. So Tn is always an odd number. So an odd number can never be a square.

An odd number can never be a square ?? How about 1 or 121 ?

Also, the very basis of your argument is incorrect. First of all, d is really d^n, but even that doesn't fix it.

T(3) = 11 + 2*1000 = 2011, definitely not 1111 .

So you start off wrong, and even if you didn't, all you would be proving is that 111..1 is an odd number : a rather convoluted way to prove something that simple.

...this is the way to prove that all the positive integers where digits are 1s (except 1) is not a perfect square.

I suggest you be more sure of your math before you go about proclaiming the superiority of your faulty proof over others previously provided.

 

[Leong]

Gokul43201,
Your skill in mathematics is really impressive;it is a fact whether i envy it or not. it is good to point out other's mistakes but i am sorry i have to say that your criticism about Supundika is not nice at all.

 

[Gokul43201]

Yes Leong, you're right - I shouldn't have said that. I take back what I said, and apologize.

Nevertheless Supundika, I strongly suggest you be sure of your math before you post, lest you mislead people that come here to get doubts cleared.

 

[grabateetrap]

does an infinite string have an end?

 

[HallsofIvy]

does an infinite string have an end?

What does tihis have to do with whether 111... 111 is a perfect square?
If you have a completely different question, start a new thread- don't "hijack" one that already exists for a different question.


The answer to your question is "No, by the definitonof "infinite string".

 

[Ethereal]

Now for the units' digit of x^2 to be 1, the units digit of x must be 1 or 9.

Is this proven by noting the square of any any single digit no. would yield a 1 in the units' place only if it were 9 or 1? Is there a proof for this, or is this self-evident?

 

[matt grime]

there is a proof and it is self evident too, if you're used to modulo arithmetic, since you only need to consider the numbers 1 to 9, square them and look at the units column. why? well, it is obvious if you've done some ring theory but if not we can explain it as:

let x = 10q+r, where 0<=r<9

x^2 = 100q^2+20rq + r^2

and neither of the terms involving q can have any contribution to the units column.

 

[Ethereal]

Actually I asked the question because I was wondering how rigorous maths should be, and to what extent intuitive proof would suffice. What is ring theory?

 

[matt grime]

Intuition may help you decide what is true and why, but does not itself constitute a proof.

Rings are abstract generalizations of the integers; they are sets that have two operations addition and multiplication that obey certain rules. A very common example, almost the first one students meet that isnt' the integers, is remainder arithmetic, or modular arithmetic.

 

[mathwonk]

matt's argument is similar to yours futbol. he notes that every number of form 111111...1,
when divided by 10, has remainder of 1. hence if it is a square of some number A then since the remander of the square is the square of the remainder, the number A must have remainder 1 or -1 when divided by 10. so he checks both cases, and neither works.

i wrote this before noting there was a second page of comments.

 

[geraldmcgarvey]

Here's a little more regarding Matt Grime's proof:

"let x = 10q+r, where 0<=r<9

x^2 = 100q^2+20rq + r^2"

and as he noted, r must be 1 or 9

If it is 1 there is no carry, so the ten's value will be 2rq, which is even and not 1.
If it is 9 then r^2 = 81, so carry the 8. But then 8 + 2rq needs to end in 1, but is even.

 

[mikeyflex]

Follow up

Hi, I'm new to the forum and came across this thread, I'm aware that the topic is a few years old. I have an interesting question which seems to be an offspring of the topic at hand.

If p is a prime number greater than 3, evaluate p^2(mod 12)

I know this is closely tied to the arthmetic modular or the so-called clock arthemtic. How do you go about proving this. When you work out the numbers every prime square mod 12 beomes 1. Therefore, this is an equivlanence relationship. What I found most interesting is when I tried the primes less than 3 (2, 3) they were not equal to 1. Any thoughts and comments would greatly be appreciated.

Mike

 

[Dick]

If you work out p(mod 12) for p a prime, what are the possibilities? Excepting 2 and 3. What does this imply for p^2(mod 12)?

 

[Edgardo]

Hi all,

I think I have found another proof on futbol's problem. I'm grateful if anyone could check it:

Proof by contradiction

Suppose there exists an integer x such that x^2 = 111...11.
x must be an odd integer because an even x can't have the ending digit 1.
Let's write x in the form x=2k+1 where k is an integer number.

Thus,
x^2 = 111...11
=> (2k+1)^2 = 111...11
=> 4k^2 + 4k + 1 = 111...11
=> 4k^2 + 4k = 111...10
=> 4k^2 + 4k = 111...1 * 10
=> 4k(k+1) = 111...1 * 10
=> 2 k(k+1) = 111...1 *5
=> 2 k(k+1) = 555...5

The left hand side is an even number and the right hand side an odd number
which is a contradiction. Therefore there exists no integer x such that
x^2 = 111...11.

 

[mikeyflex]

If you work out p(mod 12) for p a prime, what are the possibilities? Excepting 2 and 3. What does this imply for p^2(mod 12)?

Hi Dick,

The problem stated that p is a prime number that is greater than 3. When you work out p^2(mod 12) you get a remainder of 1 for every prime number square that is greater than 3. This is not the case with 2 or 3. I was wondering if there was some proof behind this.

 

[robert Ihnot]

A more advanced problem deals with when a REPUNIT is prime. This happens for 2, which is 11. Now since we can use the form \frac{10^n-1}{9}, it is clear that for n=ab, that 10^a-1 is a factor. (And if n is greater than 1, this divides out more than just the 9.) So that looking for primes we only need consider repunits of prime length 2,3,5,7...etc.

What is the next prime repunit after 11? Well, my TI-89 will handle the case for n=17 and finds it not prime with smallest factor of 2071723. Checking with Wolfram under repunits, I see that n=19 is the smallest case after n=2, which gives us a prime.

Of course, the whole matter can be generalized and examined for bases other than 10, as Wolfram indicates. A Mersenne number is the famous case of base 2.

 

[matt grime]

For mikeyflex: every prime not equal to 2 or 3 is of the form 6m+1 or 6m-1 for some integer m. This completely explains your observation.

 

[Dick]

Hi Dick,

The problem stated that p is a prime number that is greater than 3. When you work out p^2(mod 12) you get a remainder of 1 for every prime number square that is greater than 3. This is not the case with 2 or 3. I was wondering if there was some proof behind this.

A direct way to see this is that p(mod 12) can only be 1,5,7 or 11. Otherwise it is divisible by 2 or 3. But 1^2, 5^2, 7^2 and 11^2 are all equal to 1 mod 12. This is what matt was alluding to when he said all primes are equal to +1 or -1 mod 6.

 

[mikeyflex]

For mikeyflex: every prime not equal to 2 or 3 is of the form 6m+1 or 6m-1 for some integer m. This completely explains your observation.

Thanks Matt,

The problem stated that when p is a prime greater than 3, to evaluate p^2 mod 12.

When you plug in prime values for p^2 mod 12 you get 1, which is an equivalence relation. Is there any significance to this problem as pertaining to a proof of sorts?

Thanks again Matt

 

[matt grime]

Did you actually read what I wrote and think about it for any length of time. Your observation is completely proven by what I wrote.

 

[matticus]

to explain what matt's saying much farther than what needs to be done:
every prime greater than 3 can be written as 6k +1 or 6k -1 (which can be shown easily by a proof by cases).

(6k+1)^2 = 36k^2 + 12k + 1 = 12(3k^2 + k) + 1 = 1mod12
(6k - 1)^2 = 36k^2 -12k +1 = 12(3k^2 - k) + 1 = 1mod12

that's really all the "signifigance" involved in the observation

 

[mikeyflex]

to explain what matt's saying much farther than what needs to be done:
every prime greater than 3 can be written as 6k +1 or 6k -1 (which can be shown easily by a proof by cases).

(6k+1)^2 = 36k^2 + 12k + 1 = 12(3k^2 + k) + 1 = 1mod12
(6k - 1)^2 = 36k^2 -12k +1 = 12(3k^2 - k) + 1 = 1mod12

that's really all the "signifigance" involved in the observation

Ahh, now I get it, and that must be the reason why this proof doesn't work for prime numbers 2 and 3. Thank you for the observation. Number theory never seems light the bulb for me.