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

supundika

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.

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.

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

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

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

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

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 [tex]\frac{10^n-1}{9}[/tex], 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.