Square Numbers, All Digits the Same

Date: 07/10/2003 at 04:47:07
From: Elena
Subject: About square numbers

Is there any square number with all the same digits?
Like 111, 222, etc.

Date: 07/11/2003 at 09:56:22
From: Doctor Rob
Subject: Re: About square numbers

Thanks for writing to Ask Dr. Math, Elena.

Aside from the one-digit squares 0, 1, 4, and 9, there are none.  The
proof goes like this.

Any square number must end in 0, 1, 4, 5, 6, or 9.  0, 1, 4, and 9 are 
squares, but 5 and 6 are not. Now assume that the number has at least 
two digits. The possible two-digit endings are 00, X1, X4, 25, Y6, or 
X9, where X is an even digit and Y is an odd digit. That eliminates 
all but 00 and 44 as same-digit two-digit endings (since X cannot be 1 
or 9 and Y cannot be 6). 00 and 44 are not squares. Now assume that 
the number has at least three digits. Then the possible three-digit 
endings are 000 and Y44, where Y is an odd digit. That leaves only 
00...000 as a possibility, which is also impossible.

I leave it to you to figure out why what I said above is true.

Feel free to write again if I can help further.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 06/17/2003 at 02:28:26
From: Lucy
Subject: Square Numbers

A positive integer N is a square number and all its digits are the 
same. Find all such numbers N.

The most difficult part for me is how to prove my answers. I'm pretty 
sure the only possible answers are 1, 4 and 9. But these answers only 
contain 1 digit and therefore every digit is of course the same. If 
these answers are right, but how do I prove there are no more answers?

Date: 06/17/2003 at 03:58:51
From: Doctor Jacques
Subject: Re: Square Numbers

Hi Lucy,

There are indeed no other possible answers.

To see this, note first that the last digit of the square of a number 
only depends on the last digit of the number itself. For example:

  7^2 =   49
 17^2 =  289
 37^2 = 1369

and so on. The square of a number that ends in 7 always ends in 9. 
This is because, if

  n^2 = A

then

 (10+n)^2 = 100 + 20n + n^2
          = 10(10+2n) + A

i.e. when you add 10 to a number, you add a multiple of 10 to its 
square, which means that you don't change the last digit.

In the same way, the last two digits of the square of a number only 
depend on the last two digits of the number, for example:

  14^2 =   196
 314^2 = 98596

and so on.. the square of a number that ends in 14 always ends in 96.

You can read more details about this in the Dr. Math archives:

   Perfect Squares with Congruences
   http://mathforum.org/library/drmath/view/56081.html 

Let us now assume that there is a square number, N = a^2, with more 
than one digit, and that all its digits are the same. In particular, 
the last two digits will be the same.

If you look at the last two digits of the squares of all the numbers 
up to 100 (actually, you only need to check up to 50), you will find 
that the only possibilities for the last two digits are those given 
above. In particular, the only case when the last two digits are the 
same is when these digits are 00 or 44.

If all the digits of N are the same, it means that N consists only of 
4's:

  N = a^2 = 44...44

As this is an even number, "a" itself must be even (if "a" were odd, 
its square would be odd also). So we may write a = 2b, and we have:

  N = (2b)^2 = 44...44
    4*b^2 = 44...44

and, if we divide by 4, we get:

  b^2 = 11...11

which means that b^2 is a square number that ends in 11, and we have 
seen that this is not possible. This shows that there cannot be a 
square number with at least two digits where all digits are the same.

Does this help?  Write back if you'd like to talk about this 
some more, or if you have any other questions.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 06/18/2003 at 01:37:54
From: Lucy
Subject: Thank you (Square Numbers)

Thanks a lot for your time and effort! It has helped me greatly.