Odd Digits of Square Numbers

Date: 02/07/99 at 12:42:55
From: Kevin Bryceland
Subject: Odd square numbers

Dear Dr. Math,

I cannot work out why there are no square numbers other than 1 and 9 
that consist entirely of odd digits. Please help me.

Yours thankfully, 
Kevin Bryceland

Date: 02/08/99 at 17:30:48
From: Doctor Nick
Subject: Re: Odd square numbers

Hi Kevin -

The reason that these are the only two squares that have only odd 
digits is that a square number with more than 1 digit will have at 
least one even digit in the first two (that is, the right two) places. 

If n is even, then the right-most digit of n^2 is even.  

If n is odd, then the right-most digit of n^2 is odd, but the next
digit is always even! 

A simple way to prove this is to use the fact that the right two digits 
of n^2 depend entirely on the right two digits of n. That means that we 
only have to check the squares of 1, 3, 5, 7, 9, 11, ..., 99 to see 
that they have an even number in the second place. You can cut the 
effort in half by noting that 1^2 and 99^2 end in the same two digits, 
3^2 and 97^2 end in the same two digits, and so on, so you really only 
need to check the squares of 1, 3, 5, 7, ..., 49, which isn't too bad.

In fact, you can cut it down even further. Notice that

  (50-m)^2 = 50^2 - 100m + m^2

so if 0 < m < 50, then m^2 and (50-m)^2 end in the same two digits. 
That means you only have to check 1, 3, 5, 7, ..., 25.

Notice that the fact that we're using base 10 is very important here. 
In other bases there could be squares that have only odd digits.

Have fun,

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

Date: 04/01/01 at 13:17:04
From: Doctor Schwa
Subject: Re: Odd square numbers

To continue Dr. Nick's excellent response, you can cut down the checking 
even more by looking at

   (10a + b)^2 = 100a^2 + 20ab + b^2

The first term, 100a^2, doesn't affect the tens place at all; 20ab only 
changes the tens place by an even number, so you only need to check the 
squares of the one-digit numbers to verify that the tens place or the ones 
place is always even.

As Dr. Nick already pointed out, if the ones place (b) is even, then b^2
will end with an even digit as well.

So now there are only five numbers that need to be checked.

--Dr. Schwa
  http://mathforum.org/dr.math/