Yes.

Any odd number can be expressed as 10`a` + 2`b` + 1,

- where
`b`is 0, 1, 2, 3, or 4 - where
`a`is any whole number

(10`a` + 2`b` + 1)^{2}

= 100`a`^{2} + 20`a``b` + 10`a` + 20`a``b` + 4`b`^{2} + 2`b` + 10`a` + 2`b` + 1

= 100`a`^{2} + 40`a``b` + 20`a` + 4`b`^{2} + 4`b` + 1

100`a`^{2} + 40`a``b` + 20`a` always has an even second-to-last digit and a zero last digit, therefore 100`a`^{2} + 40`a``b` + 20`a` + 4`b`^{2} + 4`b` + 1 has an even second-to-last digit if 4`b`^{2} + 4`b` + 1 has an even second-to-last digit.

- If
`b`= 0 then 4`b`^{2}+ 4`b`+ 1 = 01 - If
`b`= 1 then 4`b`^{2}+ 4`b`+ 1 = 09 - If
`b`= 2 then 4`b`^{2}+ 4`b`+ 1 = 25 - If
`b`= 3 then 4`b`^{2}+ 4`b`+ 1 = 49 - If
`b`= 4 then 4`b`^{2}+ 4`b`+ 1 = 81

Since for all possible values of `b` there is an even second-to-last digit, for all odd numbers, the square has an even second-to-last digit in base 10.

Any odd number in an even base can be expressed as `c``a` + 2`b` + 1,

- where
`c`is even - where
`b`is any whole number 0 thru`c`/ 2 - 1 - where
`a`is any whole number

(`c``a` + 2`b` + 1)^{2}

= `c`^{2}`a`^{2} + 2`c``a``b` + `c``a` + 2`c``a``b` + 4`b`^{2} + 2`b` + `c``a` + 2`b` + 1

= `c`^{2}`a`^{2} + 4`c``a``b` + 2`c``a` + 4`b`^{2} + 4`b` + 1

`c`^{2}`a`^{2} + 4`c``a``b` + 2`c``a` always has an even second-to-last digit and a zero last digit, therefore `c`^{2}`a`^{2} + 4`c``a``b` + 2`c``a` + 4`b`^{2} + 4`b` + 1 has an even second-to-last digit if 4`b`^{2} + 4`b` + 1 has an even second-to-last digit.

If `c` = 2 then

- If
`b`= 0 then 4`b`^{2}+ 4`b`+ 1 = 01_{2}

If `c` = 4 then

- If
`b`= 0 then 4`b`^{2}+ 4`b`+ 1 = 01_{4} - If
`b`= 1 then 4`b`^{2}+ 4`b`+ 1 = 21_{4}

If `c` = 6 then

- If
`b`= 0 then 4`b`^{2}+ 4`b`+ 1 = 01_{6} - If
`b`= 1 then 4`b`^{2}+ 4`b`+ 1 = 13_{6} - If
`b`= 2 then 4`b`^{2}+ 4`b`+ 1 = 41_{6}

If `c` = 8 then

- If
`b`= 0 then 4`b`^{2}+ 4`b`+ 1 = 01_{8} - If
`b`= 1 then 4`b`^{2}+ 4`b`+ 1 = 11_{8} - If
`b`= 2 then 4`b`^{2}+ 4`b`+ 1 = 31_{8} - If
`b`= 3 then 4`b`^{2}+ 4`b`+ 1 = 61_{8}

If `c`=12 then

- If
`b`= 0 then 4`b`^{2}+ 4`b`+ 1 = 01_{12} - If
`b`= 1 then 4`b`^{2}+ 4`b`+ 1 = 09_{12} - If
`b`= 2 then 4`b`^{2}+ 4`b`+ 1 = 21_{12} - If
`b`= 3 then 4`b`^{2}+ 4`b`+ 1 = 41_{12} - If
`b`= 4 then 4`b`^{2}+ 4`b`+ 1 = 69_{12} - If
`b`= 5 then 4`b`^{2}+ 4`b`+ 1 = A1_{12}

And so on…

Since for all possible values of `b` in bases 2, 4, and 12 there is an even second-to-last digit, for all odd numbers in these bases, the square has an even second-to-last digit in these bases.

Thus, the answer to the third question is "It is the same in some, but not all".

From the answers to the first and third questions we see that it is the same for bases 2, 4, 10, and 12, and not for bases 6 and 8.

In base 3, 3_{10}^{2} = 1211_{3} = 14_{5} = 12_{7} = 10_{9} = 23_{11}

Therefore, it is not the same in bases 3, 5, 7, 9, and 11.

Consider that ( `x` + 2 )^{2} < 2`x`^{2} for `x` = 5

(`x` + 2)^{2} < 2`x`^{2}

(`x` + 2)^{2} + 2`x` < 2`x`^{2} + 3`x` + 2

(`x` + 2)^{2} + 2`x` + 5 < 2`x`^{2} + 3`x` + 2 + `x`

`x`^{2} + 4`x` + 4 + 2`x` + 5 < 2`x`^{2} + 4`x` + 2

`x`^{2} + 6`x` + 9 < 2(`x`^{2} + 2`x` + 1)

(`x` + 3)^{2} < 2(`x` + 1)^{2}

[(`x` + 1) + 2]^{2} < 2(`x` + 1)^{2}

Thus, for all odd whole numbers `x` equals 5 or greater, the square of the next odd number is less than half the square of the current odd number. This means that if there is no second-to-last digit in a square of an odd number, then the second-to-last digit of the square of the next odd number can be no larger than 1 in any base greater than 12.

Therefore, bases 2, 4, 10, and 12 are the only bases for which the squares of odd numbers always have an even second-to-last digit.

Q. E. D.

Last Updated: 2009-05-02

The author, Marq Thompson, wished the content of this website to be uncopyrighted after his death.