last digit y  20xy+y^2  value(s) of x that make last 2 digits of 20xy+y^2 the same
 ---------------------------------------------------------
      0            0        any (trivial case -- the square of any integer with last digit 0 has last 2 digits 00)
      1         20x+1       none
      2         40x+4       1 or 6 (last 2 digits "44")
      3         60x+9       none
      4         80x+16      none
      5        100x+25      none
      6        120x+36      none
      7        140x+49      none
      8        160x+64      3 or 8 (last 2 digits "44")
      9        180x+81      none

Let m = the number formed by the last two (like) digits only.
Let n be the number formed by k^2 with the last two digits chopped off.
Then k^2 = 100n+m, where m = 00,11,22,33,44,55,66,77,88, or 99

Since 100n is divisible by 4, then the remainder when k^2 is divided by 4
is the same as the remainder when m is divided by 4.

k^2 must end with same last digit as the last digit of the square of the last
digit of k.

0^2=0, 1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64, 9^2=81

So we can rule out m as 22,33,77,88

So m is one of these:  00,11,44,55,66,99

If m and k^2 are even, then m is one of these: 00, 44, or 66.  But an even
number is divisible by 4, but 66 isn't divisible by 4, so 66 is ruled out as a
possibility for m.

That leaves only 00,11,44,55,99

If m and k^2 are odd, then m is 11, 55 or 99.  Since k^2 is odd, k is also odd
and k = 2q+1 for some nonnegative integer q.

Then k^2 = (2q+1)^2 = 4q^2+4q+1 = 4(q^2+q)+1 

so when k^2 or m are divided by 4, the remainder must be 1. However, when 11 is
divided by 4, the quotient is 2 and the remainder is 3.  When 55 is divided by
4, the quotient is 13 and the remainder is also 3. When 99 is divided by 4, the
quotient is 24 and the remainder is also 3.  But the remainder must be 1, not 3.

So 11,33,99 are ruled out.

Therefore, m = 00 or 44.  So the last digit of k^2 is 0 or 4 and so the last
digit of k is 0, 2, or 8.

Edwin