Is there an 8-digit square that is still a square when a 1 is written in front of it?

There are two solutions to the first puzzle:

        4875^2 = 23765625; 100000000 + 23765625 = 123765625 = 11125^2
        7500^2 = 56250000; 100000000 + 56250000 = 156250000 = 12500^2

And one solution to puzzle number 2:

        8658^2 = 74960964; 74960964 * 10 + 1 = 749609641 = 27379^2

-- Dave Dodson <dodson@convex.COM>

[ Editor: Timothy Vaughan <tvaughan@buphyk.bu.edu> and Dror Lubin
  <lubin@fy.chalmers.se> posted the same answers.  The problem was posed
  by Harm Derksen <hderksen@sci.kun.nl>. ]

[ Editor: further material on this puzzle, edited separately from other collected postings (batch 18); it may repeat some of the above. ]

[ Editor: problem as stated there: ]
Find a 200-digit square, not ending in 0, which is still a square when
a 1 is put in front of it.

Also, consider a square S which is still a square if you put an extra 1
at the end, instead of at the beginning.  Example of 78 digits:

235540881893069525243891342785898289629985519673634907026646425473868923222500
and
2355408818930695252438913427858982896299855196736349070266464254738689232225001

are both squares.  How many of these squares 0 <= S <= googol are there?

[ Editor: problem statement reconstructed from the postings. ]


In article J8z@sci.kun.nl, hoeij@sci.kun.nl (Mark van Hoeij) writes:
>For 200 digits there is a solution which does not end on a zero.
>Can you find it?

The numbers are:

[(5^143) - (2^198)*(5^57)]^2  and  [(5^143) + (2^198)*(5^57)]^2

The difference is 10^200, they are odd, therefore do not end in 0, and have 200
digits.

-- Dror Lubin <lubin@fy.chalmers.se>

Well done! I have a different puzzle: consider a square S
which is still a square if you put an extra 1 at the end, instead of at the
beginning.

[...]

Now the question is: How many of these squares 0<= S <= googol are there?

-- Mark van Hoeij <hoeij@sci.kun.nl>

[ Editor: no answer to the second question was found in the collected
postings. ]