Answer :
The other number
that answers all the requirements
of the puzzle is 9,801.
If we divide this
in the middle into two numbers
and add them together we get 99, which, multiplied by itself, produces
9,801.
It is true that 2,025 may be treated in the same way, only this
number is excluded by the condition which requires that no two figures
should be alike.
The general
solution is curious. Call the number
of figures in each half of the torn label n.
Then, if we add 1 to each
of the exponents of the prime factors (other than 3) of 10^{n}  1
(1 being regarded as a factor with the constant exponent, 1), their
product will be the number of solutions.
Thus, for a label of six
figures, n = 3. The factors of 10^{n}  1
are 1^{1} × 37^{1}
(not considering the 3^{3}),
and the product of
2 × 2 = 4, the number of
solutions.
This always includes the special cases
98  01, 00  01,
998  01, 000  001, etc.
The
solutions are obtained as follows:—Factorize 10^{3}  1
in all possible ways, always keeping the powers of 3 together, thus,
37 × 27,
999 × 1. Then solve the equation
37x = 27y + 1.
Here
x = 19 and y = 26.
Therefore,
19 × 37 = 703, the square
of which gives one label, 494,209.
A complementary
solution (through
27x = 37x + 1) can at once be found
by 10^{n}  703 = 297,
the square of which gives 088,209 for second label.
(These
nonsignificant noughts to the left must be included, though they lead
to peculiar cases like 00238  04641 = 4879^{2},
where 0238  4641 would not work.)
The special case
999 × 1 we can write at once 998,001,
according to the law shown above, by adding nines on one half and
noughts on the other, and its complementary will be 1 preceded by five
noughts, or 000001.
Thus we get the squares of 999 and 1. These are the
four solutions.
