Palindromic Squares
Do you know any numbers besides 14641 where both the number and its square root read the same left to right as right to left?
A number whose digits read the same forward or backward is called a
palindrome, or a palindromic number. You are asking about palindromes
whose squares are also palindromes.
This is a very interesting problem. There are other solutions - in
fact, there are infinitely many. For example:
1^1 = 1,
2^2 = 4,
3^2 = 9.
Then there are this kind:
11^2 = 121,
101^2 = 10201,
1001^2 = 1002001,
10001^2 = 100020001, etc.,
22^2 = 484,
202^2 = 40804,
2002^2 = 4008004,
20002^2 = 400080004, etc.
By the way, this shows that there are infinitely many. Then there
are:
111^2 = 12321,
1111^2 = 1234321,
11111^2 = 123454321,
111111^2 = 12345654321,
1111111^2 = 1234567654321,
11111111^2 = 123456787654321,
111111111^2 = 12345678987654321,
but this is where this sequence ends.
Then there are the kind like
1111000001111^2 = 1234321002468642001234321,
where the number of ones can be from 1 to 4 and any number of zeroes.
Finally,
212^2 = 44944.
This is all there are with the root less than 1000.
You can prove that there are no roots starting with 4, 5, 6, 7, 8, or
9, in the following way. If the root began with 7, it would end with
7. Its square would end with 9, so would begin with 9. If a square
begins with 9, its root must have initial digits between Sqrt[9] = 3
and Sqrt[10] = 3.162278 (and so begin with 3), or between Sqrt[90] =
9.486833 and Sqrt[100] = 10 (and so begin with 9), which are both
impossible since it begins with 7.
Similarly, a root beginning with 6 would imply that it began with 2,
7, or 8, all impossible. A root beginning with 5 would imply that it
began with 2 or 7, both impossible.
A root beginning with 4 would imply that it began with 2, 7, or 8.
A root beginning with 8 would imply that it began with 2, 6, or 7.
A root beginning with 9 would imply that it began with 1, 3, or 4.
A more complicated case: A root beginning with 3 would imply the root
ending in 3, and hence the square ending in 9, and hence beginning
with 9. Squares beginning with 9 must have roots which begin with 9
(impossible) or 3 (possible). Those squares are in the range 9*100^n
to 10*100^n, for some n >= 0, so the root is in the range 3*10^n to
Sqrt[10]*10^n = 3.162278*10^n, and must end in 3. For each n there
are possibilities. For n = 0, you have 3 alone, which works (see
above). For n = 1, there are none. For n = 2, 303 and 313 have to be
checked, but neither works. For n = 3, 3003 and 3113 have to be
checked, and neither works. For n = 4, 30003, 30103, 30203, 30303,
30403, 30503, 30603, 30703, 30803, 30903, 31013, 31113, 31213, 31313,
31413, 31513, and 31613 have to be checked. The first ten all have
squares ending in 09, so beginning with 90, so roots less than
Sqrt[910000000] = 30166.21, so only the first two are possible, and
neither works. The last seven all have squares ending in 69, so
beginning with 96, so roots between Sqrt[960000000] = 30983.87 and
Sqrt[970000000] = 31144.82, so only the first two are possible, and
neither works. For n = 5, a similar argument means that only four
possible roots have to be checked: 300003, 301103, 310013, and
311113; and none of them work. This exhausts the possible roots up to
1000000 beginning with 3.
The cases where the root begins with 2 or 1 are still more
complicated. You can try to work out the details. Again you will be
considering two-digit endings for squares, three-digit endings for
squares, and so on.
-Doctor Rob, The Math Forum
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