Square of Small-Digit Palindromic Number is Palindromic

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Theorem

Let $n$ be an integer such that the sum of the squares of the digits of $n$ in decimal representation is less than $10$.


Let $n$ be palindromic.


Then $n^2$ is also palindromic.


The sequence of such numbers begins:

$0, 1, 2, 3, 11, 22, 101, 111, 121, 202, 212, 1001, 1111, \dots$

This sequence is A057135 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Proof

Let $\ds n = \sum_{k \mathop = 0}^m a_k 10^k$ be a number satisfying the conditions above.

Then:

\(\text {(1)}: \quad\) \(\ds \sum_{k \mathop = 0}^m a_k^2\) \(<\) \(\ds 10\)
\(\text {(2)}: \quad\) \(\ds a_k\) \(=\) \(\ds a_{m - k}\) \(\ds \forall k: 0 \le k \le m\)


Consider $\ds n^2 = \paren {\sum_{k \mathop = 0}^m a_k 10^k}^2$.

From definition of Multiplication of Polynomials, the coefficient of $10^l$ in the product is:

\(\ds \) \(\) \(\ds \sum_{\substack {j \mathop + k \mathop = l \\ j, k \mathop \in \Z} } a_j a_k\)
\(\ds \) \(\le\) \(\ds \sqrt {\paren {\sum_{j \mathop = 0}^l a_j^2} \paren {\sum_{k \mathop = 0}^l a_k^2} }\) Cauchy's Inequality
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 0}^l a_j^2\)
\(\ds \) \(\le\) \(\ds \sum_{j \mathop = 0}^m a_j^2\)
\(\ds \) \(<\) \(\ds 10\) From (1)

so no carries occur in the multiplication, and this form satisfies Basis Representation Theorem.


Moreover:

\(\ds \sum_{\substack {j \mathop + k \mathop = 2 m \mathop - l \\ j, k \mathop \in \Z} } a_j a_k\) \(=\) \(\ds \sum_{\substack {m \mathop - j \mathop + m \mathop - k \mathop = 2 m \mathop - l \\ m \mathop - j, m \mathop - k \mathop \in \Z} } a_{m - j} a_{m - k}\)
\(\ds \) \(=\) \(\ds \sum_{\substack {j \mathop + k \mathop = l \\ j, k \mathop \in \Z} } a_{m - j} a_{m - k}\)
\(\ds \) \(=\) \(\ds \sum_{\substack {j \mathop + k \mathop = l \\ j, k \mathop \in \Z} } a_j a_k\) From (2)

so the coefficient of $10^{2 m - l}$ is equal to the coefficient of $10^l$ in the expansion of $n^2$.

This shows that $n^2$ is palindromic.

$\blacksquare$


Examples

Square of $11$

$11^2 = 121$


Square of $22$

$22^2 = 484$


Square of $111$

$111^2 = 12321$


Square of $121$

$121^2 = 14641$


Square of $212$

$212^2 = 44944$


Square of $1111$

$1111^2 = 1234321$


Square of $11 \, 111$

$11 \, 111^2 = 123 \, 454 \, 321$


Square of $111 \, 111$

$111 \, 111^2 = 12 \, 345 \, 654 \, 321$


Sources

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