# Square of Small-Digit Palindromic Number is Palindromic

## 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:

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

## 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 $111 \, 111$

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