Difference between Two Squares equal to Repdigit

From ProofWiki
Jump to navigation Jump to search

Theorem

Some differences of two squares that each make a repdigit number include:

\(\ds 6^2 - 5^2\) \(=\) \(\ds 11\)
\(\ds 56^2 - 45^2\) \(=\) \(\ds 1111\)
\(\ds 556^2 - 445^2\) \(=\) \(\ds 111 \, 111\)
\(\ds \) \(:\) \(\ds \)


\(\ds 7^2 - 4^2\) \(=\) \(\ds 33\)
\(\ds 67^2 - 34^2\) \(=\) \(\ds 3333\)
\(\ds 667^2 - 334^2\) \(=\) \(\ds 333 \, 333\)
\(\ds \) \(:\) \(\ds \)


\(\ds 8^2 - 3^2\) \(=\) \(\ds 55\)
\(\ds 78^2 - 23^2\) \(=\) \(\ds 5555\)
\(\ds 778^2 - 223^2\) \(=\) \(\ds 555 \, 555\)
\(\ds \) \(:\) \(\ds \)


\(\ds 9^2 - 2^2\) \(=\) \(\ds 77\)
\(\ds 89^2 - 12^2\) \(=\) \(\ds 7777\)
\(\ds 889^2 - 112^2\) \(=\) \(\ds 777 \, 777\)
\(\ds \) \(:\) \(\ds \)


Proof

Let $a, b$ be integers with $1 \le b < a \le 8$ and $a + b = 9$.

Then:

\(\ds \paren {1 + \sum_{k \mathop = 0}^n a 10^k}^2 - \paren {1 + \sum_{k \mathop = 0}^n b 10^k}^2\) \(=\) \(\ds \paren {1 + \sum_{k \mathop = 0}^n a 10^k - 1 - \sum_{k \mathop = 0}^n b 10^k} \paren {1 + \sum_{k \mathop = 0}^n a 10^k + 1 + \sum_{k \mathop = 0}^n b 10^k}\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \paren {\sum_{k \mathop = 0}^n \paren {a - b} 10^k} \paren {2 + \sum_{k \mathop = 0}^n 9 \times 10^k}\) $a + b = 9$
\(\ds \) \(=\) \(\ds \paren {\sum_{k \mathop = 0}^n \paren {a - b} 10^k} \paren {1 + 10^{n + 1} }\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n \paren {a - b} 10^k + \sum_{k \mathop = 0}^n \paren {a - b} 10^{k + n + 1}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n \paren {a - b} 10^k + \sum_{k \mathop = n + 1}^{2 n + 1} \paren {a - b} 10^k\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^{2 n + 1} \paren {a - b} 10^k\)

which is a repdigit number.


The examples above are instances with $\tuple {a, b} = \tuple {5, 4}, \tuple {6, 3}, \tuple {7, 2}, \tuple {8, 1}$.

$\blacksquare$


Sources