Difference between Two Squares equal to Repunit/Corollary 2
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Theorem
The sequence of differences of two squares that each make a repunit begins:
| \(\ds 6^2 - 5^2\) | \(=\) | \(\ds 11\) | ||||||||||||
| \(\ds 56^2 - 45^2\) | \(=\) | \(\ds 1111\) | ||||||||||||
| \(\ds 5056^2 - 5045^2\) | \(=\) | \(\ds 111 \, 111\) | ||||||||||||
| \(\ds \) | \(:\) | \(\ds \) |
and in general for integer $n$:
- $R_{2 n} = {\underbrace{5050 \ldots 56}_{n - 1 \ 5 \text{'s} } }^2 - {\underbrace{5050 \ldots 45}_{n - 1 \ 5 \text{'s} } }^2$
that is:
- $\ds \sum_{k \mathop = 0}^{2 n - 1} 10^k = \left({\sum_{k \mathop = 1}^{n - 1} 5 \times 10^{2 k - 1} + 6}\right)^2 - \left({\sum_{k \mathop = 1}^{n - 1} 5 \times 10^{2 k - 1} - 5}\right)^2$
Proof
From Difference between Two Squares equal to Repunit, $R_{2n} = x^2 - y^2$ exactly when $R_{2n} = a b$ where $x = \dfrac {a + b} 2$ and $y = \dfrac {a - b} 2$.
By the Basis Representation Theorem
| \(\ds R_{2n}\) | \(=\) | \(\ds \sum_{0 \mathop \le k \mathop < 2 n} 10^k\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{\substack {0 \mathop \le k \mathop < 2 n \\ k \text { even} } } 10^k + \sum_{\substack {0 \mathop \le k \mathop < 2 n \\ k \text { odd} } } 10^k\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{\substack {0 \mathop \le k \mathop < 2 n \\ k \text { even} } } 10^k + 10 \times \sum_{\substack {0 \mathop \le k \mathop < 2 n \\ k \text { even} } } 10^k\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^{n - 1} 10^{2 k} + 10 \times \sum_{k \mathop = 0}^n 10^{2 k}\) | change of indices | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^{n - 1} 11 \times 10^{2 k}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 11 \sum_{k \mathop = 0}^{n - 1} 10^{2 k}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 11 \times \underbrace {10101 \ldots 01}_{n \ 1 \text{'s} }\) |
Thus, let:
- $a = \ds \sum_{k \mathop = 0}^{n - 1} \times 10^{2 k}$
- $b = 11$
So:
| \(\ds a + b\) | \(=\) | \(\ds \sum_{k \mathop = 0}^{n - 1} 10^{2 k} + 11\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^{n - 1} 10^{2 k} + 12\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \dfrac {a + b} 2\) | \(=\) | \(\ds \sum_{k \mathop = 1}^{n - 1} \frac {10^{2 k} } 2 + 6\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^{n - 1} 5 \times 10^{2 k - 1} + 6\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \underbrace {5050 \ldots 50}_{n - 1 \ 5 \text{'s} } + 6\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \underbrace {5050 \ldots 56}_{n - 1 \ 5 \text{'s} }\) |
Similarly:
| \(\ds a - b\) | \(=\) | \(\ds \sum_{k \mathop = 0}^{n - 1} \times 10^{2 k} - 11\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^{n - 1} 10^{2 k} - 10\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \dfrac {a - b} 2\) | \(=\) | \(\ds \sum_{k \mathop = 1}^{n - 1} \frac {10^{2 k} } 2 - 5\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^{n - 1} 5 \times 10^{2 k - 1} - 5\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \underbrace {5050 \ldots 50}_{n - 1 \ 5 \text{'s} } - 5\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \underbrace {5050 \ldots 45}_{n - 1 \ 5 \text{'s} }\) |
Hence the result.
$\blacksquare$