Difference between Two Squares equal to Repunit/Corollary 1

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Theorem

The sequence of differences of two squares that each make a repunit begins:

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

and in general for integer $n$:

$R_{2 n} = {\underbrace {55 \ldots 56}_{\text {$n - 1$ $5$'s} } }^2 - {\underbrace {44 \ldots 45}_{\text {$n - 1$ $4$'s} } }^2$

that is:

$\ds \sum_{k \mathop = 0}^{2 n - 1} 10^k = \paren {\sum_{k \mathop = 1}^{n - 1} 5 \times 10^k + 6}^2 - \paren {\sum_{k \mathop = 1}^{n - 1} 4 \times 10^k + 5}^2$


Proof

From Difference between Two Squares equal to Repunit, $R_{2 n} = x^2 - y^2$ exactly when $R_{2 n} = a b$ where $x = \dfrac {a + b} 2$ and $y = \dfrac {a - b} 2$.

By the Basis Representation Theorem

\(\ds R_{2n}\) \(=\) \(\ds \sum_{k \mathop = 0}^{2 n - 1} 10^k\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^{n - 1} 10^k + \sum_{k \mathop = n}^{2 n - 1} 10^k\) splitting the summation into two
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^{n - 1} 10^k + 10^n \sum_{k \mathop = 0}^{n - 1} 10^k\) factoring $10^n$ out of the second part
\(\ds \) \(=\) \(\ds \paren {10^n + 1} \sum_{k \mathop = 0}^{n - 1} 10^k\)
\(\ds \) \(=\) \(\ds \underbrace {100 \ldots 01}_{\text {$n - 1$ $0$'s} } \times \underbrace {111 \ldots 1}_{\text {$n$ $1$'s} }\)


Thus, let:

$a = \paren {10^n + 1}$
$b = \ds \sum_{k \mathop = 0}^{n - 1}$

So:

\(\ds a + b\) \(=\) \(\ds \sum_{k \mathop = 0}^n 10^k + 1\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n 10^k + 2\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {a + b} 2\) \(=\) \(\ds \sum_{k \mathop = 1}^n \frac {10^k} 2 + 1\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n 5 \times 10^{k - 1} + 1\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^{n - 1} 5 \times 10^k + 1\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^{n - 1} 5 \times 10^k + 6\)
\(\ds \) \(=\) \(\ds \underbrace {55 \ldots 50}_{\text {$n - 1$ $5$'s} } + 6\)
\(\ds \) \(=\) \(\ds \underbrace {55 \ldots 56}_{\text {$n - 1$ $5$'s} }\)


Similarly:

\(\ds a - b\) \(=\) \(\ds \paren {10^n + 1} - \sum_{k \mathop = 0}^{n - 1} 10^k\)
\(\ds \) \(=\) \(\ds \paren {\sum_{k \mathop = 0}^{n - 1} 9 \times 10^k + 1} + 1 - \sum_{k \mathop = 0}^{n - 1} 10^k\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^{n - 1} 8 \times 10^k + 2\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {a - b} 2\) \(=\) \(\ds \sum_{k \mathop = 0}^{n - 1} 4 \times 10^k + 1\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^{n - 1} 4 \times 10^k + 5\)
\(\ds \) \(=\) \(\ds \underbrace {44 \ldots 45}_{\text {$n - 1$ $4$'s} }\)

Hence the result.

$\blacksquare$


Sources

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