Repdigit Triangular Numbers

From ProofWiki
Jump to navigation Jump to search

Theorem

The only repdigit numbers which are also triangular are:

$55, 66, 666$

This sequence is A045914 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Proof

\(\ds 55\) \(=\) \(\ds \dfrac {10 \times 11} 2\)
\(\ds 66\) \(=\) \(\ds \dfrac {11 \times 12} 2\)
\(\ds 666\) \(=\) \(\ds \dfrac {36 \times 37} 2\)


Let $\dfrac {d \paren {10^j - 1}} 9$ be a $j$-digit repdigit number.

Suppose it is triangular.


For $d = 1$, write:

\(\ds \frac {k \paren {k + 1} } 2\) \(=\) \(\ds \frac {10^j - 1} 9\)
\(\ds 9 k \paren {k + 1}\) \(=\) \(\ds 2 \times 10^j - 2\)
\(\ds \paren {3 k + 1} \paren {3 k + 2}\) \(=\) \(\ds 2^{j + 1} \times 5^j\)

Since $3 k + 1 \perp 3 k + 2$, one of them is equal to $2^{j + 1}$ while the other is equal to $5^j$.

This gives $2^{j + 1} \pm 1 = 5^j$.

Since $j \ge 1$, taking modulo $4$:

$0 \pm 1 \equiv 1^j \equiv 1 \pmod 4$

so we have:

$2^{j + 1} + 1 = 5^j$


By 1 plus Power of 2 is not Perfect Power except 9, the above equation has no solutions for $j > 1$.

For $j = 1$ we have equality.

Hence $1$ is the only triangular repunit.

$\Box$


By Odd Square is Eight Triangles Plus One:

$x = 1 + \dfrac {8 d \paren {10^j - 1}} 9$ is a perfect square.

For $d = 2, 4, 7, 9$, $x$ ends in $2$ or $3$, which is not a perfect square by Square Modulo 5/Corollary.

$\Box$


For $d = 3, 8$, $x$ ends in $5$.

Since it is a perfect square, $x$ must end in $25$.

This is satisfied only when $d = 3, j = 1$.

For $d = 3, j > 1$ and $d = 8, j = 1$, $x$ ends in $65$.

For $d = 8, j > 1$, $x$ ends in $05$.

This corresponds to $\dfrac {x - 1} 8 = 3$.

$\Box$


For $d = 5$, $x$ is in the form $44 \dots 41$.

We show that $x$ is square only when $x = 441$.


This needs considerable tedious hard slog to complete it.
In particular: see talk
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Finish}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.


This corresponds to $\dfrac {x - 1} 8 = 55$.

$\Box$


For $d = 6$, $x$ is either:

$49$ for $j = 1$
$53 \dots 329$ for $j > 1$, with $j - 1$ $3$'s

we show that $x$ is square only when $x = 49, 529, 5329$.


This needs considerable tedious hard slog to complete it.
In particular: see talk
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Finish}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.


These correspond to $\dfrac {x - 1} 8 = 6$, $66$ and $666$.

$\Box$

We have hence demonstrated that the only triangular repdigit numbers are:

$1, 3, 6, 55, 66, 666$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Repdigit_Triangular_Numbers&oldid=478621"