Cool Repeating Digits

Number Theory Level 3

The first ten solutions to the equation $m^{2}=n^{3}+n^{2}$ where $m$ and $n$ are positive integers are listed below.

$\begin{array} {rrrrr} 3^{3}&+&3^{2}&=&\color{#D61F06}6\color{#333333}^{2} \\ 8^{3}&+&8^{2}&=&2\color{#D61F06}4\color{#333333}^{2} \\ 15^{3}&+&15^{2}&=&6\color{#D61F06}0\color{#333333}^{2} \\ 24^{3}&+&24^{2}&=&12\color{#D61F06}0\color{#333333}^{2} \\ 35^{3}&+&35^{2}&=&21\color{#D61F06}0\color{#333333}^{2} \\ 48^{3}&+&48^{2}&=&33\color{#3D99F6}6\color{#333333}^{2} \\ 63^{3}&+&63^{2}&=&50\color{#3D99F6}4\color{#333333}^{2} \\ 80^{3}&+&80^{2}&=&72\color{#3D99F6}0\color{#333333}^{2} \\ 99^{3}&+&99^{2}&=&99\color{#3D99F6}0\color{#333333}^{2} \\ 120^{3}&+&120^{2}&=&132\color{#3D99F6}0\color{#333333}^{2} \end{array}$

Is the highlighted pattern occurring within the unit digit for all $m$ ever disrupted?

Yes No

1 solution

Chew-Seong Cheong
Oct 22, 2018

Given that $m^2 = n^3+n^2 = n^2(n+1)$ , we note that solution exist when $n+1$ is a perfect square. Let $a^2 = n+1$ for $a \ge 2$ , then $n=k^2 - 1$ and $m^2 = (a^2-1)^2a^2 = \big((a-1)a(a+1)\big)^2$ . Therefore, $m$ is the product of three consecutive integers. Let $m_k = k(k+1)(k+2)$ , where $k$ is a positive integer. Let $i \in [1, 10]$ and $j$ , a positive integer. We note that $m_{10j+i} \equiv (10j+i)(10j+i+1)(10j+i+2) \equiv (i)(i+1)(i+2) \equiv m_i \text{ (mod 10)}$ . Therefore, the pattern of the unit digit repeats and has a period of 5. No , the pattern is not disrupted.

Out of curiosity, do you also know how to show the reason for the relationship between $m$ and $n$ being able to be expressed as the following?

$m_{a}=n_{a}\times{(a+1)}$

Where $n_{a}$ is the $a$ th $n$ term and $m_{a}$ is the $a$ th $m$ term

Aidan Poor - 2 years, 7 months ago

Cool Repeating Digits

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