SAMO Problem 6: Diophantine equation $20^m - 10m^2 + 1 = 19^n$

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SAMO Problem 6: Diophantine equation $20^m - 10m^2 + 1 = 19^n$

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Source: South African Mathematics Olympiad 2019, Problem 6

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DylanN

#1 h Jul 25, 2019, 12:57 PM • 2 Y

Y by Adventure10, Mango247

Determine all pairs $(m, n)$ of non-negative integers that satisfy the equation
$20^m - 10m^2 + 1 = 19^n.$

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test20

#2 h Jul 25, 2019, 2:44 PM • 4 Y

Y by Pluto1708, DylanN, Infin1ty_, Adventure10

Modulo $10$ : As LHS $\equiv1$ , we get that $n$ is even.
Modulo $4$ : As RHS $\equiv1$ , we get that $m$ is even.
Hence $20^m$ and $19^n$ both are squares.
We write $m=2p$ .

Since LHS is a square strictly smaller than $(20^p)^2$ ,
we conclude that LHS is at most $(20^p-1)^2$ .
Hence $20^{2p} - 10(2p)^2 + 1 ~\le~ (20^p-1)^2 ~=~ 20^{2p}-2\cdot20^p+1$ ,
which simplifies to $p^2\ge20^{p-1}$ . (*)

This last inequality (*) holds true only for $p=1$ .
This yields $m=2$ and $n=2$ as the only solution.

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338 posts

#3 h Jul 13, 2023, 2:42 AM

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DylanN wrote:

Determine all pairs $(m, n)$ of non-negative integers that satisfy the equation
$20^m - 10m^2 + 1 = 19^n.$

$\color{blue}\boxed{\textbf{Answer: (2,2)}}$
$\color{blue}\boxed{\textbf{Proof:}}$
$\color{blue}\rule{24cm}{0.3pt}$
$20^m-10m^2+1=19^n...(\alpha)$ $\Rightarrow (-1)^m-m^2+1\equiv 1\pmod{3}$ $\Rightarrow m^2\equiv (-1)^m \pmod{3}$ If $m\equiv 1 \pmod{2}:$
$\Rightarrow m^2\equiv 2 \pmod{3} (\Rightarrow \Leftarrow)$ If $m\equiv 0 \pmod{2}:$
In $(\alpha):$
$\Rightarrow 0-0+1\equiv (-1)^n \pmod{5}$ $\Rightarrow n \text{ is even}$ $\Rightarrow 20^m \text{ and } 19^n \text{ are perfect squares}$ In $(\alpha):$
$\Rightarrow 19^n<20^m$ $m=2t:$
$\Rightarrow 19^n\le (20^t-1)^2$ $\Rightarrow 20^m-10m^2+1\le 20^m-2\times 20^t +1$ $\Rightarrow 20^t<5m^2$ $\Rightarrow 20^t<20t^2$ $\Rightarrow 20^{t-1}<t^2$ $\Rightarrow t\le 1$ $\Rightarrow m\in \{ 0,2 \}$ If $m=0:$
In $(\alpha):$
$\Rightarrow 2=19^n (\Rightarrow \Leftarrow)$ If $m=2:$
In $(\alpha):$
$\Rightarrow 400-40+1=19^n$ $\Rightarrow n=2$ $\Rightarrow (2,2) \text{ is the only solution}_\blacksquare$ $\color{blue}\rule{24cm}{0.3pt}$

This post has been edited 1 time. Last edited by hectorleo123, Jul 13, 2023, 2:42 AM

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