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I found this question in my Math Challenge II Number Theory packet: Find all positive integers $n$ that satisfy $323|20^n+16^n-3^n-1$. I don't even have any idea how to approach this question. Any suggestions?

Bill Dubuque
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Jason Chen
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2 Answers2

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$323 = 17\cdot 19$. So $323$ divides $20^n+16^n-3^n-1$ if and only if both its prime factors, $17$ and $19$, divide it. We have

$$20^n + 16^n - 3^n - 1 \equiv 3^n + (-1)^n - 3^n - 1 \pmod{17}$$

and

$$20^n + 16^n - 3^n - 1 \equiv 1^n + (-3)^n - 3^n - 1 \pmod{19}.$$

From that, it is easy to find the desired $n$.

Daniel Fischer
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  • One value of n is 1, but are there any others? – Jason Chen May 09 '14 at 04:36
  • $n = 1$ doesn't work, $20^1+16^1-3^1-1 = 32$. $n=2$ does work, $20^2+16^2-3^2-1 = 400 + 256 - 9 - 1 = 656 - 10 = 646 = 2\cdot 323$. If you look at the congruences modulo $17$ and modulo $19$, there is something in either case that you can simplify. – Daniel Fischer May 09 '14 at 09:15
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Slightly more work yields a much more general result. Let $\,o_n(x) :=$ order of $\,x\pmod{\! n}$

Theorem $\ $ If $\,\ \color{#0a0}{m\mid \color{#90f}a\!-\!c},\,\ \color{#c00}{\overline m \mid \color{#90f}a\!-\!d}\ $ and $\ b\:\!\color{#0af}{\text{ is coprime to } \,\color{#0a0}m,\color{#c00}{\bar m}}\,$ then

$\qquad\qquad\ \ \ {\rm lcm}(\color{#0a0}m,\color{#c00}{\overline m})\mid e = \color{#90f}a^n\!+\color{c00}b^n\!-\color{#0a0}c^n\!-\color{#c00}d^n\!\iff {\rm lcm}(o_m(d/b),o_{\overline m}(c/b))\mid n$

Proof $\:\! $ by below $\,m,\bar m\mid e\!\iff\! o_m(d/b),o_{\bar m}(c/b))\mid n,\,$ so apply $\,x,y\,|\,z\!\iff\! {\rm lcm}(x,y)\,|\, z$

$\qquad\ \ \begin{align}{\rm mod}\ \color{#0a0}{m\!:\ \color{#90f}a\equiv c}\ \ \:\!&\ {\rm so}\,\ \ e\equiv b^n\!-d^n\!\equiv 0\!\!\overset{\color{#0af}{(b,m)=1}\!\!}\iff (d/b)^n\equiv 1\!\iff o_m(d/b)\mid n\\[.1em] {\rm mod}\ \color{#c00}{\overline m\!:\ \color{#90f}a\equiv d}\ \ &\ {\rm so}\,\ \ e\equiv b^n\!-c^n\equiv 0\!\!\overset{\color{#0af}{(b,\bar m)=1}\!\!}\iff (c/b)^n\equiv 1\iff o_{\overline m}(c/b)\mid n \end{align}$

$\begin{eqnarray}\text{By the Theorem}\ \ \, \color{#0a0}m\,\color{#c00}{\overline m}\,&\mid&\ \ \color{#90f}a^n\, &+&\ \ b^n&-&\color{#0a0}c^n&-&\color{#c00}d^n&\!\!\!\iff& {\rm lcm}(o_m(d/b),o_{\overline m}(c/b))\mid n\\[.2em] {\rm i.e.}\ \ \ 323= \color{#0a0}{17}\cdot\color{#c00}{19}\,&\mid& \color{#90f}{20}^n&+&16^n&-&\color{#0a0}3^n&-&\color{#c00}{\bf 1}^n&\!\!\!\iff& {\rm lcm}(\color{#c00}2,\color{#0a0}2)=2\mid n,\ \rm since \end{eqnarray}$

$\qquad\ \ $ you have $\ \ \color{#0a0}{\overset{\Large 17\ \mid\ \color{#90f}{20}-3}{m\mid\ \color{#90f}a\!-\!c}},\, $ and $\ \ \overset{\Large \color{#c00}{19}\ \mid\ \color{#90f}{20}-\color{#c00}{\bf 1}\! }{\color{#c00}{\overline m}\mid\: \color{#90f}a\!-\!\color{#c00}d },\,$ and $\overset{\Large 16}b$ is coprime to $\overset{\Large 17,\,19}{\,m,\,\overline m},\ $ and

$\qquad\ \ \begin{align}{\rm mod}\ m=\color{#0a0}{17}\!:\,\ &\color{#c00}d/b = 1/16\equiv 1/{-}1\equiv -1\rm\ \ has\ order\ \ \color{#c00}2\\[.3em] {\rm mod}\ \overline m=\color{#c00}{19}\!:\,\ &\:\!\color{#0a0}c/b = 3/16\equiv 3/{-}3\equiv -1\rm\ \ has\ order \ \ \color{#0a0}2\end{align}$

Bill Dubuque
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