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I'm unclear what is the best method to teach this with minimum math experience.

Brian M. Scott
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user61646
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  • Exploit innate $\rm\color{#c00}{symmetry}$! as in linked dupe. $\tag{}$ $\qquad\phantom{\Rightarrow}\ \ { 20^2,\ \ \ \ 16^2}\ \equiv, {3^2,\ \ \ 1^2}\ \ \ \ {\rm mod},\ 17,19,\ \ $
    $\tag    {}$ $\qquad\Rightarrow\ {20^{2k},\ \ 16^{2k}} \equiv {3^{2k},\ ,1^{2k}}\ \ \ {\rm mod},\ 17,19,\ \ $ by the Congruence Power Rule
    $\tag*{}$ $\qquad\Rightarrow\ \ 20^{2k} + 16^{2k}\ \ \equiv \ 3^{2k}+ 1^{2k}\ \ \ \ {\rm mod},\ 17,19,\ \ $ so also $,{\rm mod}\ 323 = {\rm lcm}(17,19)$ by CCRT.$\ \ $     – Bill Dubuque Sep 04 '23 at 19:32

5 Answers5

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We have $n=2k$. Hence, $$f(k) = 400^k + 256^k - 9^k - 1$$You can in fact prove that $646$ divides $f(k)$.

Note that $646 = 2 \cdot 19 \cdot 17$. In general, to prove that $abc \mid n$, where $a$, $b$, and $c$ are mutually relatively prime, it suffices to prove that $a \mid n$, $b \mid n$ and $c \mid n$.

First it is easy to prove that $f(k)$ is even, since $$f(k) = 400^k + 256^k - 9^k - 1 = \text{even} + \text{even} - \text{odd} - \text{odd} = \text{even}$$ Hence, $2 \mid f(k)$.

Now note that $(400-9) \mid (400^k - 9^k)$, i.e., $391 \mid (400^k-9^k)$ and $17 \mid 391$.

Similarly, $(256-1) \mid (256^k-1)$, i.e., $255 \mid (256^k-1)$ and $17 \mid 255$.

Hence, $17 \mid f(k)$.

Now note that $(400-1) \mid (400^k - 1)$, i.e., $399 \mid (400^k-1)$ and $19 \mid 399$

Similarly, $(256-9) \mid (256^k-9^k)$, i.e., $247 \mid (256^k-9^k)$ and $19 \mid 247 $

Hence, $19 \mid f(k)$.

Hence, $(2 \cdot 17 \cdot 19) \mid f(k)$

vonbrand
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$$20^{2m}+16^{2m}-3^{2m}-1=400^m-1+256^m-9^m$$

Now, $400^m-1$ is divisible by $400-1=399$ and $256^m-9^m$ is divisible by $256-9=247$

Now, $(247,399)=(247,152)=(95,152)=19(5,8)=19\implies 19\mid (20^{2m}+16^{2m}-3^{2m}-1)$

Again, $$20^{2m}+16^{2m}-3^{2m}-1=400^m-9^m+256^m-1$$

Now, $400^m-9^m$ is divisible by $400-9=391$ and $256^m-1$ is divisible by $256-1=255$

Now, $(255,391)=(136,255)=(136,119)=17\implies 17\mid (20^{2m}+16^{2m}-3^{2m}-1)$

So, lcm$(19,17)\mid (20^{2m}+16^{2m}-3^{2m}-1)$

But lcm$(19,17)=19\cdot17=323$ as $(17,19)=1$

Alternatively, as $323=19\cdot17$ and $(17,19)=1,$

using the congruence property#10 here $a\equiv b\pmod m\implies a^n\equiv b^n\pmod m$ where $n\ge0, a,b,m$ are integers,

$20^n+16^n-3^n-1\equiv3^n+(-1)^n-3^n-1\pmod{17}\equiv(-1)^n-1$ which is $\equiv0$ if $n$ is even, then $17\mid (20^n+16^n-3^n-1)$

Again, $20^n+16^n-3^n-1\equiv1^n+(-3)^n-3^n-1\pmod{19}\equiv 3^n\{(-1)^n-1\}$ which is $\equiv0$ if $n$ is even, then $19\mid (20^n+16^n-3^n-1)$.

lab bhattacharjee
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We have $323 = 17 \cdot 19$ with $\gcd(17, 19) = 1$. Also, $19 = 20-1 \mid 20^n - 1, 19 = 16 + 3 \mid 16^n - 3^n \Rightarrow 19 \mid 20^n+16^n-3^n-1$ and $17 = 20 - 3 \mid 20^n - 3^n$, $17 = 16+1 \mid 16^n +1 \Rightarrow 17 \mid 20^n+16^n-3^n-1$

$\gcd(17,19)=1 \Rightarrow 323 = 17 \cdot 19 \mid 20^n+16^n-3^n-1$

vonbrand
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Xeing
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Hint $\ $ Below, put $\rm\, m=17,\ n=19,\ a=20,\ b=16,\ c=3,\ d = 1$

$\rm mod\ m\!:\ a\equiv c,\, b\equiv -d\:\Rightarrow\:a^n + b^n - c^n - d^n \equiv\, c^n + (-d)^n - c^n -d^n \equiv\, 0$

$\rm mod\ \,n\!:\ a\equiv d,\, b\equiv -c\:\Rightarrow\: a^n + b^n - c^n - d^n \equiv\, d^n + (-c)^n - c^n -d^n \equiv\, 0$

Math Gems
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Since $n$ is even:

$20^n + 16^n - 3^n - 1 \equiv 1^n + (-3)^n - 3^n - 1 = 1 + 3^n - 3^n - 1 = 0 \bmod 19$

Also:

$20^n + 16^n - 3^n - 1 \equiv 3^n + (-1)^n - 3^n - 1 = 3^n + 1 - 3^n - 1 = 0 \bmod 17$

So the quantity is divisible by $17\times 19 = 323$

fretty
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