20250915 Senior Varsity Training

Problem 1

Find the smallest $n$ such that $2013^n$ ends in $001$.

Solution

It is easy to prove that the problem is equivalent to finding the smallest $n>0$ such that $2013^n\equiv 1(\mathrm{mod}1000)$.

$$\begin{array}{rl}2013^n\equiv 1(\mathrm{mod}1000)& ⟺2013^n\equiv 1(\mathrm{mod}{2}^3)\wedge 2013^n\equiv 1(\mathrm{mod}{5}^3)\\ & ⟺5^n\equiv 1(\mathrm{mod}{2}^3)\wedge 13^n\equiv 1(\mathrm{mod}{5}^3)\\ & ⟺2^3\mid 5^n-1^n\wedge 5^3\mid 13^n-1^n\\ & ⟺{\nu }_2(5^n-1^n)\ge 3\wedge {\nu }_5(13^n-1^n)\ge 3\end{array}$$

If $n\equiv 0(\mathrm{mod}2)$, applying the Lifting-the-Exponent Lemma yields

$$\begin{array}{rl}{\nu }_2(5^n-1^n)& ={\nu }_2(4)+{\nu }_2(6)+{\nu }_2(n)-1\\ & =2+1+{\nu }_2(n)-1\\ & \ge 3\end{array}$$

If $n\equiv 1(\mathrm{mod}2)$, applying the Lifting-the-Exponent Lemma yields

$$\begin{array}{rl}{\nu }_2(5^n-1^n)& ={\nu }_2(4)\\ & =2\\ & <3\end{array}$$

Therefore,

$${\nu }_2(5^n-1^n)\ge 3⟺2\mid n$$ $$\begin{array}{rl}{\nu }_5(13^n-1^n)\ge 3& ⟹3^n\equiv 1(\mathrm{mod}5)\\ & ⟺{\delta }_5(3)\mid n\\ & ⟺4\mid n\end{array}$$

Therefore,

$$\begin{array}{rl}{\nu }_5(13^n-1^n)\ge 3& ⟺4\mid n\wedge {\nu }_5(13^n-1^n)\ge 3\\ & ⟺\exists k\in \mathbb{Z},n=4k\wedge {\nu }_5(13^n-1^n)\ge 3\\ & ⟺\exists k\in \mathbb{Z},n=4k\wedge {\nu }_5((13^4{)}^k-(1^4{)}^k)\ge 3\end{array}$$

Applying the Lifting-the-Exponent Lemma yields

$$\begin{array}{rl}{\nu }_5((13^4{)}^k-(1^4{)}^k)& ={\nu }_5(13^4-1^4)+{\nu }_5(k)\\ & ={\nu }_5(13-1)+{\nu }_5(13+1)+{\nu }_5(13^2+1^2)+{\nu }_5(k)\\ & =1+{\nu }_5(k)\end{array}$$

Therefore,

$$\begin{array}{rl}{\nu }_5(13^n-1^n)\ge 3& ⟺\exists k\in \mathbb{Z},n=4k\wedge 1+{\nu }_5(k)\ge 3\\ & ⟺\exists k\in \mathbb{Z},n=4k\wedge 25\mid k\\ & ⟺100\mid n\end{array}$$

Therefore,

$$\begin{array}{rl}2013^n\equiv 1(\mathrm{mod}1000)& ⟺2\mid n\wedge 100\mid n\\ & ⟺100\mid n\end{array}$$

The smallest $n>0$ such that $100\mid n$ is $100$.

Problem 2

Let $n$ be the least positive integer such that $149^n-2^n$ is divisible by $3^3{5}^5{7}^7$. Find the number of positive divisors of $n$.

Solution

$$\begin{array}{rl}{3}^3{5}^5{7}^7\mid (149^n-2^n)& ⟺3^3\mid 149^n-2^n\wedge 5^5\mid 149^n-2^n\wedge 7^7\mid 149^n-2^n\\ & ⟺{\nu }_3(149^n-2^n)\ge 3\wedge {\nu }_5(149^n-2^n)\ge 5\wedge {\nu }_7(149^n-2^n)\ge 7\end{array}$$

Applying the Lifting-the-Exponent Lemma yields

$$\begin{array}{rl}{\nu }_3(149^n-2^n)& ={\nu }_3(147)+{\nu }_3(n)\\ & =1+{\nu }_3(n)\end{array}$$

Therefore,

$$\begin{array}{rl}{\nu }_3(149^n-2^n)\ge 3& ⟺1+{\nu }_3(n)\ge 3\\ & ⟺3^2\mid n\end{array}$$ $${\nu }_5(149^n-2^n)\ge 5⟹149^n\equiv 2^n(\mathrm{mod}5)$$

Since

$$149^n\equiv \{\begin{array}{ll}1(\mathrm{mod}5),& n\equiv 0(\mathrm{mod}2)\\ 4(\mathrm{mod}5),& n\equiv 1(\mathrm{mod}2)\end{array}$$ $$2^n\equiv \{\begin{array}{ll}1(\mathrm{mod}5),& n\equiv 0(\mathrm{mod}4)\\ 2(\mathrm{mod}5),& n\equiv 1(\mathrm{mod}4)\\ 4(\mathrm{mod}5),& n\equiv 2(\mathrm{mod}4)\\ 3(\mathrm{mod}5),& n\equiv 3(\mathrm{mod}4)\end{array}$$

it follows that

$$149^n\equiv 2^n(\mathrm{mod}5)⟺4\mid n$$

Therefore,

$$\begin{array}{rl}{\nu }_5(149^n-2^n)\ge 5& ⟺4\mid n\wedge {\nu }_5(149^n-2^n)\ge 5\\ & ⟺\exists k\in \mathbb{Z},n=4k\wedge {\nu }_5(149^n-2^n)\ge 5\\ & ⟺\exists k\in \mathbb{Z},n=4k\wedge {\nu }_5((149^4{)}^n-(2^4{)}^n)\ge 5\end{array}$$

Applying the Lifting-the-Exponent Lemma yields

$$\begin{array}{rl}{\nu }_5((149^4{)}^k-(2^4{)}^k)& ={\nu }_5(149^4-2^4)+{\nu }_5(k)\\ & ={\nu }_5(149-2)+{\nu }_5(149+2)+{\nu }_5(149^2+2^2)+{\nu }_5(k)\\ & =1+{\nu }_5(k)\end{array}$$

Therefore,

$$\begin{array}{rl}{\nu }_5(149^n-2^n)\ge 5& ⟺\exists k\in \mathbb{Z},n=4k\wedge 1+{\nu }_5(k)\ge 5\\ & ⟺\exists k\in \mathbb{Z},n=4k\wedge 5^4\mid k\\ & ⟺4\cdot 5^4\mid n\end{array}$$

Applying the Lifting-the-Exponent Lemma yields

$$\begin{array}{rl}{\nu }_7(149^n-2^n)& ={\nu }_7(147)+{\nu }_7(n)\\ & =2+{\nu }_7(n)\end{array}$$

Therefore,

$$\begin{array}{rl}{\nu }_7(149^n-2^n)\ge 7& ⟺2+{\nu }_7(n)\ge 7\\ & ⟺7^5\mid n\end{array}$$

Therefore,

$$\begin{array}{rl}{3}^3{5}^5{7}^7\mid 149^n-2^n& ⟺3^2\mid n\wedge 4\cdot 5^4\mid n\wedge 7^5\mid n\\ & ⟺2^2{3}^2{5}^4{7}^5\mid n\end{array}$$

The smallest $n>0$ such that $2^2{3}^2{5}^4{7}^5\mid n$ is $2^2{3}^2{5}^4{7}^5$, it has $(2+1)(2+1)(4+1)(5+1)=270$ divisors.

Problem 3

Find the largest $k$ such that

$$1991^k\mid 1990^{1991^{1992}}+1992^{1991^{1990}}$$

Solution

$$\begin{array}{rl}1991^k\mid 1990^{1991^{1992}}+1992^{1991^{1990}}& ⟺11^k\mid 1990^{1991^{1992}}+1992^{1991^{1990}}\wedge 181^k\mid 1990^{1991^{1992}}+1992^{1991^{1990}}\\ & ⟺{\nu }_{11}(1990^{1991^{1992}}+1992^{1991^{1990}})\ge k\wedge {\nu }_{181}(1990^{1991^{1992}}+1992^{1991^{1990}})\ge k\end{array}$$ $$\begin{array}{rl}1990^{1991^{1992}}+1992^{1991^{1990}}& =1990^{1991^{1990}1991^2}+1992^{1991^{1990}}\\ & =(1990^{1991^2}{)}^{1991^{1990}}+1992^{1991^{1990}}\end{array}$$

Therefore,

$${\nu }_{11}(1990^{1991^{1992}}+1992^{1991^{1990}})={\nu }_{11}((1990^{1991^2}{)}^{1991^{1990}}+1992^{1991^{1990}})$$

Applying the Lifting-the-Exponent Lemma yields

$$\begin{array}{rl}{\nu }_{11}((1990^{1991^2}{)}^{1991^{1990}}+1992^{1991^{1990}})& ={\nu }_{11}(1990^{1991^2}+1992)+1990{\nu }_{11}(1991)\\ & ={\nu }_{11}(\sum _{n=0}^{1991^2}(\genfrac{}{}{0ex}{}{1991^2}{n})1991^n(-1{)}^{1991^2-n}+1992)+1990\\ & ={\nu }_{11}(\sum _{n=1}^{1991^2}(\genfrac{}{}{0ex}{}{1991^2}{n})1991^n(-1{)}^{1991^2-n}+1991)+1990\\ & ={\nu }_{11}(1991)+{\nu }_{11}(\sum _{n=1}^{1991^2}(\genfrac{}{}{0ex}{}{1991^2}{n})1991^{n-1}(-1{)}^{1991^2-n}+1)+1990\\ & =1+0+1990\\ & =1991\end{array}$$

Similarly, it follows that

$${\nu }_{181}(1990^{1991^{1992}}+1992^{1991^{1990}})=1991$$

Therefore,

$$\begin{array}{rl}1991^k\mid 1990^{1991^{1992}}+1992^{1991^{1990}}& ⟺1991\ge k\wedge 1991\ge k\\ & ⟺1991\ge k\end{array}$$

The largest $k$ such that $1991\ge k$ is $1991$.

Problem 4

If $m$ is an integer such that

$$3^m\mid 7^{3^{527}}-1$$

find the maximum possible value of $m$.

Solution

$$3^m\mid 7^{3^{527}}-1⟺{\nu }_3(7^{3^{527}}-1^{3^{527}})\ge m$$

Applying the Lifting-the-Exponent Lemma yields

$$\begin{array}{rl}{\nu }_3(7^{3^{527}}-1^{3^{527}})& ={\nu }_3(6)+{\nu }_3(3^{527})\\ & =1+527\\ & =528\end{array}$$

Therefore,

$$3^m\mid 7^{3^{527}}-1⟺528\ge m$$

The largest integer $m$ such that $528\ge m$ is $528$.

Problem 5

Find the sum of all the divisors $d$ of $N=19^{88}-1$ which are of the form $d=2^a{3}^b$ with $a,b\in \mathbb{N}$.

Solution

$$\begin{array}{rl}{2}^a{3}^b\mid 19^{88}-1& ⟺2^a\mid 19^{88}-1\wedge 3^b\mid 19^{88}-1\\ & ⟺{\nu }_2(19^{88}-1^{88})\ge a\wedge {\nu }_3(19^{88}-1^{88})\ge b\end{array}$$

Applying the Lifting-the-Exponent Lemma yields

$$\begin{array}{rl}{\nu }_2(19^{88}-1^{88})& ={\nu }_2(19-1)+{\nu }_2(19+1)+{\nu }_2(88)-1\\ & =1+2+3-1\\ & =5\end{array}$$

Applying the Lifting-the-Exponent Lemma yields

$$\begin{array}{rl}{\nu }_3(19^{88}-1^{88})& ={\nu }_3(19-1)+{\nu }_3(88)\\ & =2\end{array}$$

Therefore,

$$2^a{3}^b\mid 19^{88}-1⟺5\ge a\wedge 2\ge b$$

The sum of all the divisors $d$ of the form $d=2^a{3}^b$ is

$$\begin{array}{rl}\sum _{a=0}^5\sum _{b=0}^2{2}^a{3}^b& =(\sum _{a=0}^5{2}^a)(\sum _{b=0}^2{3}^b)\\ & =63\cdot 13\\ & =819\end{array}$$

Problem 6

Let $k$ be a positive integer. Find all positive integers $n$ such that $3^k\mid 2^n-1$.

Solution

$$\begin{array}{rl}{3}^k\mid 2^n-1& ⟹3\mid 2^n-1\\ & ⟺2^n\equiv 1(\mathrm{mod}3)\\ & ⟺{\delta }_3(2)\mid n\\ & ⟺2\mid n\end{array}$$

Therefore,

$$\begin{array}{rl}{3}^k\mid 2^n-1& ⟺2\mid n\wedge 3^k\mid 2^n-1\\ & ⟺\exists m\in \mathbb{Z},n=2m\wedge 3^k\mid 2^n-1\\ & ⟺\exists m\in \mathbb{Z},n=2m\wedge 3^k\mid 4^m-1\\ & ⟺\exists m\in \mathbb{Z},n=2m\wedge {\nu }_3(4^m-1^m)\ge k\end{array}$$

Applying the Lifting-the-Exponent Lemma yields

$$\begin{array}{rl}{\nu }_3(4^m-1^m)& ={\nu }_3(4-1)+{\nu }_3(m)\\ & =1+{\nu }_3(m)\end{array}$$

Therefore,

$$\begin{array}{rl}{3}^k\mid (2^n-1)& ⟺\exists m\in \mathbb{Z},n=2m\wedge 1+{\nu }_3(m)\ge k\\ & ⟺\exists m\in \mathbb{Z},n=2m\wedge 3^{k-1}\mid m\\ & ⟺2\cdot 3^{k-1}\mid m\end{array}$$

Problem 7

Find all primes $p,q$ such that

$$\frac{(5^p-2^p)(5^q-2^q)}{pq}$$

is an integer.

Solution

Wlog, assume $p\le q$.

$$\frac{(5^p-2^p)(5^q-2^q)}{pq}\in \mathbb{Z}⟺pq\mid (5^p-2^p)(5^q-2^q)$$

Since

$$(5^p-2^p)(5^q-2^q)\equiv 1(\mathrm{mod}2)$$

it follows that

$$pq\mid (5^p-2^p)(5^q-2^q)⟹p\ne 2\wedge q\ne 2$$ $$\begin{array}{rl}pq\mid (5^p-2^p)(5^q-2^q)& ⟹p\mid (5^p-2^p)(5^q-2^q)\\ & ⟺(5^p-2^p)(5^q-2^q)\equiv 0(\mathrm{mod}p)\end{array}$$

Applying Fermat’s Little Theorem yields

$$5^p-2^p\equiv 3(\mathrm{mod}p)$$

Therefore, if $p\ne 3$,

$$\begin{array}{rl}(5^p-2^p)(5^q-2^q)\equiv 0(\mathrm{mod}p)& ⟺5^q-2^q\equiv 0(\mathrm{mod}p)\\ & ⟺{\left(\frac{5}{2}\right)}^q\equiv 1(\mathrm{mod}p)\\ & ⟺{\delta }_p\left(\frac{5}{2}\right)\mid q\\ & ⟺{\delta }_p\left(\frac{5}{2}\right)=q\end{array}$$

Applying Fermat’s Little Theorem yields

$${\left(\frac{5}{2}\right)}^{p-1}\equiv 1(\mathrm{mod}p)$$ $${\left(\frac{5}{2}\right)}^{p-1}\equiv 1(\mathrm{mod}p)⟺{\delta }_p\left(\frac{5}{2}\right)\mid p-1$$

Therefore,

$$\begin{array}{rl}(5^p-2^p)(5^q-2^q)\equiv 0(\mathrm{mod}p)& ⟹q\mid p-1\\ & ⟹q<p\end{array}$$

By contradiction, if follows that $p=3$.

Therefore,

$$\begin{array}{rl}pq\mid (5^p-2^p)(5^q-2^q)& ⟺39(5^q-2^q)\equiv 0(\mathrm{mod}q)\\ & ⟺q\mid 117\\ & ⟺q=3\vee q=13\end{array}$$

Therefore, all primes $p,q$ such that

$$\frac{(5^p-2^p)(5^q-2^q)}{pq}$$

is an integer are $⟨p,q⟩=⟨3,3⟩$, $⟨p,q⟩=⟨3,13⟩$ and $⟨p,q⟩=⟨13,3⟩$.

Problem 8

Find all positive integers $n$ such that $3^n-1$ is divisible by $2^n$.

Solution

$$2^n\mid 3^n-1⟺{\nu }_2(3^n-1^n)\ge n$$

Applying the Lifting-the-Exponent Lemma yields

$$\begin{array}{rl}{\nu }_2(3^n-1^n)& =\{\begin{array}{ll}{\nu }_2(3-1)+{\nu }_2(3+1)+{\nu }_2(n)-1,& n\equiv 0(\mathrm{mod}2)\\ {\nu }_2(3-1),& n\equiv 1(\mathrm{mod}2)\end{array}\\ & =\{\begin{array}{ll}{\nu }_2(n)+2,& n\equiv 0(\mathrm{mod}2)\\ 1,& n\equiv 1(\mathrm{mod}2)\end{array}\end{array}$$

Therefore,

$$\begin{array}{rl}{\nu }_2(3^n-1^n)\ge n& ⟺(n\equiv 0(\mathrm{mod}2)\wedge {\nu }_2(n)+2\ge n)\vee (n\equiv 1(\mathrm{mod}2)\wedge 1\ge n)\\ & ⟺(n\equiv 0(\mathrm{mod}2)\wedge 2^{n-2}\mid n)\vee n=1\\ & ⟺n=1\vee n=2\vee n=4\end{array}$$

Therefore,

$$2^n\mid 3^n-1⟺n=1\vee n=2\vee n=4$$

Problem 9

Find all positive integers $a$ such that $\frac{5^a+1}{3^a}$ is a positive integer.

Solution

$$\begin{array}{rl}\frac{5^a+1}{3^a}\in {\mathbb{Z}}_{+}& ⟺3^a\mid 5^a+1\\ & ⟹5^a+1\equiv 0(\mathrm{mod}3)\\ & ⟺2^a\equiv 2(\mathrm{mod}3)\\ & ⟺a\equiv 1(\mathrm{mod}2)\end{array}$$

Therefore,

$$\begin{array}{rl}{3}^a\mid 5^a+1& ⟺a\equiv 1(\mathrm{mod}2)\wedge 3^a\mid 5^a+1\\ & ⟺a\equiv 1(\mathrm{mod}2)\wedge {\nu }_3(5^a+1^a)\ge a\end{array}$$

Applying the Lifting-the-Exponent Lemma yields that, if $a\equiv 1(\mathrm{mod}2)$,

$$\begin{array}{rl}{\nu }_3(5^a+1^a)& ={\nu }_3(6)+{\nu }_3(a)\\ & =1+{\nu }_3(a)\end{array}$$

Therefore,

$$\begin{array}{rl}{3}^a\mid 5^a+1& ⟺1+{\nu }_3(a)\ge a\\ & ⟺3^{a-1}\mid a\\ & ⟺a=1\end{array}$$

Therefore, the only positive integer $a$ such that $\frac{5^a+1}{3^a}$ is a positive integer is $a=1$.

Problem 10

Let $p>3$ be a prime and $k\ge 0$ an integer. Find the multiplicity of $X-1$ in the factorization of

$$f(X)=X^{3p^k-1}+X^{3p^k-2}+\cdots +X+1$$

modulo $p$.

Solution

$$\begin{array}{rl}f(X)& \equiv X^{3p^k-1}+X^{3p^k-2}+\cdots +X+1(\mathrm{mod}p)\\ & \equiv \frac{X^{3p^k}-1}{X-1}(\mathrm{mod}p)\\ & \equiv \frac{(X^3{)}^{p^k}-1^{p^k}}{X-1}(\mathrm{mod}p)\\ & \equiv \frac{(X^3-1{)}^{p^k}}{X-1}(\mathrm{mod}p)\\ & \equiv (X-1{)}^{p^k-1}(X^2+X+1{)}^{p^k}(\mathrm{mod}p)\end{array}$$

Since when $X=1$, $X-1\equiv 0(\mathrm{mod}p)$, $X^2+X+1\not\equiv 0(\mathrm{mod}p)$, it follows that the multiplicity of $X-1$ in the factorization of $f(X)$ is $p^k-1$, modulo $p$.

Problem 11

Find all positive integers $n$ such that for every positive odd integer $a$, we have

$$2^{2017}\mid a^n-1$$

Solution

$$\forall a,2^{2017}\mid a^n-1⟺\forall a,{\nu }_2(a^n-1^n)\ge 2017$$

Applying the Lifting-the-Exponent Lemma yields

$${\nu }_2(a^n-1^n)=\{\begin{array}{ll}{\nu }_2(a-1)+{\nu }_2(a+1)+{\nu }_2(n)-1,& n\equiv 0(\mathrm{mod}2)\\ {\nu }_2(a-1),& n\equiv 1(\mathrm{mod}2)\end{array}$$

Therefore,

$$\begin{array}{rl}\forall a,2^{2017}\mid a^n-1& ⟺(n\equiv 0(\mathrm{mod}2)\wedge \forall a,{\nu }_2(a-1)+{\nu }_2(a+1)+{\nu }_2(n)-1\ge 2017)\vee (n\equiv 1(\mathrm{mod}2)\wedge \forall a,{\nu }_2(a-1)\ge 2017)\\ & ⟺n\equiv 0(\mathrm{mod}2)\wedge \forall a,{\nu }_2(n)\ge 2018-{\nu }_2(a-1)-{\nu }_2(a+1)\\ & ⟺n\equiv 0(\mathrm{mod}2)\wedge {\nu }_2(n)\ge 2018-\underset{a}{\min }\{{\nu }_2(a-1)+{\nu }_2(a+1)\}\\ & ⟺n\equiv 0(\mathrm{mod}2)\wedge {\nu }_2(n)\ge 2015\\ & ⟺2^{2015}\mid n\end{array}$$

Problem 12

Let $m,k\ge 0,p=2^{2^m}+1$ a prime. Prove:

1. $2^{2^{m+1}{p}^k}\equiv 1(\mathrm{mod}{p}^{k+1})$;

   Solution

   $$\begin{array}{rl}{2}^{2^{m+1}{p}^k}\equiv 1(\mathrm{mod}{p}^{k+1})& ⟺p^{k+1}\mid 2^{2^{m+1}{p}^k}-1\\ & ⟺{\nu }_p((2^{2^{m+1}}{)}^{p^k}-1^{p^k})\ge k+1\end{array}$$

   Applying the Lifting-the-Exponent Lemma yields

   $$\begin{array}{rl}{\nu }_p((2^{2^{m+1}}{)}^{p^k}-1^{p^k})& ={\nu }_p(2^{2^{m+1}}-1)+k{\nu }_p(p)\\ & ={\nu }_p(2^{2^m}-1)+{\nu }_p(2^{2^m}+1)+k\\ & =k+1\end{array}$$

   Therefore,

   $$2^{2^{m+1}{p}^k}\equiv 1(\mathrm{mod}{p}^{k+1})$$

2. The smallest $n$ such that $2^n\equiv 1(\mathrm{mod}{p}^{k+1})$ is $2^{m+1}{p}^k$.

   Solution

   If ${\delta }_{p^{k+1}}(2)<2^{m+1}{p}^k$,

   $$\begin{array}{rl}{2}^{2^{m+1}{p}^k}\equiv 1(\mathrm{mod}{p}^{k+1})& ⟺{\delta }_{p^{k+1}}(2)\mid 2^{m+1}{p}^k\\ & ⟺{\delta }_{p^{k+1}}(2)\mid 2^m{p}^k\vee (k>0\wedge {\delta }_{p^{k+1}}(2)\mid 2^{m+1}{p}^{k-1})\\ & ⟺2^{2^m{p}^k}\equiv 1(\mathrm{mod}{p}^{k+1})\vee (k>0\wedge 2^{2^{m+1}{p}^{k-1}}\equiv 1(\mathrm{mod}{p}^{k+1}))\end{array}$$ $$\begin{array}{rl}p=2^{2^m}+1& ⟺2^{2^m}\equiv -1(\mathrm{mod}p)\\ & ⟹2^{2^m{p}^k}\equiv -1(\mathrm{mod}p)\\ & ⟺p\mid 2^{2^m{p}^k}+1\\ & ⟹p\nmid 2^{2^m{p}^k}-1\\ & ⟹p^{k+1}\nmid 2^{2^m{p}^k}-1\\ & ⟺2^{2^m{p}^k}\not\equiv 1(\mathrm{mod}{p}^{k+1})\end{array}$$

   If $k>0$,

   $$\begin{array}{rl}{2}^{2^{m+1}{p}^{k-1}}\equiv 1(\mathrm{mod}{p}^{k+1})& ⟺p^{k+1}\mid 2^{2^{m+1}{p}^{k-1}}-1\\ & ⟺{\nu }_p((2^{2^{m+1}}{)}^{p^{k-1}}-1^{p^{k-1}})\ge k+1\\ & ⟺k\ge k+1\end{array}$$

   Therefore,

   $$2^{2^{m+1}{p}^{k-1}}\not\equiv 1(\mathrm{mod}{p}^{k+1})$$

   Therefore,

   $$2^{2^{m+1}{p}^k}\not\equiv 1(\mathrm{mod}{p}^{k+1})$$

   By contradiction, it follows that

   $${\delta }_{p^{k+1}}(2)=2^{m+1}{p}^k$$

   Therefore, the smallest $n$ such that $2^n\equiv 1(\mathrm{mod}{p}^{k+1})$ is $2^{m+1}{p}^k$.