第25篇

\section{A Nice and Tricky Lemma (Lifting the Exponent)}%25
\markboth{Articles}{A Nice and Tricky Lemma (Lifting the Exponent)}

\vspace{4.2cm}

This article presents a powerful lemma which is useful in solving olympiad problems.

{\bf Lemma 1.}
{\it Let $p$ be an odd prime.
For two different integers $a$ (with $p\nmid a$) and $b$ with
$a\equiv b\pmod p$
and a positive integer $n$, the exponent of $p$ in $a^n-b^n$
is equal to the sum of the exponent of $p$ in $a-b$ and the exponent of $p$ in $n$.
}

We write $p^a\|n$ if and only if $p^a\mid n$ and $p^{a+1}\nmid n$,
where $a,p,n\in \mathbb{Z}$.
It allows us to state the lemma as follows.

Let $p$ be an odd prime and let $a,b,n\in \mathbb{Z}$ with $p\nmid a$.
Then $p^\alpha \|a-b$, $\alpha \ge 1$ and $p^\beta \|n$ implies
$p^{\beta +\alpha }\|a^n-b^n$.

{\bf Proof.}
Let us prove that if $a\equiv b\pmod p$ and $p^\beta \|n$ then
$p^\beta \|\ds\f{a^n-b^n}{a-b}$.
It is clear that the lifting lemma will follow, because with the condition
$p^\alpha \|a-b$ we have $p^{\alpha +\beta }\|a^n-b^n$.
Assume $n=p^\beta k$ with $n\nmid k$.
We fix $k$ and proceed by mathematical induction on $\beta$.
The base case is $\beta =0$.
It follows that $p\nmid n$ and we have
$$\begin{align*} a^k & \equiv b^k\pmod p\\ a^k b^{n-k-1} & \equiv b^{n-1}\pmod p\\ \sum_{k=0}^{n-1}a^k b^{n-k-1} & \equiv \sum_{k=0}^{n-1}b^{n-1}\pmod p\\ & \equiv nb^{n-1}\pmod p\\ & \not\equiv 0\pmod p. \end{align*}$$

Because $\ds\f{a^n-b^n}{a-b}=\ds\sum_{i=0}^{n-1}a^{n-i-1}b^i$
we get
$\ds\f{a^n-b^n}{a-b}$ is not a multiple of $p$.

Assume that
$p^\beta \|\ds\f{a^n-b^n}{a-b}$.
We want to prove that
$p\|\ds\f{a^{np}-b^{np}}{a^n-b^n}$.
As $p\mid a-b$, we have
$a=b+xp$ and $a^k\equiv b^k+kb^{k-1}xp\pmod {p^2}$
$$\begin{align*} \ds\f{a^{np}-b^{np}}{a^n-b^n} & =\sum_{i=0}^{p-1}a^{ni}b^{n(p-i-1)} \equiv \sum_{i=0}^{p-1}(b^{ni}+nixpb^{ni-1})b^{n(p-i-1)}\\ & \equiv pb^{n(p-1)}+\ds\f{p^2(p-1)nx}{2}b^{n(p-1)-1} \equiv pb^{n(p-1)}\pmod {p^2}. \end{align*}$$
$$p\|\ds\f{a^{np}-b^{np}}{a^n-b^n}.$$
Finally,
$$p^\beta \cdot p\|\ds\f{a^n-b^n}{a-b}\cdot \ds\f{a^{np}-b^{np}}{a^n-b^n} \Lr p^{\beta +1}\|\ds\f{a^{np}-b^{np}}{a-b}.$$

The lemma is proven.

{\bf Lemma 2.}
{\it Now we show the corresponding case for $p=2$.
Let $a,b,n\in \mathbb{Z}$ with $a,b$ odd, such that
$2^\alpha \|\ds\f{a^2-b^2}{2}$ and $2^\beta \|n$ with $\beta \ge 1$.
Then
$$2^{\beta +\alpha }\|a^n-b^n.$$

}

{\bf Proof.}
Again it is enough to prove that if
$2\|\ds\f{a^2-b^2}{2}$ and $2^\beta \|n$, $\beta \ge 1$, then
$$2^{\beta -1}\|\ds\f{a^n-b^n}{a^2-b^2}.$$

Assume $n=2^\beta m$, where $m$ is odd.
We fix $m$ and proceed by mathematical induction on $\beta$.
The base case is $\beta =1$ or $n=2m$.
From $2\mid \ds\f{a^2-b^2}{2}$ we get $2\mid a-b$.
Therefore
$$\begin{align*} a & \equiv b\pmod 2\\ a^{2m-2i-2}b^{2i} & b^{2m-2}\pmod 2\\ \sum_{i=0}^{2m-2}a^{2m-2i-2}b^{2i} & \equiv mb^{m-1}\pmod 2\\ & \equiv 1\pmod 2 \end{align*}$$

Because $\ds\f{a^{2m}-b^{2m}}{a^2-b^2}=\ds\sum_{i=0}^{2m-2}a^{2m-2i-2}b^{2i}$
we get
$\ds\f{a^{2m}-b^{2m}}{a^2-b^2}$
is an odd number that is equivalent to
$2^0\|\ds\f{a^{2m}-b^{2m}}{a^2-b^2}$.
Assume that
$$2^{\beta -1}\|\ds\f{a^n-b^n}{a^2-b^2}.$$

We know that $a$ and $b$ are odd, and $n$ is even, thus
$$\begin{align*} a^n & \equiv 1\pmod 4\\ b^n & \equiv 1\pmod 4\\ a^n+b^n & \equiv 2\pmod 4 \end{align*}$$

It follows that 2 is the greatest power of 2 that divides $a^n+b^n$ or $2\|a^n+b^n$.
Multiplying this result with the induction hypothesis we obtain
$$2^\beta \|\ds\f{a^n-b^n}{a^2-b^2}(a^n+b^n) =\ds\f{a^{2n}-b^{2n}}{a^2-b^2}(a^n+b^n).$$

The extended case of the lemma is proven.

{\bf Remark 1.}
Note that if $\beta =0$ this version of the lemma is only true if $4\mid a-b$.

We continue with problems that exemplify the use of this lemma.

{\bf Problem 1.}
{\it Find the least positive integer $n$ satisfying:
$2^{2007}\mid 17^n-1$.
}

{\bf Solution.}
We have
$2^4\|\ds\f{17^2-1}{2}$.
Suppose $2^\alpha \|n$.
By lemma, $2^{4+\alpha }\|17^n-1$.
We want to have
$\alpha +4\ge 2007\Ri \alpha \ge 2003$.
This means that $2^{2003}\mid n$ which implies that $n\ge 2^{2003}$.
Using our lemma we obtain
$2^{2007}\mid 17^{2^{2003}}-1$.
Thus the minimum value of $n$ is $2^{2003}$.

{\bf Problem 2.}
(Russia 1996)
{\it Let $a^n+b^n=p^k$ for positive integers $a,b$ and $k$, where $p$ is an odd prime
and $n>1$ is an odd integer.
Prove that $n$ must be a power of $p$.
}

{\bf Solution.}
We can factor
$p^k=a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+\ldots -ab^{n-2}+b^{n-1})$,
because $n$ is odd.
Therefore $a+b=p^r$ for some positive integer $r$ less or equal to $k$.
Since $a$ and $b$ are positive integers we have $r\ge 1$.
Now suppose that $p^\beta \|n$.
Using our lemma we get
$p^{r+\beta }\|a^n-(-b)^n=a^n+b^n=p^k$.
This last result is equivalent to $p^{r+\beta }\|p^k\Ri \beta =k-r$.
This means that we have to take the least integer $n$ such that $p^\beta \|n$
in order to have $a^n+b^n=p^k$,
because $a^m+b^m\ge a^n+b^n$ for $m>n$.
The least positive integer $n$ such that $p^\beta \|n$ is $p^\beta$.
Thus $n$ must be a power of $p$ and we are done.

{\bf Problem 3.}
(IMO 1990)
{\it Find all positive integers $n$ such that $2^n\mid 2^n+1$.
}

{\bf Solution.}
Note that $n$ must be odd because $2^n+1$ is always odd.
Let $p_1$ be the smallest prime divisor of $n$.
We have $2^{2n}\equiv 1\pmod {p_1}$.
Now let $d=\ord_{p_1}2$.
Clearly $d<p_1$, $d\mid 2n$ and $\gcd(n,d)=1$,
because $p_1$ is the least prime that divides $n$.
Knowing that, we obtain $d\mid 2$, which implies that $d=1$ or $d=2$.
If $d=1$ we get $p_1\mid 1$ which is absurd.
Thus $d=2$ and $p_1\mid 3\Ri p_1=3$.

Let us apply our lemma:
$3\|2-(-1)$
and we suppose that $3^\beta \|n$.
Therefore
$3^{\beta +1}\|2^n-(-1)^n=2^n+1^n$.
We have
$3^{2\beta }\mid 3^{\beta +1}\|2^n+1$.
This means that
$2\beta \le \beta +1$ or equivalently $\beta \le 1$.
Thus $3\|n$ and we can write $n=3n'$ with $\gcd(3,n')=1$.
Let $p_2$ be the smallest prime that divides $n'$.
We have $2^{6n'}\equiv 1\pmod {p_2}$.
Letting $d_2=\ord_{p_2}2$ we get $d_2<p_2$ and $d_2\mid 6n'$.
But $\gcd(d_2,n')=1$, thus $d_2\mid 6$.
Clearly $d_2$ can't be 1 or 2 as we proved before.
It follows that $d_2=3$ or $d_2=6$.
If $d_2=3$ we have $p_2\mid 7\Ri p_2=7$.
If $d_2=6$ we have $p_2\mid 63=7\cdot 9\Ri p_2\mid 7$;
hence $p_2=7$.
Note that $2^3\equiv 1\pmod 7\Ri 2^{k+3}\equiv 2^k\pmod 7$.
Observe that $2^1\equiv 2\pmod 7$ and $2^2\equiv 4\pmod 7$.
This means
$2^k\equiv -1\pmod 7$ does not have solution in integers.
Thus
$7\nmid 2^{7k}+1$ for any $k$, and $p_2$ does not exist.
Finally we obtain that the only prime divisor of $n$ is 3 and
$3\|n\Ri n=3$.
It follows that the only solution are $n=1$ and $n=3$.

{\bf Problem 4.}
(IMO 2000)
{\it Does there exist a positive integer $n$ such that $n$ has exactly $2000$
prime divisors and $n$ divides $2^n+1$?
}

{\bf Solution.}
We will prove by induction on $k$ that there exists $n$ with exactly $k$ prime
divisors such that $n\mid 2^n+1$.
Before we start the induction, we observe that the divisors of $n$ will be odd,
because $2^n+1$ is odd for all positive integers $n$.
The base case is $k=1$.
Nine has just one prime divisor and $9\mid 2^9+1=513$.
It also happens that $19\mid 513$.
Suppose that for $k=t$ there is $n_t$ such that $n_t\mid 2^{n_t}+1$ and there exists $p_t$
such that $p_t\mid 2^{n_t}+1$, $\gcd(n_t,p_t)=1$.
We will prove that there exist $n_{t+1}$ with $t+1$ prime divisors such that
$n_{t+1}\mid 2^{n_{t+1}}+1$ and that there is also a prime $p_{t+1}$ such that
$p_{t+1}\mid 2^{n_{t+1}}+1$ and $\gcd(n_{t+1},p_{t+1})=1$.
We will also prove that $n_{t+1}=n_t p_t$.
As $\gcd(n,p_t)=1\Ri n_t p_t\mid 2^{n_t}+1\mid 2^{n_t p_t}+1$, thus $n_{t+1}=n_t p_t$ works.
We will apply our lemma to prove that $p_{t+1}$ exists.
Let $q$ be a prime divisor of $n_t$.
Suppose $q^\alpha \|2^{n_t}+1$ and we have $q^0\|p_t$.
The lemma tells us that $q^\alpha \|2^{n_{t+1}}+1$.
Now suppose that $p_t^\beta \|2^{n_t}+1$ and we know $p_t\|p_t$.
The lemma tells us that
$p_t^{\beta +1}\|2^{n_{t+1}}+1$.
This means that all we have to prove is
$$p_t(2^{n_t}+1)<2^{n_{t+1}}+1=(2^{n_t}+1)(2^{n_t(p_t-1)}-2^{n_t(p_t-2)}+\ldots - 2^{n_t}+1).$$

This is equivalent to
$$p_t<2^{n_t(p_t-1)}-2^{n_t(p_t-2)}+\ldots -2^{n_t}+1.$$
We have
$$2^{n_t(p_t-1)}-2^{n_t(p-2)}+\ldots -2^{n_t}+1>\ds\f{p_t-3}{2}2^{n_t}+1 \ge 2^8(p_t-3)+1>p_t.$$

This means that there exists a prime $p_{t+1}$ such that $(n_{t+1},p_{t+1})=1$
and $p_{t+1}\mid 2^{n_{t+1}}$ because $p_t>3$.
The problem is solved.

{\bf Problem 5.}
{\it Let $a\ge 3$ be an integer.
Prove that there exists an integer $n$ with exactly $2007$ prime divisors such that
$n\mid a^n-1$.
}

We use mathematical induction on the number of divisors.
This problem is interesting, because we need to combine both lemmas.
For the base case we have to prove there exists a prime $p$ such that $p\mid a-1$
(we will take 2 as the first prime if $a$ is odd).
We will prove that there is a power of $p$ such that
$a^{p^k}-1$
has another prime divisor $q$ that is not $p$.
If $a$ is even we can apply the lemma directly.
We have that the exponent of $p$ in $a^p-1$ is the exponent
of $p$ in $a-1$ plus one.
Thus we need
$p=\ds\sum_{i=0}^{p-1}a^i>p$,
which is impossible.
If $a$ is odd then we have two cases.
If $a > 3$ we have that
$2\mid a^2-1=(a-1)(a+1)$
and there is one odd divisor of $a^2-1$ because
$\gcd(a-1,a+1)=2$.
If $a=3$ we have that $4\mid 3^4-1$ and 5 does also divide it.
This completes the base case.

Suppose that $n_k$ has exactly $k$ prime divisors such that $n_k\mid a^{n_k}-1$ and
there exists $p_k$ such that
$p_k\mid a^{n_k}-1$ with $\gcd(n_k,p_k)=1$.
This means that
$a_kp_k\mid a^{n_k}-1\mid a^{n_kp_k}-1$.
We say that $n_{k+1}=n_kp_k$.
Now we have to prove that there exists a prime $p_{k+1}$ such that
$\gcd(n_{k+1},p_{k+1})=1$ and $p_{k+1}\mid a^{n_{k+1}}-1$.
Let us use our lemma.
Because we have taken 2 as the first prime (when it was possible)
we have no problems with the exponent of 2, as for $k\ge 2$ the
exponent of 2 does not increase (from the second case of the lemma).
Now for any odd prime divisor of $n_k$ the exponents of $p$ in
$a^{n_k}-1$ and $a^{n_{k+1}}-1$
are equal except for $p_k$, whose exponent has increased by one.
We have
$$a^{n_{k+1}}-1=a^{n_kp_k}-1=(a^{n_k}-1)(a^{n_k(p_k-1)}+a^{n_k(p_k-2)}+\ldots +a^{n_k}+1).$$

Thus $p_{k+1}$ does not exist whenever
$p_k=(a^{n_k(p_k-1)}+a^{n_k(p_k-2)}+\ldots +a^{n_k}+1)$
(because the exponent of $p_k$ has increased just by one).
But the last equation
can not hold, because $a > 1$ and the right-hand side has $p_k$ terms added
together, and all except the last term are greater than 1.
Thus the right-hand
side is greater than the left-hand side, which proves the existence of $p_k+1$ and
we are done.

\bigskip
\hfill
{\Large Santiago Cuellar, Bogota, Colombia}

\hfill
{\Large Jose Alejandro Samper, Bogota, Colombia}

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