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.
{f Lemma 1.}
{it Let $p$ be an odd prime.
For two different integers $a$ (with $p
mid a$) and $b$ with
$aequiv bpmod p$
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 $p^amid n$ and $p^{a+1}
mid n$,
where $a,p,nin mathbb{Z}$.
It allows us to state the lemma as follows.
Let $p$ be an odd prime and let $a,b,nin mathbb{Z}$ with $p
mid a$.
Then $p^alpha |a-b$, $alpha ge 1$ and $p^eta |n$ implies
$p^{eta +alpha }|a^n-b^n$.
{f Proof.}
Let us prove that if $aequiv bpmod p$ and $p^eta |n$ then
$p^eta |dsf{a^n-b^n}{a-b}$.
It is clear that the lifting lemma will follow, because with the condition
$p^alpha |a-b$ we have $p^{alpha +eta }|a^n-b^n$.
Assume $n=p^eta k$ with $n
mid k$.
We fix $k$ and proceed by mathematical induction on $eta $.
The base case is $eta =0$.
It follows that $p
mid n$ and we have
egin{align*}
a^k & equiv b^kpmod 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\
&
otequiv 0pmod p.
end{align*}
Because $dsf{a^n-b^n}{a-b}=dssum_{i=0}^{n-1}a^{n-i-1}b^i$
we get
$dsf{a^n-b^n}{a-b}$ is not a multiple of $p$.
Assume that
$p^eta |dsf{a^n-b^n}{a-b}$.
We want to prove that
$p|dsf{a^{np}-b^{np}}{a^n-b^n}$.
As $pmid a-b$, we have
$a=b+xp$ and $a^kequiv b^k+kb^{k-1}xppmod {p^2}$
egin{align*}
dsf{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)}+dsf{p^2(p-1)nx}{2}b^{n(p-1)-1}
equiv pb^{n(p-1)}pmod {p^2}.
end{align*}
$$p|dsf{a^{np}-b^{np}}{a^n-b^n}.$$
Finally,
$$p^eta cdot p|dsf{a^n-b^n}{a-b}cdot dsf{a^{np}-b^{np}}{a^n-b^n}
Lr p^{eta +1}|dsf{a^{np}-b^{np}}{a-b}.$$
The lemma is proven.
{f Lemma 2.}
{it Now we show the corresponding case for $p=2$.
Let $a,b,nin mathbb{Z}$ with $a,b$ odd, such that
$2^alpha |dsf{a^2-b^2}{2}$ and $2^eta |n$ with $eta ge 1$.
Then
$$2^{eta +alpha }|a^n-b^n.$$
}
{f Proof.}
Again it is enough to prove that if
$2|dsf{a^2-b^2}{2}$ and $2^eta |n$, $eta ge 1$, then
$$2^{eta -1}|dsf{a^n-b^n}{a^2-b^2}.$$
Assume $n=2^eta m$, where $m$ is odd.
We fix $m$ and proceed by mathematical induction on $eta $.
The base case is $eta =1$ or $n=2m$.
From $2mid dsf{a^2-b^2}{2}$ we get $2mid a-b$.
Therefore
egin{align*}
a & equiv bpmod 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 1pmod 2
end{align*}
Because $dsf{a^{2m}-b^{2m}}{a^2-b^2}=dssum_{i=0}^{2m-2}a^{2m-2i-2}b^{2i}$
we get
$dsf{a^{2m}-b^{2m}}{a^2-b^2}$
is an odd number that is equivalent to
$2^0|dsf{a^{2m}-b^{2m}}{a^2-b^2}$.
Assume that
$$2^{eta -1}|dsf{a^n-b^n}{a^2-b^2}.$$
We know that $a$ and $b$ are odd, and $n$ is even, thus
egin{align*}
a^n & equiv 1pmod 4\
b^n & equiv 1pmod 4\
a^n+b^n & equiv 2pmod 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
$$2^eta |dsf{a^n-b^n}{a^2-b^2}(a^n+b^n)
=dsf{a^{2n}-b^{2n}}{a^2-b^2}(a^n+b^n).$$
The extended case of the lemma is proven.
{f Remark 1.}
Note that if $eta =0$ this version of the lemma is only true if $4mid a-b$.
We continue with problems that exemplify the use of this lemma.
{f Problem 1.}
{it Find the least positive integer $n$ satisfying:
$2^{2007}mid 17^n-1$.
}
{f Solution.}
We have
$2^4|dsf{17^2-1}{2}$.
Suppose $2^alpha |n$.
By lemma, $2^{4+alpha }|17^n-1$.
We want to have
$alpha +4ge 2007Ri alpha ge 2003$.
This means that $2^{2003}mid n$ which implies that $nge 2^{2003}$.
Using our lemma we obtain
$2^{2007}mid 17^{2^{2003}}-1$.
Thus the minimum value of $n$ is $2^{2003}$.
{f 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 $p$.
}
{f 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 $rge 1$.
Now suppose that $p^eta |n$.
Using our lemma we get
$p^{r+eta }|a^n-(-b)^n=a^n+b^n=p^k$.
This last result is equivalent to $p^{r+eta }|p^kRi eta =k-r$.
This means that we have to take the least integer $n$ such that $p^eta |n$
in order to have $a^n+b^n=p^k$,
because $a^m+b^mge a^n+b^n$ for $m>n$.
The least positive integer $n$ such that $p^eta |n$ is $p^eta $.
Thus $n$ must be a power of $p$ and we are done.
{f Problem 3.}
(IMO 1990)
{it Find all positive integers $n$ such that $2^nmid 2^n+1$.
}
{f 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 $2^{2n}equiv 1pmod {p_1}$.
Now let $d=ord_{p_1}2$.
Clearly $d<p_1$, $dmid 2n$ and $gcd(n,d)=1$,
because $p_1$ is the least prime that divides $n$.
Knowing that, we obtain $dmid 2$, which implies that $d=1$ or $d=2$.
If $d=1$ we get $p_1mid 1$ which is absurd.
Thus $d=2$ and $p_1mid 3Ri p_1=3$.
Let us apply our lemma:
$3|2-(-1)$
and we suppose that $3^eta |n$.
Therefore
$3^{eta +1}|2^n-(-1)^n=2^n+1^n$.
We have
$3^{2eta }mid 3^{eta +1}|2^n+1$.
This means that
$2eta le eta +1$ or equivalently $eta le 1$.
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 $2^{6n'}equiv 1pmod {p_2}$.
Letting $d_2=ord_{p_2}2$ we get $d_2<p_2$ and $d_2mid 6n'$.
But $gcd(d_2,n')=1$, thus $d_2mid 6$.
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 $p_2mid 7Ri p_2=7$.
If $d_2=6$ we have $p_2mid 63=7cdot 9Ri p_2mid 7$;
hence $p_2=7$.
Note that $2^3equiv 1pmod 7Ri 2^{k+3}equiv 2^kpmod 7$.
Observe that $2^1equiv 2pmod 7$ and $2^2equiv 4pmod 7$.
This means
$2^kequiv -1pmod 7$ does not have solution in integers.
Thus
$7
mid 2^{7k}+1$ for any $k$, and $p_2$ does not exist.
Finally we obtain that the only prime divisor of $n$ is 3 and
$3|nRi n=3$.
It follows that the only solution are $n=1$ and $n=3$.
{f Problem 4.}
(IMO 2000)
{it Does there exist a positive integer $n$ such that $n$ has exactly $2000$
prime divisors and $n$ divides $2^n+1$?
}
{f Solution.}
We will prove by induction on $k$ that there exists $n$ with exactly $k$ prime
divisors such that $nmid 2^n+1$.
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 $9mid 2^9+1=513$.
It also happens that $19mid 513$.
Suppose that for $k=t$ there is $n_t$ such that $n_tmid 2^{n_t}+1$ and there exists $p_t$
such that $p_tmid 2^{n_t}+1$, $gcd(n_t,p_t)=1$.
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 $gcd(n,p_t)=1Ri n_t p_tmid 2^{n_t}+1mid 2^{n_t p_t}+1$, 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 $p_t^eta |2^{n_t}+1$ and we know $p_t|p_t$.
The lemma tells us that
$p_t^{eta +1}|2^{n_{t+1}}+1$.
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
$$2^{n_t(p_t-1)}-2^{n_t(p-2)}+ldots -2^{n_t}+1>dsf{p_t-3}{2}2^{n_t}+1
ge 2^8(p_t-3)+1>p_t.$$
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.
{f Problem 5.}
{it Let $age 3$ be an integer.
Prove that there exists an integer $n$ with exactly $2007$ prime divisors such that
$nmid a^n-1$.
}
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 $pmid a-1$
(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
$p=dssum_{i=0}^{p-1}a^i>p$,
which is impossible.
If $a$ is odd then we have two cases.
If $a > 3$ we have that
$2mid a^2-1=(a-1)(a+1)$
and there is one odd divisor of $a^2-1$ because
$gcd(a-1,a+1)=2$.
If $a=3$ we have that $4mid 3^4-1$ and 5 does also divide it.
This completes the base case.
Suppose that $n_k$ has exactly $k$ prime divisors such that $n_kmid a^{n_k}-1$ and
there exists $p_k$ such that
$p_kmid a^{n_k}-1$ with $gcd(n_k,p_k)=1$.
This means that
$a_kp_kmid a^{n_k}-1mid a^{n_kp_k}-1$.
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 $kge 2$ 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.
igskip
hfill
{Large Santiago Cuellar, Bogota, Colombia}
hfill
{Large Jose Alejandro Samper, Bogota, Colombia}
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