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T h i s i s a n o l d r e v i s i o n o f t h i s p a g e , a s e d i t e d b y S t o w g u l l ( t a l k | c o n t r i b s ) at 2 0 : 2 8 , 1 6 M a r c h 2 0 2 4 . T h e p r e s e n t a d d r e s s ( U R L ) i s a p e r m a n e n t l i n k t o t h i s r e v i s i o n , w h i c h m a y d i f f e r s i g n i f i c a n t l y f r o m t h e c u r r e n t r e v i s i o n .
( d i f f ) ← P r e v i o u s r e v i s i o n | L a t e s t r e v i s i o n ( d i f f ) | N e w e r r e v i s i o n → ( d i f f )
In elementary number theory , the lifting-the-exponent lemma (LTE lemma ) provides several formulas for computing the p-adic valuation
ν
p
{\displaystyle \nu _{p}}
of special forms of integers. The lemma is named as such because it describes the steps necessary to "lift" the exponent of
p
{\displaystyle p}
in such expressions. It is related to Hensel's lemma .
Background
The exact origins of the LTE lemma are unclear; the result, with its present name and form, has only come into focus within the last 10 to 20 years.[1] However, several key ideas used in its proof were known to Gauss and referenced in his Disquisitiones Arithmeticae .[2] Despite chiefly featuring in mathematical olympiads , it is sometimes applied to research topics, such as elliptic curves .[3] [4]
Statements
For any integers
x
,
y
{\displaystyle x,y}
, a positive integer
n
{\displaystyle n}
, and a prime number
p
{\displaystyle p}
such that
p
∤
x
{\displaystyle p\nmid x}
and
p
∤
y
{\displaystyle p\nmid y}
, the following statements hold:
When
p
{\displaystyle p}
is odd:
If
p
∣
x
−
y
{\displaystyle p\mid x-y}
, then
ν
p
(
x
n
−
y
n
)
=
ν
p
(
x
−
y
)
+
ν
p
(
n
)
{\displaystyle \nu _{p}(x^{n}-y^{n})=\nu _{p}(x-y)+\nu _{p}(n )}
.
If
n
{\displaystyle n}
is odd and
p
∣
x
+
y
{\displaystyle p\mid x+y}
, then
ν
p
(
x
n
+
y
n
)
=
ν
p
(
x
+
y
)
+
ν
p
(
n
)
{\displaystyle \nu _{p}(x^{n}+y^{n})=\nu _{p}(x+y)+\nu _{p}(n )}
.
When
p
=
2
{\displaystyle p=2}
:
If
2
∣
x
−
y
{\displaystyle 2\mid x-y}
and
n
{\displaystyle n}
is even, then
ν
2
(
x
n
−
y
n
)
=
ν
2
(
x
−
y
)
+
ν
2
(
x
+
y
)
+
ν
2
(
n
)
−
1
{\displaystyle \nu _{2}(x^{n}-y^{n})=\nu _{2}(x-y)+\nu _{2}(x+y)+\nu _{2}(n )-1}
.
If
2
∣
x
−
y
{\displaystyle 2\mid x-y}
and
n
{\displaystyle n}
is odd, then
ν
2
(
x
n
−
y
n
)
=
ν
2
(
x
−
y
)
{\displaystyle \nu _{2}(x^{n}-y^{n})=\nu _{2}(x-y)}
. (Follows from the general case below.)
Corollaries:
If
4
∣
x
−
y
{\displaystyle 4\mid x-y}
, then
ν
2
(
x
+
y
)
=
1
{\displaystyle \nu _{2}(x+y)=1}
and thus
ν
2
(
x
n
−
y
n
)
=
ν
2
(
x
−
y
)
+
ν
2
(
n
)
{\displaystyle \nu _{2}(x^{n}-y^{n})=\nu _{2}(x-y)+\nu _{2}(n )}
.
If
2
∣
x
+
y
{\displaystyle 2\mid x+y}
and
n
{\displaystyle n}
is even, then
ν
2
(
x
n
+
y
n
)
=
1
{\displaystyle \nu _{2}(x^{n}+y^{n})=1}
.
If
2
∣
x
+
y
{\displaystyle 2\mid x+y}
and
n
{\displaystyle n}
is odd, then
ν
2
(
x
n
+
y
n
)
=
ν
2
(
x
+
y
)
{\displaystyle \nu _{2}(x^{n}+y^{n})=\nu _{2}(x+y)}
.
For all
p
{\displaystyle p}
:
If
gcd
(
n
,
p
)
=
1
{\displaystyle \gcd(n,p)=1}
and
p
∣
x
−
y
{\displaystyle p\mid x-y}
, then
ν
p
(
x
n
−
y
n
)
=
ν
p
(
x
−
y
)
{\displaystyle \nu _{p}(x^{n}-y^{n})=\nu _{p}(x-y)}
.
If
gcd
(
n
,
p
)
=
1
{\displaystyle \gcd(n,p)=1}
,
p
∣
x
+
y
{\displaystyle p\mid x+y}
and
n
{\displaystyle n}
odd, then
ν
p
(
x
n
+
y
n
)
=
ν
p
(
x
+
y
)
{\displaystyle \nu _{p}(x^{n}+y^{n})=\nu _{p}(x+y)}
.
Outline of proof
Base case
The base case
ν
p
(
x
n
−
y
n
)
=
ν
p
(
x
−
y
)
{\displaystyle \nu _{p}(x^{n}-y^{n})=\nu _{p}(x-y)}
when
gcd
(
n
,
p
)
=
1
{\displaystyle \gcd(n,p)=1}
is proven first. Because
p
∣
x
−
y
⟺
x
≡
y
(
mod
p
)
{\displaystyle p\mid x-y\iff x\equiv y{\pmod {p}}}
,
x
n
−
1
+
x
n
−
2
y
+
x
n
−
3
y
2
+
⋯
+
y
n
−
1
≡
n
x
n
−
1
≢
0
(
mod
p
)
(
1
)
{\displaystyle x^{n-1}+x^{n-2}y+x^{n-3}y^{2}+\dots +y^{n-1}\equiv nx^{n-1}\not \equiv 0{\pmod {p}}\ (1 )}
The fact that
x
n
−
y
n
=
(
x
−
y
)
(
x
n
−
1
+
x
n
−
2
y
+
x
n
−
3
y
2
+
⋯
+
y
n
−
1
)
{\displaystyle x^{n}-y^{n}=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^{2}+\dots +y^{n-1})}
completes the proof. The condition
ν
p
(
x
n
+
y
n
)
=
ν
p
(
x
+
y
)
{\displaystyle \nu _{p}(x^{n}+y^{n})=\nu _{p}(x+y)}
for odd
n
{\displaystyle n}
is similar.
General case (odd p )
Via the binomial expansion , the substitution
y
=
x
+
k
p
{\displaystyle y=x+kp}
can be used in (1 ) to show that
ν
p
(
x
p
−
y
p
)
=
ν
p
(
x
−
y
)
+
1
{\displaystyle \nu _{p}(x^{p}-y^{p})=\nu _{p}(x-y)+1}
because (1 ) is a multiple of
p
{\displaystyle p}
but not
p
2
{\displaystyle p^{2}}
.[1] Likewise,
ν
p
(
x
p
+
y
p
)
=
ν
p
(
x
+
y
)
+
1
{\displaystyle \nu _{p}(x^{p}+y^{p})=\nu _{p}(x+y)+1}
.
Then, if
n
{\displaystyle n}
is written as
p
a
b
{\displaystyle p^{a}b}
where
p
∤
b
{\displaystyle p\nmid b}
, the base case gives
ν
p
(
x
n
−
y
n
)
=
ν
p
(
(
x
p
a
)
b
−
(
y
p
a
)
b
)
=
ν
p
(
x
p
a
−
y
p
a
)
{\displaystyle \nu _{p}(x^{n}-y^{n})=\nu _{p}((x^{p^{a}})^{b}-(y^{p^{a}})^{b})=\nu _{p}(x^{p^{a}}-y^{p^{a}})}
.
By induction on
a
{\displaystyle a}
,
ν
p
(
x
p
a
−
y
p
a
)
=
ν
p
(
(
(
…
(
x
p
)
p
…
)
)
p
−
(
(
…
(
y
p
)
p
…
)
)
p
)
(exponentiation used
a
times per term)
=
ν
p
(
x
−
y
)
+
a
{\displaystyle {\begin{aligned}\nu _{p}(x^{p^{a}}-y^{p^{a}})&=\nu _{p}(((\dots (x^{p})^{p}\dots ))^{p}-((\dots (y^{p})^{p}\dots ))^{p})\ {\text{(exponentiation used }}a{\text{ times per term)}}\\&=\nu _{p}(x-y)+a\end{aligned}}}
A similar argument can be applied for
ν
p
(
x
n
+
y
n
)
{\displaystyle \nu _{p}(x^{n}+y^{n})}
.
General case (p = 2)
The proof for the odd
p
{\displaystyle p}
case cannot be directly applied when
p
=
2
{\displaystyle p=2}
because the binomial coefficient
(
p
2
)
=
p
(
p
−
1
)
2
{\displaystyle {\binom {p}{2}}={\frac {p(p-1)}{2}}}
is only an integral multiple of
p
{\displaystyle p}
when
p
{\displaystyle p}
is odd.
However, it can be shown that
ν
2
(
x
n
−
y
n
)
=
ν
2
(
x
−
y
)
+
ν
2
(
n
)
{\displaystyle \nu _{2}(x^{n}-y^{n})=\nu _{2}(x-y)+\nu _{2}(n )}
when
4
∣
x
−
y
{\displaystyle 4\mid x-y}
by writing
n
=
2
a
b
{\displaystyle n=2^{a}b}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are integers with
b
{\displaystyle b}
odd and noting that
ν
2
(
x
n
−
y
n
)
=
ν
2
(
(
x
2
a
)
b
−
(
y
2
a
)
b
)
=
ν
2
(
x
2
a
−
y
2
a
)
=
ν
2
(
(
x
2
a
−
1
+
y
2
a
−
1
)
(
x
2
a
−
2
+
y
2
a
−
2
)
⋯
(
x
2
+
y
2
)
(
x
+
y
)
(
x
−
y
)
)
=
ν
2
(
x
−
y
)
+
a
{\displaystyle {\begin{aligned}\nu _{2}(x^{n}-y^{n})&=\nu _{2}((x^{2^{a}})^{b}-(y^{2^{a}})^{b})\\&=\nu _{2}(x^{2^{a}}-y^{2^{a}})\\&=\nu _{2}((x^{2^{a-1}}+y^{2^{a-1}})(x^{2^{a-2}}+y^{2^{a-2}})\cdots (x^{2}+y^{2})(x+y)(x-y))\\&=\nu _{2}(x-y)+a\end{aligned}}}
because since
x
≡
y
≡
±
1
(
mod
4
)
{\displaystyle x\equiv y\equiv \pm 1{\pmod {4}}}
, each factor in the difference of squares step in the form
x
2
k
+
y
2
k
{\displaystyle x^{2^{k}}+y^{2^{k}}}
is congruent to 2 modulo 4.
The stronger statement
ν
2
(
x
n
−
y
n
)
=
ν
2
(
x
−
y
)
+
ν
2
(
x
+
y
)
+
ν
2
(
n
)
−
1
{\displaystyle \nu _{2}(x^{n}-y^{n})=\nu _{2}(x-y)+\nu _{2}(x+y)+\nu _{2}(n )-1}
when
2
∣
x
−
y
{\displaystyle 2\mid x-y}
is proven analogously.[1]
In competitions
Example problem
The LTE lemma can be used to solve 2020 AIME I #12:
Let
n
{\displaystyle n}
be the least positive integer for which
149
n
−
2
n
{\displaystyle 149^{n}-2^{n}}
is divisible by
3
3
⋅
5
5
⋅
7
7
.
{\displaystyle 3^{3}\cdot 5^{5}\cdot 7^{7}.}
Find the number of positive integer divisors of
n
{\displaystyle n}
.[5]
Solution. Note that
149
−
2
=
147
=
3
⋅
7
2
{\displaystyle 149-2=147=3\cdot 7^{2}}
. Using the LTE lemma, since
3
∤
149
{\displaystyle 3\nmid 149}
and
2
{\displaystyle 2}
but
3
∣
147
{\displaystyle 3\mid 147}
,
ν
3
(
149
n
−
2
n
)
=
ν
3
(
147
)
+
ν
3
(
n
)
=
ν
3
(
n
)
+
1
{\displaystyle \nu _{3}(149^{n}-2^{n})=\nu _{3}(147)+\nu _{3}(n )=\nu _{3}(n )+1}
. Thus,
3
3
∣
149
n
−
2
n
⟺
3
2
∣
n
{\displaystyle 3^{3}\mid 149^{n}-2^{n}\iff 3^{2}\mid n}
. Similarly,
7
∤
149
,
2
{\displaystyle 7\nmid 149,2}
but
7
∣
147
{\displaystyle 7\mid 147}
, so
ν
7
(
149
n
−
2
n
)
=
ν
7
(
147
)
+
ν
7
(
n
)
=
ν
7
(
n
)
+
2
{\displaystyle \nu _{7}(149^{n}-2^{n})=\nu _{7}(147)+\nu _{7}(n )=\nu _{7}(n )+2}
and
7
7
∣
149
n
−
2
n
⟺
7
5
∣
n
{\displaystyle 7^{7}\mid 149^{n}-2^{n}\iff 7^{5}\mid n}
.
Since
5
∤
147
{\displaystyle 5\nmid 147}
, the factors of 5 are addressed by noticing that since the residues of
149
n
{\displaystyle 149^{n}}
modulo 5 follow the cycle
4
,
1
,
4
,
1
{\displaystyle 4,1,4,1}
and those of
2
n
{\displaystyle 2^{n}}
follow the cycle
2
,
4
,
3
,
1
{\displaystyle 2,4,3,1}
, the residues of
149
n
−
2
n
{\displaystyle 149^{n}-2^{n}}
modulo 5 cycle through the sequence
2
,
2
,
1
,
0
{\displaystyle 2,2,1,0}
. Thus,
5
∣
149
n
−
2
n
{\displaystyle 5\mid 149^{n}-2^{n}}
iff
n
=
4
k
{\displaystyle n=4k}
for some positive integer
k
{\displaystyle k}
. The LTE lemma can now be applied again:
ν
5
(
149
4
k
−
2
4
k
)
=
ν
5
(
(
149
4
)
k
−
(
2
4
)
k
)
=
ν
5
(
149
4
−
2
4
)
+
ν
5
(
k
)
{\displaystyle \nu _{5}(149^{4k}-2^{4k})=\nu _{5}((149^{4})^{k}-(2^{4})^{k})=\nu _{5}(149^{4}-2^{4})+\nu _{5}(k )}
. Since
149
4
−
2
4
≡
(
−
1
)
4
−
2
4
≡
−
15
(
mod
25
)
{\displaystyle 149^{4}-2^{4}\equiv (-1)^{4}-2^{4}\equiv -15{\pmod {25}}}
,
ν
5
(
149
4
−
2
4
)
=
1
{\displaystyle \nu _{5}(149^{4}-2^{4})=1}
. Hence
5
5
∣
149
n
−
2
n
⟺
5
4
∣
k
⟺
4
⋅
5
4
∣
n
{\displaystyle 5^{5}\mid 149^{n}-2^{n}\iff 5^{4}\mid k\iff 4\cdot 5^{4}\mid n}
.
Combining these three results, it is found that
n
=
2
2
⋅
3
2
⋅
5
4
⋅
7
5
{\displaystyle n=2^{2}\cdot 3^{2}\cdot 5^{4}\cdot 7^{5}}
, which has
(
2
+
1
)
(
2
+
1
)
(
4
+
1
)
(
5
+
1
)
=
270
{\displaystyle (2+1)(2+1)(4+1)(5+1)=270}
positive divisors.
References
^ Gauss, C. (1801) Disquisitiones arithmeticae. Results shown in Articles 86–87. https://gdz.sub.uni-goettingen.de/id/PPN235993352?tify={%22pages%22%3A%5B70%5D}
^ Geretschläger, R. (2020). Engaging Young Students in Mathematics through Competitions – World Perspectives and Practices. World Scientific. https://books.google.com/books?id=FNPkDwAAQBAJ&pg=PP1
^ Heuberger, C. and Mazzoli, M. (2017). Elliptic curves with isomorphic groups of points over finite field extensions. Journal of Number Theory, 181 , 89–98. https://doi.org/10.1016/j.jnt.2017.05.028
^ 2020 AIME I Problems. (2020). Art of Problem Solving. Retrieved July 11, 2020, from https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems
R e t r i e v e d f r o m " https://en.wikipedia.org/w/index.php?title=Lifting-the-exponent_lemma&oldid=1214075467 "
C a t e g o r y :
● L e m m a s i n n u m b e r t h e o r y
H i d d e n c a t e g o r i e s :
● A r t i c l e s w i t h s h o r t d e s c r i p t i o n
● S h o r t d e s c r i p t i o n m a t c h e s W i k i d a t a
● T h i s p a g e w a s l a s t e d i t e d o n 1 6 M a r c h 2 0 2 4 , a t 2 0 : 2 8 ( U T C ) .
● T h i s v e r s i o n o f t h e p a g e h a s b e e n r e v i s e d . B e s i d e s n o r m a l e d i t i n g , t h e r e a s o n f o r r e v i s i o n m a y h a v e b e e n t h a t t h i s v e r s i o n c o n t a i n s f a c t u a l i n a c c u r a c i e s , v a n d a l i s m , o r m a t e r i a l n o t c o m p a t i b l e w i t h t h e C r e a t i v e C o m m o n s A t t r i b u t i o n - S h a r e A l i k e L i c e n s e .
● P r i v a c y p o l i c y
● A b o u t W i k i p e d i a
● D i s c l a i m e r s
● C o n t a c t W i k i p e d i a
● C o d e o f C o n d u c t
● D e v e l o p e r s
● S t a t i s t i c s
● C o o k i e s t a t e m e n t
● M o b i l e v i e w