Énoncé Partie I. Outils. Partie II. Inégalités. Partie III. Diviseurs

n0n≥n0Kn, 2nJ
p m vp(m)∈Np
m p m
vp(m)≥1vp(mm0) = vp(m) + vp(m0)m m0
n∈N\ {0,1,2}
J1, nKπ(n)
J1, nKPn
Kn, 2nKRnRn= 1
π(8) = 4, P8= 2 ×3×5×7 = 210, P2n=PnRn
π(n)n≤14 n≥17
J17, nK
∀n≥14, π(n)≤n
2−1
a b m
a m
b m
a∧b= 1
⇒ab m
b2xc − 2bxc∈{0,1}xb2xc= 2bxc
{x}=x− bxcT
∀x∈]0,+∞[, T (x)=4x
3√x(2x)−√x
2
ln(T(x)) +∞
ln(T(x)) T+∞
n!n
2n
n2n
n2
3n−√2n
k
(1 + 1)2k+1 2k+1
k≤4k
pJk+ 2,2(k+ 1)Kp2k+1
k
Jk+ 2,2k+ 1K
4k
∀n∈N\ {0,1}, Pn≤4n
λn2(n+1)
n+1 =λn2n n
√1−x≤1−x
2x≤1
∀n∈N\ {0,1},2n
n>4n
2√n
n!
p n
max {vp(m), m ∈J2, nK}=bln n
ln pc
Vp(n)
Vp(2n)−Vp(n)∈ {0,1}εp(n)
q∈K1, nKqJ1, nKbn
qc
i∈J1, Vp(n)K
m∈J2, nKvp(m) = i

m∈J2, nKvp(m) = i
∀n∈N\ {0,1}, vp(n!) =
Vp(n)
X
j=1 bn
pjc
v2(100!) v5(100!)
100!
2n
n
p2n
vp(2n
n)≤
Vp(n)
X
j=1 b2n
pjc − 2bn pjc+εp(n)≤Vp(2n)
pvp((2n
n))≤2n
√2n≤p⇒vp(2n
n)≤1
p∈K2
3n, nK⇒vp(2n n)=0
RnKn, 2nK2n
n
Qn∈N2n
n=QnRn
n≥5Qn2
3n
∀n≥5, Qn≤(2n)π(√2n)42n
3
∀n≥98, Rn≥(2n)−√n
24n
3√n
2,3,5,7,11,13,17
π
n
π(n)
n
2−1
n≥17 J17, nK2k+ 1
17 ≤2k+ 1 ≤n⇔8≤k≤n−1
2
bn−1
2c − 7 2
15 16 π(n)
π(n)≤π(14) + J17, nK
≤6 + bn−1
2c − 7=bn−1 2c − 1≤n 2−1
α β u v
m=αa =βb ua +vb = 1
α b α
α=uaα +vbα =um +vbα =uβb +vbα = (uβ +vα)b⇒m=αa = (uβ +vα)ab
x−1<bxc ≤ x× −2
2x−1<b2xc ≤ 2x× −2
−1<b2xc − 2bxc<2
b2xc − 2bxc ∈ Zb2xc − 2bxc ∈ {0,1}
{x}
2x= 2bxc+ 2 {x}2bxc=b2xc ⇔ 2{x} ∈ [0,1[⇔ {x} ∈ [0,1
2[

ln(T(x))
+∞+∞
ln(T(x)) = ln 4
3x−1
√2√xln x−ln 2 √2√x+1
2ln x
ln(T(x)) ∼ln 4
3x T +∞+∞
m≤n m
n!m≤n
n
2n
n(2n)! (2n)! = (n!)22n np2n n
(2n)! p≤2n
2
3n−√2n
4n2−18n= 2n(2n−9)
n= 3 4 n≥5
(1 + 1)2k+1
2k+1
k  2k+1 2k+1−k
22k+ 1
k=2k+ 1 k+2k+ 1 2k+ 1 −k≤(1 + 1)2k+1 = 22k+1 = 2 ×4k
⇒2k+ 1
k≤4k
2k+ 1
k=
k
z }| {
(2k+ 1)(2k)···
k!=(2k+ 1) ···(k+ 2) k!
2k+ 1 −k+ 1 = k+ 2 (2k+ 1) ···(k+ 2) 2k+1
k
Jk+ 2,2k+ 1K(k+ 2) ···(2k+ 1)
2k+1
k
Jk+2,2k+1K2k+1
k
2k+1
k 2k+1 k
(In) : Pn≤4nn
P2= 2 ≤42= 16, P3= 6 ≤43, P4= 6 ≤44,···
(In)⇒(In+1)
n= 2k+ 1 n+ 1 Pn+1 =Pn≤
4n≤4n+1
n= 2k n + 1 = 2k+ 1
k+ 1 Jk+ 2,2k+ 1K
Pn+1 =P2k+1 ≤Pk+1 4k≤42k+1 = 4n+1
Pk+1
λn=(2n+ 2)!
((n+ 1)!)2
(n!)2
(2n)! =(2n+ 1)(2n+ 2) (n+ 1)2=2(2n+ 1) n+ 1
f]− ∞,1]
f(x) = √1−x−1 + x
2
]− ∞,1] ] − ∞,1[
∀x∈]0,1[, f0(x) = 1
2−1
√1−x+ 1<0 0 <1−x < 1
[0,1]
]− ∞,0] f(0) = 0 ] − ∞,1]
In:2n
n≥4n
2√nn= 2
2n
n=4 2= 6 4n
2√n=42
2√2= 4√2⇒ I262= 36 ≥(4√2)2= 32

In⇒ In+1
2(n+ 1)
n+ 1 =2(2n+ 1) n+ 1 2n n≥2(2n+ 1) n+ 1
4n
2√n
In+1
Tn=
2(2n+1)
n+1
4n
2√n
4n+1
2√n+1 ≥1
Tn=2(2n+ 1)√n+ 1
(n+ 1)4√n=2n+ 1
2p(n+ 1)n=1 + 1
2n
q1 + 1
n
Tn≥1f(−1
n)≤0
n!
pJ2, nKVp(n)vp(m)m∈J2, nK
m vp(m) = Vp(n)m
pVp(n)m⇒pVp(n)≤m≤n⇒Vp(n) ln p≤ln n⇒Vp(n)≤ln n
ln p
⇒Vp(n)≤ bln n
ln pc
k=bln n
ln pc ∈ Npk≤n m =pkvp(m) = k
Vp(n)≥k=bln n
ln pc
Vp(n)≤Vp(2n) = bln 2 + ln n
ln pc ≤ 1 + bln n ln pc=Vp(n)+1
0<ln 2
ln p≤1
q∈J2, nKqJ1, nKkq
q≤kq ≤n⇔1≤k≤n
q⇔1≤k≤ bn qc
≤ bn
qc
i1≤i≤Vp(n)pi≤n
m∈J2, nKvp(m) = i pim pi+1
m
J1, nKMipiMi+1
pi+1
vp(m) = i⇔m∈ Mim /∈ Mi+1
Mi+1 ⊂ Mi
]{m∈J1, nKvp(m) = i}=]Mi−]Mi+1 =bn
qic−b n qi+1 c
]Ω Ω
vp(n!)
vp(n!) =
n
X
m=2
vp(m)
m vp(m) 1 Vp(n)
i m vp(m) = i
vp(n!) =
Vp(n)
X
i=0
i×(m vp(m) = i) =
Vp(n)
X
i=0
ibn
pic−b n pi+1 c
vp(n!) =
Vp(n)
X
i=0
ibn
pic −
Vp(n)+1
X
i=1
(i−1)bn
pic=
Vp(n)
X
i=1 bn
pic
i= 0 i
i=Vp(n)+1 pi> n
V2(100) i2i≤100 6
26= 64 v2(100) v5(100!)
i
2i
b100
2ic
i
2i
b100
2ic

v2(100!) = 50 + 25 + 12 + 6 + 3 + 1 = 97 v5(200!) = 20 + 4 = 24
100!
24
2n
n
2n
n=(2n)! (n!)2⇒vp(2n n) = vp((2n)!) −2vp(n!)
Vp(2n) = Vp(n)Vp(n)+1
vp(2n
n) =
Vp(n)
X
i=1 b2n
pic − 2bn pic+
0Vp(2n) = Vp(n)
b2n
pVp(2n)cVp(2n) = Vp(n)+1
k=Vp(n)Vp(2n) = k+ 1
0 1
k=Vp(n)⇒pk≤n<pk+1 ⇒2n < 2pk+1 ⇒2n
pk+1 <2⇒ b 2n pk+1 c∈{0,1}
Vp(n)
X
i=1 b2n
pic − 2bn pic
| {z }
= 0 1
+εp(n)≤Vp(n) + Vp(2n)−Vp(n) = Vp(2n)
Vp(n)
pvp((2n
n))≤pVp(2n)≤pbln(2n)
ln pc≤pln(2n) ln p= 2n
√2n≤p⇒1
2ln(2n)≤ln p⇒ln n ln p≤2−ln 2 ln p⇒Vp(n) = bln n ln pc ≤ 1
⇒vp(2n
n)≤Vp(n)≤1
2
3n<p⇒ln 2 + ln n≤ln p+ ln 3 ⇒ln n ln p≤ln(3
2)
ln p<1⇒Vp(n)=0⇒vp(2n n) = 0
n!2n
n= (2n)(2n−1) ···(n+ 1)
Jn+ 1,2nK(2n)(2n−1) ···(n+ 1)
n!2n
nRn
n≥5√2n≤2
3n
n√2n
QnQn
n
J2
3n, nK
Qn2
3n
Qn
J2,√2nKp
pvp((2n
n))≤2n
π(√2n)
≤(2n)π(√2n)
J√2n, nKp
p≤2
3n vp(2n n)=1
≤Pb2
3nc≤42
3n