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{SECT 0 {EXCHG {PARA 261 "" 0 "" {TEXT 261 34 "TP 5 : INSTRUCTION IF F
I - CORRIG\311" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 "1" }}}{EXCHG 
{PARA 256 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 122 "
for k from 0 to 1000 do n:=k*(k+1)+1 :\nif is(root(n,3),integer) then \+
\nprintf(`%a.%a + 1 = %a^3\\n`,k,k+1,root(n,3)) fi od:\n" }}{PARA 6 "
" 1 "" {TEXT -1 13 "0.1 + 1 = 1^3" }}{PARA 6 "" 1 "" {TEXT -1 15 "18.1
9 + 1 = 7^3" }}}{EXCHG {PARA 260 "" 0 "" {TEXT 262 88 "La d\351monstra
tion du fait qu'il n'y a pas d'autres possibilt\351s est extr\352memen
t difficile." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 32 "2  Nombres parfaits et abondants" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 75 "with(numtheory):\nfor n to 1000 do if sigma(n)
 = 2*n then print(n) fi od : \n" }{TEXT -1 35 "Nombres parfaits entre \+
1 et 1000 : " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"'" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#\"#G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"$'\\" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "for n by 2 while sigma(n) <
= 2*n do od : n;\n" }{TEXT -1 25 "Premier abondant impair :" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#\"$X*" }}}{PARA 0 "" 0 "" {TEXT -1 139 "Rema
rque : on sait que si un nombre parfait impair existe, il a au moins 3
5 chiffres, mais il n'a pas \351t\351 prouv\351 qu'il n'y en avait pas
...." }}{EXCHG {PARA 260 "" 0 "" {TEXT 257 30 "3  Nombres premiers jum
eaux :\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "p:=2: while p<=100 do i
f isprime(p+2) then\nprintf(`%a et %a ; `,p,p+2) fi ; p:=nextprime(p) \+
  od:" }}{PARA 6 "" 1 "" {TEXT -1 84 "3 et 5 ; 5 et 7 ; 11 et 13 ; 17 \+
et 19 ; 29 et 31 ; 41 et 43 ; 59 et 61 ; 71 et 73 ; " }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 100 "for p from 1000 to 1100 do p:=nextprime(
p) ; if isprime(p+2) then\nprintf(`%a et %a ; `,p,p+2) fi od:" }}
{PARA 6 "" 1 "" {TEXT -1 75 "1019 et 1021 ; 1031 et 1033 ; 1049 et 105
1 ; 1061 et 1063 ; 1091 et 1093 ; " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 102 "for p from 10000 to 10100 do p:=nextprime(p) ; if is
prime(p+2) then\nprintf(`%a et %a ; `,p,p+2) fi od:" }}{PARA 6 "" 1 "
" {TEXT -1 68 "10007 et 10009 ; 10037 et 10039 ; 10067 et 10069 ; 1009
1 et 10093 ; " }}}{EXCHG {PARA 260 "" 0 "" {TEXT 258 120 "On conjectur
e qu'il y a une infinit\351 de couples de nombres premiers jumeaux, ma
is c'est \340 l'heure actuelle non d\351montr\351." }}{PARA 260 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 260 "" 0 "" 
{TEXT 259 15 "4  Les REPUNITS" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 182 "for a from 5 to 10 do printf(`Sont premiers en base %a :\\n`,a)
;\n\011for i to 20 do \n\011\011if isprime(i) then b:=(a^i-1)/(a-1);\n
\011\011\011if isprime(b) then  printf(`M%a = %a\\n`,i,b) fi fi\n\011o
d \nod;" }}{PARA 6 "" 1 "" {TEXT -1 25 "Sont premiers en base 5 :" }}
{PARA 6 "" 1 "" {TEXT -1 7 "M3 = 31" }}{PARA 6 "" 1 "" {TEXT -1 10 "M7
 = 19531" }}{PARA 6 "" 1 "" {TEXT -1 14 "M11 = 12207031" }}{PARA 6 "" 
1 "" {TEXT -1 15 "M13 = 305175781" }}{PARA 6 "" 1 "" {TEXT -1 25 "Sont
 premiers en base 6 :" }}{PARA 6 "" 1 "" {TEXT -1 6 "M2 = 7" }}{PARA 
6 "" 1 "" {TEXT -1 7 "M3 = 43" }}{PARA 6 "" 1 "" {TEXT -1 10 "M7 = 559
87" }}{PARA 6 "" 1 "" {TEXT -1 25 "Sont premiers en base 7 :" }}{PARA 
6 "" 1 "" {TEXT -1 9 "M5 = 2801" }}{PARA 6 "" 1 "" {TEXT -1 17 "M13 = \+
16148168401" }}{PARA 6 "" 1 "" {TEXT -1 25 "Sont premiers en base 8 :
" }}{PARA 6 "" 1 "" {TEXT -1 7 "M3 = 73" }}{PARA 6 "" 1 "" {TEXT -1 
25 "Sont premiers en base 9 :" }}{PARA 6 "" 1 "" {TEXT -1 26 "Sont pre
miers en base 10 :" }}{PARA 6 "" 1 "" {TEXT -1 7 "M2 = 11" }}{PARA 6 "
" 1 "" {TEXT -1 25 "M19 = 1111111111111111111" }}}{EXCHG {PARA 260 "" 
0 "" {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT 260 26 "5   Fractions \+
\351gyptiennes\n" }}{PARA 0 "" 0 "" {TEXT -1 136 "Ci-dessous un progra
mme permettant de chercher toutes les d\351compositions en somme de 1,
 2 ou 3 fractions \351gyptiennes d'un rationnel x ; " }}{PARA 0 "" 0 "
" {TEXT -1 85 "il utilise le fait que si x = 1/a+ 1/b + 1/c  avec a <=
 b <= c, alors 1/x <= a <= 3/x" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 285 "
for x in \{1/3+1/7\} do\nfor a from ceil(1/x) to 3/x do y:=x-1/a:\nif \+
y=0 then printf(`%a=1/%a\\n`,x,a) else\nfor b from max(ceil(1/y),a) to
 2/y do c:=y-1/b :\nif c=0 then printf(`%a=1/%a+1/%a\\n`,x,a,b) \nelif
 c > 0 and is(1/c,integer) then printf(`%a=1/%a+1/%a+%a\\n`,x,a,b,c) f
i  od fi od od:" }}{PARA 6 "" 1 "" {TEXT -1 13 "10/21=1/3+1/7" }}
{PARA 6 "" 1 "" {TEXT -1 18 "10/21=1/3+1/8+1/56" }}{PARA 6 "" 1 "" 
{TEXT -1 19 "10/21=1/3+1/14+1/14" }}{PARA 6 "" 1 "" {TEXT -1 18 "10/21
=1/4+1/7+1/12" }}{PARA 6 "" 1 "" {TEXT -1 17 "10/21=1/6+1/6+1/7" }}}
{EXCHG {PARA 262 "" 1 "" {TEXT -1 137 "Je cherche maintenant la liste \+
des nombres de la forme 3/n avec n de 4 \340 100 qui ne sont pas somme
 de une ou deux fractions \351gyptiennes :" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 197 "liste:=\{seq(k,k=4..100)\}:\nfor k in liste do\nx:
=3/k :for a from ceil(1/x) to 2/x do y:=x-1/a:if y=0 then liste:=liste
 minus \{k\} else b:=1/y :if is(b,integer) then liste:=liste minus\{k
\} fi fi od  od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "liste;" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#</\"\"(\"#8\"#>\"#J\"#P\"#V\"#\\\"#h
\"#n\"#t\"#z\"#\"*\"#(*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "Je co
nstate que ces nombres sont tous de la forme 3p +1 :en effet, on peut \+
\351crire pour les autres  : " }{XPPEDIT 18 0 "3/(3*p-1) = 1/p+1/p/(3*
p-1);" "6#/*&\"\"$\"\"\",&*&F%F&%\"pGF&F&F&!\"\"F*,&*&F&F&F)F*F&*(F&F&
F)F*,&*&F%F&F)F&F&F&F*F*F&" }}{PARA 0 "" 0 "" {TEXT -1 8 "et m\352me \+
" }{XPPEDIT 18 0 "a/(a*p-1) = 1/p+1/p/(a*p-1);" "6#/*&%\"aG\"\"\",&*&F
%F&%\"pGF&F&F&!\"\"F*,&*&F&F&F)F*F&*(F&F&F)F*,&*&F%F&F)F&F&F&F*F*F&" }
{TEXT -1 38 "ce qui montre que MIN(a/ (ap-1)) = 2. " }}}{EXCHG {PARA 
262 "" 1 "" {TEXT -1 143 "Je cherche maintenant la liste des nombres d
e la forme 4/n avec n de 5 \340 100 qui ne sont pas somme de une deux \+
ou trois fractions \351gyptiennes :" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 233 "liste:=\{seq(k,k=5..10)\}:for k in liste do\nx:=4/k \+
:for a from ceil(1/x) to 3/x do y:=x-1/a:if y<>0 then \nfor b from max
(ceil(1/y),a) to 2/y do c:=y-1/b :\nif c=0 or (c > 0 and is(1/c,intege
r)) then liste:=liste minus\{k\} fi od fi od od:" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 6 "liste;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<\"" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 " Il n'y en a pas !  On sait que  l
es " }{XPPEDIT 18 0 "4/(4*p-1);" "6#*&\"\"%\"\"\",&*&F$F%%\"pGF%F%F%!
\"\"F)" }{TEXT -1 153 " se d\351composent en deux fractions \351gyptie
nnes avec la m\351thode ci-dessus, mais, bien qu'on n'ait jamais trouv
\351 d'exemple o\371 l'on ne puisse pas d\351composer " }{XPPEDIT 18 
0 "4/(4*p+1);" "6#*&\"\"%\"\"\",&*&F$F%%\"pGF%F%F%F%!\"\"" }{TEXT -1 
122 " sous la forme de 3 fractions \351gyptiennes au plus, on  n'a pas
 \340 l'heure actuelle d\351montr\351 que c'\351tait toujours possible
 !" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 24 "6 Sommes de trois carr\351s" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 207 "for x in \{seq(k,k=95..100)\} do\nfor a from 0 to ev
alf(root(x/3,2)) do y:=x-a^2 : for b from a to evalf(root(y/2,2)) do :
z:=y-b^2:c:=iroot(z,2):\nif c^2=z then printf(`%a=%a^2+%a^2+%a^2\\n`,x
,a,b,c) fi od od od:" }}{PARA 6 "" 1 "" {TEXT -1 14 "96=4^2+4^2+8^2" }
}{PARA 6 "" 1 "" {TEXT -1 14 "97=0^2+4^2+9^2" }}{PARA 6 "" 1 "" {TEXT 
-1 14 "97=5^2+6^2+6^2" }}{PARA 6 "" 1 "" {TEXT -1 14 "98=0^2+7^2+7^2" 
}}{PARA 6 "" 1 "" {TEXT -1 14 "98=1^2+4^2+9^2" }}{PARA 6 "" 1 "" 
{TEXT -1 14 "98=3^2+5^2+8^2" }}{PARA 6 "" 1 "" {TEXT -1 14 "99=1^2+7^2
+7^2" }}{PARA 6 "" 1 "" {TEXT -1 14 "99=3^2+3^2+9^2" }}{PARA 6 "" 1 "
" {TEXT -1 14 "99=5^2+5^2+7^2" }}{PARA 6 "" 1 "" {TEXT -1 16 "100=0^2+
0^2+10^2" }}{PARA 6 "" 1 "" {TEXT -1 15 "100=0^2+6^2+8^2" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 72 "Voici maintenant la liste des entiers de \+
1 \340 100 non somme de 3 carr\351s :" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 190 "A:=\{seq(k,k=0..100)\}:for x in A do\nfor a from 0 t
o evalf(root(x/3,2)) do y:=x-a^2 : for b from a to evalf(root(y/2,2)) \+
do :z:=y-b^2:c:=iroot(z,2):\nif c^2=z then A:=A minus \{x\} fi od od o
d:A;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<1\"\"(\"#:\"#B\"#G\"#J\"#R\"#
Z\"#b\"#g\"#j\"#r\"#z\"#()\"##*\"#&*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 41 "B:=select(is,A,odd);C:=select(is,A,even);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"BG<.\"\"(\"#:\"#B\"#J\"#R\"#Z\"#b\"#j\"#
r\"#z\"#()\"#&*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"CG<%\"#G\"#g\"#
#*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 137 "Je remarque que B est form
\351 des entiers < 100 de la fotrme 8k-1  et que C est form\351 des pr
oduits de ces derniers par des puissances de 4." }}{PARA 0 "" 0 "" 
{TEXT -1 22 "Donc je conjecture :  " }{XPPEDIT 18 0 "n<>a^2+b^2+c^2" "
6#0%\"nG,(*$%\"aG\"\"#\"\"\"*$%\"bGF(F)*$%\"cGF(F)" }{TEXT -1 5 " <=> \+
" }{TEXT 263 1 "n" }{TEXT -1 17 " est de la forme " }{XPPEDIT 18 0 "4^
q*(8k-1" "6#*&)\"\"%%\"qG\"\"\",&*&\"\")F'%\"kGF'F'F'!\"\"F'" }{TEXT 
-1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 18 "Ceci a \351t\351 prouv\351." 
}}}{EXCHG {PARA 258 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
4 "7 ) " }{XPPEDIT 18 0 "(a+b)^2 = ab;" "6#/*$,&%\"aG\"\"\"%\"bGF'\"\"
#%#abG" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 120 "for k from 2 to 4 do for a to 10^k-1 do for b to 10^
k-1 do\nif (a+b)^2=(10^k)*a+b then print(a,b,(10^k)*a+b) fi od od od:
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6%\"#?\"#D\"%D?" }}{PARA 11 "" 1 "" {XPPMATH 20 "6%\"#I\"#D\"%DI" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6%\"#)*\"\"\"\"%,)*" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6%\"#))\"$4#\"&4#))" }}{PARA 11 "" 1 "" {XPPMATH 20 "6%
\"$%\\\"$4#\"'4U\\" }}{PARA 11 "" 1 "" {XPPMATH 20 "6%\"$)**\"\"\"\"',
!)**" }}{PARA 11 "" 1 "" {XPPMATH 20 "6%\"$%\\\"%H<\"(H<%\\" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6%\"$W(\"%%)>\"(%)>W(" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6%\"%]C\"%+D\")+D]C" }}{PARA 11 "" 1 "" {XPPMATH 20 "6%\"
%]D\"%+D\")+D]D" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 157 "Nicolas Karec
ki (PCSI 2005) a remarqu\351 la famille infinie : (99...98 + 00...01)
\262 = 99...9800...01, et JB Sizes (PCSI 2007) a remarqu\351 les 2 aut
res familles :" }}{PARA 0 "" 0 "" {TEXT -1 115 "(2500...005000..00+250
0..00)\262 = 2500...0050..002500..000 et (2499..995000..00+2500..00)
\262 = 2499..995000..002500..00" }}{PARA 0 "" 0 "" {TEXT -1 60 "( \351
crire ceci avec des puissances de 10  pour le justifier)." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "(25005000+25000000)^2;(24995000+250
00000)^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"1+++D+]+D" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#\"1+++D+]*\\#" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 34 }{VIEWOPTS 1 1 0 1 1 1803 1 1 
1 1 }{PAGENUMBERS 0 1 2 33 1 1 }