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diff --git a/TATA54/TENTOR/tenta-TATA54-20240601-losning.tex b/TATA54/TENTOR/tenta-TATA54-20240601-losning.tex
index 114f1e6ba8cc04aba9a309960d28803fb0edaff9..ed0c3b14ea86eb7cde2ee43c6af1e3d8a5c72d4d 100644
--- a/TATA54/TENTOR/tenta-TATA54-20240601-losning.tex
+++ b/TATA54/TENTOR/tenta-TATA54-20240601-losning.tex
@@ -1,4 +1,4 @@
-\documentclass[12pt,a4paper]{article}
+\documentclass[11pt,a4paper]{article}
%\usepackage{times}
%\usepackage{euler}
\usepackage{a4wide}
@@ -21,7 +21,7 @@
\pagestyle{empty}
\begin{document}
\noindent{
-{\Huge Lösningsförslag} \\
+{\Huge Lösningsförslag} \\ \\
Talteori 6hp, Kurskod TATA54, Provkod TEN1 \\
1 juni 2024 \\
LINKÖPINGS UNIVERSITET\\
@@ -30,145 +30,123 @@ Examinator: Jan Snellman}
\bigskip
-
-Alla problem ger maximalt 3 poäng. Full poäng kräver fullständig lösning.
-8p räcker för betyg 3, 11p för betyg 4, 14p för betyg 5.
-
-\medskip
-
-{\small{Med \(\mu\) avses Möbiusfunktionen, den multiplikativa funktion som uppfyller
-\(\mu(1)=1\), \(\mu(p)=-1\) för \(p\) primtal, och \(\mu(n)=0\) för alla \(n>1\) som
-delas av kvadraten av något primtal. Med \(\phi\) avses Eulers phi-funktion,
-som räknar antalet multiplikativt inverterbara kongruensklasser modulo \(n\).}}
-
-\medskip
-
\begin{enumerate}[1)]
-\item Hitta alla lösningar till
- \begin{displaymath}
- x^5 +2x +3 \equiv 0 \mod 27
- \end{displaymath}
- och bestäm \textbf{antalet} sådana lösningar modulo 81.
-
-
- \soln
- Polynomet \(f(x) = x^5 + 2x +3 \equiv x(x^4-1) \mod 3\),
+\item
+ Sätt \(f(x) = x^5 + 2x +3\). Då är \(f(x) \equiv x(x^4-1) \mod 3\),
så det har nollställena 0,1,-1 modulo 3.
Den formella derivatan är
- \(f'(x)=5x^4+2 \equiv -(x^4+1) \mod 3\) vilket är
- skiljt från noll evaluerat på de tre nollställena.
- Alla dessa tre nollställen lyfter alltså unikt till
- nollställen modulo \(3^k\) enligt Hensels lemma,
- så det finns 3 nollställen modulo 27 och modulo 81.
+ \(f'(x)=5x^4+2\), så \(f'(x) \equiv -(x^4+1) \mod 3\).
+ Alltså är \(f'(x)\) konstant noll evaluerat på \(\Z_{3}\).
+ Vidare: enligt Hensels lemma så lyfter dessa tre nollställen
+ unikt till
+ nollställen modulo \(3^k\), för \(k \ge 2\).
+ Det finns alltså 3 nollställen modulo 27 och modulo 81.
Modulo 27 är dessa 12, 22, och 26.
+ \item
+ Eftersom \(99=3^2*11\) så ges denna maximala ordning
+ av \(\mgm(\phi(9),\phi(11)) = \mgm(6,10) = 30\),
+ enligt sats 9.22 i Rosen.
+ Vi ser att redan \(\mathrm{ord}_{99}(2)=30\).
-\item Hitta alla rationella punkter på kurvan
- \begin{displaymath}
- x^2 -7y^2 - 1 = 0
- \end{displaymath}
-
- \soln Parametrisera rationellt genom
+\item Parametrisera rationellt genom
att skicka \(t \in \Reals\)
till den unika skärningspunkten mellan kurvan och linjen
- med lutning \(t\) genom \((-1,0)\). Det ger en bijektion
- mellan kurvans högra gren och ett intervall av reella linjen;
+ med lutning \(t\) genom \((-1,0)\).
+ Explicit så
+ \begin{displaymath}
+ \left(\frac{-1}{\sqrt{7}},\frac{1}{\sqrt{7}}\right) \ni t \mapsto \left(
+ \frac{1+7t^{2}}{1-7t^{2}}, \frac{2t}{1-7t^{2}}
+ \right)
+ \end{displaymath}
+ med invers
+ \begin{displaymath}
+ (x,y) \mapsto \frac{y}{x+1}.
+ \end{displaymath}
+ Det ger en bijektion
+ mellan kurvans högra gren och ett intervall av reella linjen;
denna bijektion skickar rationella \(t\) till
rationella punkter, och omvänt. Använd symmetri för att
få med vänstra grenen.
\item
- Ange explicit två olika punkter på kurvan
- \begin{displaymath}
- (x^2 -7y^2 - 1)(x^2-7y^2+1)=0
- \end{displaymath}
- vars koordinater är positiva heltal; beskriv alla övriga
- sådana punkter.
-
- \soln
- En punkt \((x,y)\) på kurvan uppfyller antingen
+ En punkt \((x,y)\) på kurvan uppfyller antingen
\(x^2-7y^2=1\) eller
- \(x^2-7y^2=-1\). Heltalspunkter är således antingen lösningar
+ \(x^2-7y^2=-1\). Positiva heltalspunkter är således antingen lösningar
till Pells ekvation med \(d=7\), eller till ``negativa Pell''.
Kedjebråksutvecklingen är
\[\sqrt{7} = [2; \overline{1,1,1,4}] \]
- Sats i Rosen ger att
+ En sats i Rosen ger att
om konvergenterna till ovanstående kedjebråksutveckling är
\(P_k/Q_k\) så ges lösningarna till Pells ekvation
av
\[(x,y) = (Q_k, \, P_k) \qquad \text{ då } Q_k^2-7P_k =1\]
och motsvarande för negativa Pell. Eftersom periodlängden 4
- är jämn, så har negativa Pell inga lösningar, medan
+ är jämn, så säger en annan sats i Rosen att negativa Pell saknar lösningar, medan
Pell har lösningarna ovan då \(k=3,7,11,15,\dots\).
För \(k=3,7\) erhålls lösningarna \((8,3)\) och
- \((127,48)\).
+ \((127,48)\).
+
+ Vi kan också få alla lösningar \((x_{k},y_{k})\) från den primitiva lösningen
+ \((x_{1},y_{1})=(8,3)\) via
+ \begin{displaymath}
+ (8 + 3\sqrt{7})^{k} = x_{k} + \sqrt{7}y_{k}·
+ \end{displaymath}
+ Speciellt så är
+ \begin{displaymath}
+ (8 + 3\sqrt{7})^{2} = 64 + 2*8*3*\sqrt{7} + 3^{2}*7 = 127 + 48\sqrt{7}.
+ \end{displaymath}
+
\item
- LÃ¥t \(\tau(n)\) beteckna antalet positiva delare
- till heltalet \(n\).
- Visa att
- \[
- \sum_{\divides{k}{n}} \tau(k)
- = \prod_{j=1}^r \binom{a_j + 2}{2}
- \]
- då \(n\) har primfaktoriseringen
- \(n=\prod_{j=1}^r p_j^{a_j}\).
- Visa även att
- \[
- \sum_{k=1}^n \tau(k) =
- \sum_{k=1}^n \left \lfloor \frac{n}{k} \right \rfloor
- \]
-
- \soln Låt \(\ett(n)=1\) för alla positiva heltal \(n\).
- Då är \(\tau = \ett * \ett\) och
+ Låt \(\ett(n)=1\) för alla positiva heltal \(n\).
+ Då är \(\tau(n) = \sum_{\divides{k}{n}}1\)
+ så \(\tau = \ett * \ett\) och
\[h(n):= \sum_{\divides{k}{n}} \tau(k)
= (\ett * \ett * \ett)(n)
\]
Eftersom \(\ett\) är multiplikativ, är även
\(h=\ett * \ett * \ett\) det, så det räcker
att verifiera likheten för fallet att \(n=p^s\) är
- en primpotens. Vi har att
+ en primpotens. Delarna till \(p^{s}\) är \(p^{a}\) med \(0 \le a \le s\),
+ så vi har att
\[
- h(p^s) = \sum_{a+b+c=s} 1 = \binom{s+3-1}{3-1}
+ h(p^s) = \sum_{a+b+c=s} 1 = \binom{s+3-1}{3-1}
+ = \binom{s+2}{2}
\]
enligt urval med återläggning.
- För den andra delen, sätt
- \(A(n,k) = \lfloor \frac{n}{k} \rfloor\)
- och notera att
- \[
- A(n,k) - A(n-1,k)
- =
- \begin{cases}
- 1 & \divides{k}{n} \\
- 0 & \text{ annars }
- \end{cases}
- \]
-
+
\item
- LÃ¥t \(p\) vara ett udda primtal. Visa att ekvationen
- \[ x^4 \equiv -1 \mod p\]
- är lösbar om och endast om \(p \equiv 1 \mod 8\).
+ Sats 9.17 i Rosen säger att kongruensen ovan har en lösning om och endast om
+ \begin{displaymath}
+ (-1)^{\phi(p)/d} \equiv 1 \mod p
+ \end{displaymath}
+ där \(d=\sgd(4,\phi(p)) = \sgd(4,p-1)\).
- \soln
- Rosen har en sats om \(n\)-ära residyer.
-
-\item Vad är den största multiplikativa ordningen
- av ett heltal modulo 99?
- Ange något sådant heltal av maximal multiplikativ ordning.
+ \(p =8k+1\): Nu blir
+ \(\frac{p-1}{d} = \frac{8k}{4} = 2k\) vilket är jämnt, så
+ \((-1)^{\phi(p)/d} \equiv 1 \mod p\).
- \soln
- Eftersom \(99=3^2*11\) så ges denna maximala ordning
- av \(\mgm(\phi(9),\phi(11)) = \mgm(6,10) = 30\),
- enligt sats i Rosen.
- Vi ser att redan \(\mathrm{ord}_{99}(2)=30\).
-\item Visa att \(n^3(n^{18}-1) \equiv 0 \mod 26\)
- för alla heltal \(n\).
+ \(p =8k+3\): Nu blir
+ \(\frac{p-1}{d} = \frac{8k+2}{2} = 4k+1\) vilket är udda, så
+ \((-1)^{\phi(p)/d} \equiv -1 \mod p\).
+
+ \(p =8k+5\): Nu blir
+ \(\frac{p-1}{d} = \frac{8k+4}{4} = 2k+1\) vilket är udda, så
+ \((-1)^{\phi(p)/d} \equiv -1 \mod p\).
+
+ \(p =8k+7\): Nu blir
+ \(\frac{p-1}{d} = \frac{8k+6}{2} = 4k+3\) vilket är udda, så
+ \((-1)^{\phi(p)/d} \equiv -1 \mod p\).
+
+
+
- \soln Taget, och förenklat, från en gammal Melkersson-tenta,
- lösning finns på kurshemsidan.
+\item Härma lösningen till första uppgiften på tentan som gick
+ 2016-06-09; finns på kurshemsidan.
\end{enumerate}
diff --git a/TATA54/TENTOR/tenta-TATA54-20240601.log b/TATA54/TENTOR/tenta-TATA54-20240601.log
index e83f5c756dd69980fde001f47fa11bcddf3bfedf..78bfee0c73500cca80ce498a904c8a2e54f72e31 100644
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diff --git a/TATA54/TENTOR/tenta-TATA54-20240601.tex b/TATA54/TENTOR/tenta-TATA54-20240601.tex
index 775f11077a2530537d4f0d5677bd747e18c36d21..ab76b3e0e6aeff4cc9f72cb6a254042d1666d85a 100644
--- a/TATA54/TENTOR/tenta-TATA54-20240601.tex
+++ b/TATA54/TENTOR/tenta-TATA54-20240601.tex
@@ -1,7 +1,7 @@
\documentclass[12pt,a4paper]{article}
%\usepackage{times}
%\usepackage{euler}
-\usepackage{a4wide}
+%\usepackage{a4wide}
\usepackage{enumerate}
\usepackage{amsmath,amssymb,amsthm}
\usepackage[utf8]{inputenc}
@@ -10,7 +10,8 @@
\newcommand{\set}[1]{\left\{{#1}\right\}}
\newcommand{\vek}[1]{\boldsymbol{#1}}
\newcommand\setsuchas[2]{\left\{\,{#1}\,\vrule\,{#2}\,\right\}}
-\newcommand{\Z}{\boldsymbol{Z}}
+\newcommand{\Z}{\mathbb{Z}}
+\newcommand{\QQ}{\mathbb{Q}}
\newcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}
\newcommand{\divides}[2]{{#1} \lvert {#2}}
\pagestyle{empty}
@@ -30,10 +31,10 @@ Alla problem ger maximalt 3 poäng. Full poäng kräver fullständig lösning.
\medskip
-{\small{Med \(\mu\) avses Möbiusfunktionen, den multiplikativa funktion som uppfyller
-\(\mu(1)=1\), \(\mu(p)=-1\) för \(p\) primtal, och \(\mu(n)=0\) för alla \(n>1\) som
-delas av kvadraten av något primtal. Med \(\phi\) avses Eulers phi-funktion,
-som räknar antalet multiplikativt inverterbara kongruensklasser modulo \(n\).}}
+{\small{\(\mu\) är Möbiusfunktionen, \(\phi\) är Eulers phi-funktion,
+som räknar antalet multiplikativt inverterbara kongruensklasser modulo \(n\),
+ \(\tau\) räknar antalet positiva delare till heltalet \(n\).
+}}
\medskip
@@ -45,10 +46,16 @@ som räknar antalet multiplikativt inverterbara kongruensklasser modulo \(n\).}}
\end{displaymath}
och bestäm \textbf{antalet} sådana lösningar modulo 81.
+\item Vad är den största multiplikativa ordningen
+ av ett heltal modulo 99?
+ Ange något sådant heltal av maximal multiplikativ ordning.
+
+
\item Hitta alla rationella punkter på kurvan
\begin{displaymath}
x^2 -7y^2 - 1 = 0
\end{displaymath}
+ det vill säga alla lösningar \((x,y)\) med \(x,y \in \QQ\).
\item
Ange explicit två olika punkter på kurvan
@@ -61,27 +68,26 @@ som räknar antalet multiplikativt inverterbara kongruensklasser modulo \(n\).}}
\item
LÃ¥t \(\tau(n)\) beteckna antalet positiva delare till heltalet \(n\).
Visa att
- \[
- \sum_{\divides{k}{n}} \tau(k)
- = \prod_{j=1}^r \binom{a_j + 2}{2}
- \]
- då \(n\) har primfaktoriseringen \(n=\prod_{j=1}^r p_j^{a_j}\).
- Visa även att
- \[
- \sum_{k=1}^n \tau(k) =
- \sum_{k=1}^n \left \lfloor \frac{n}{k} \right \rfloor
- \]
+ \[
+ h(n) =
+ \sum_{\divides{k}{n}} \tau(k)
+ = \prod_{j=1}^r \binom{a_j + 2}{2}
+ \]
+ då \(n\) har primfaktoriseringen \(n=\prod_{j=1}^r p_j^{a_j}\).
+ Summan löper förstås över \emph{positiva} delare till \(n\).
+ % Visa även att
+ % \[
+ % \sum_{k=1}^n \tau(k) =
+ % \sum_{k=1}^n \left \lfloor \frac{n}{k} \right \rfloor
+ % \]
\item
LÃ¥t \(p\) vara ett udda primtal. Visa att ekvationen
\[ x^4 \equiv -1 \mod p\]
är lösbar om och endast om \(p \equiv 1 \mod 8\).
-\item Vad är den största multiplikativa ordningen
- av ett heltal modulo 99?
- Ange något sådant heltal av maximal multiplikativ ordning.
-\item Visa att \(n^3(n^{18}-1) \equiv 0 \mod 26\)
+\item Visa att \(n^{15} - n^{3} \equiv 0 \mod 36\)
för alla heltal \(n\).
\end{enumerate}
diff --git a/TATA54/TENTOR/tentaTATA5420240601sketch.org b/TATA54/TENTOR/tentaTATA5420240601sketch.org
index b6a290e2bfd7f4b574e22fabb5b5a84427135172..189f78ada33ab8b0c64b21100f365f133d80ab2a 100644
--- a/TATA54/TENTOR/tentaTATA5420240601sketch.org
+++ b/TATA54/TENTOR/tentaTATA5420240601sketch.org
@@ -74,6 +74,31 @@ def H_L_tree(f, p: int, r: int):
Parametrisera med lutning, rationell punkt (-1,0)
+#+begin_src sage :session
+var('s,t,x,y')
+kurva = x^2 - 7*y^2-1
+linje = y - t*(x+1)
+soln = solve([kurva,linje],[x,y])
+soln
+rp = soln[1]
+rp
+soln2 = solve([linje],[t])
+soln2
+#+end_src
+
+#+RESULTS:
+: (s, t, x, y)
+: [[x == -1, y == 0], [x == -(7*t^2 + 1)/(7*t^2 - 1), y == -2*t/(7*t^2 - 1)]]
+: [x == -(7*t^2 + 1)/(7*t^2 - 1), y == -2*t/(7*t^2 - 1)]
+: [t == y/(x + 1)]
+
+#+begin_src sage :session :results file :file "u2.png"
+epsi = 10^(-1)
+parametric_plot((rp[0].rhs(),rp[1].rhs()), (t,+epsi,1/sqrt(7) -epsi) )
+#+end_src
+
+#+RESULTS:
+[[file:u2.png]]
** U3
@@ -118,6 +143,13 @@ Antingen lösning till Pell x^2-7y^2=1 eller till negativa Pell x^2-7y^2=-1.
14 20830/7873 20830 7873 -3
#+end_example
+#+begin_src sage :session
+u = 8 + 3*sqrt(7)
+(u^2).expand()
+#+end_src
+
+#+RESULTS:
+: 48*sqrt(7) + 127
** U4
#+begin_src latex
@@ -134,6 +166,23 @@ Antingen lösning till Pell x^2-7y^2=1 eller till negativa Pell x^2-7y^2=-1.
\]
#+end_src
+#+begin_src sage :session
+number_of_divisors(10)
+def h(n):
+ return sum(number_of_divisors(k) for k in divisors(n))
+[h(n) for n in range(1,20)]
+def g(s):
+ return binomial(s+2,2)
+def gg(n):
+ return product([g(v[1]) for v in factor(n)])
+
+[gg(n) for n in range(1,20)]
+#+end_src
+
+#+RESULTS:
+: 4
+: [1, 3, 3, 6, 3, 9, 3, 10, 6, 9, 3, 18, 3, 9, 9, 15, 3, 18, 3]
+: [1, 3, 3, 6, 3, 9, 3, 10, 6, 9, 3, 18, 3, 9, 9, 15, 3, 18, 3]
** U5
#+begin_src latex
@@ -156,6 +205,62 @@ Ange något sådant heltal av maximal multiplikativ ordning.
** U7
#+begin_src latex
- Visa att \(n^3(n^{18}-1) \equiv 0 \mod 26\) för alla heltal \(n\).
+ Visa att \(n^3(n^{18}-1) \equiv 0 \mod 36\) för alla heltal \(n\).
+#+end_src
+
+#+begin_src sage :session
+p,q =2,3
+a,b = 2,2
+m = p^a*q^b; m
+R.<n> = Integers()[]
+f = n^15 -n^3; f
+[(k,f(k)) for k in Integers(m)]
+f.factor()
+euler_phi(m)
#+end_src
+
+#+RESULTS:
+#+begin_example
+36
+n^15 - n^3
+[(0, 0),
+ (1, 0),
+ (2, 0),
+ (3, 0),
+ (4, 0),
+ (5, 0),
+ (6, 0),
+ (7, 0),
+ (8, 0),
+ (9, 0),
+ (10, 0),
+ (11, 0),
+ (12, 0),
+ (13, 0),
+ (14, 0),
+ (15, 0),
+ (16, 0),
+ (17, 0),
+ (18, 0),
+ (19, 0),
+ (20, 0),
+ (21, 0),
+ (22, 0),
+ (23, 0),
+ (24, 0),
+ (25, 0),
+ (26, 0),
+ (27, 0),
+ (28, 0),
+ (29, 0),
+ (30, 0),
+ (31, 0),
+ (32, 0),
+ (33, 0),
+ (34, 0),
+ (35, 0)]
+(n - 1) * (n + 1) * n^3 * (n^2 - n + 1) * (n^2 + 1) * (n^2 + n + 1) * (n^4 - n^2 + 1)
+12
+#+end_example
+
* Slut
diff --git a/TATA54/TENTOR/u2.png b/TATA54/TENTOR/u2.png
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