lösningar till tenta 2024-08-22

TATA54/TENTOR/#tenta-TATA54-20240822-skiss.org#

+#+title: TATA54 tenta 20240601, lösningskiss
+#+author: Jan Snellman
+
+* Preliminärt
+5 uppgifter, 3p för varje. Försöker täcka
+- Kongruensräkning
+- Hensellyft
+- Kvadratisk reciprocite
+- Kedjebråk
+- Pell eller Pytagoranska tripplar
+
+
+* Gemensamm kod
+#+begin_src sage :session :export none
+def H_L_tree(f, p: int, r: int):
+    vert = [(0,0)]
+    for j in range(1,r+1):
+        fj = f.change_ring(Integers(p^j))
+        fzj = fj.roots(multiplicities=False)
+        vert += [(z,j) for z in fzj]
+    return DiGraph([vert, 
+                    lambda u,v: (u[1] == 1 and v[1] == 0) 
+                    or 
+                    ( (u[1] == v[1]+1) and ((u[0] - v[0]) % p^v[1] ==0) ) 
+                   ])
+#+end_src
+
+#+RESULTS:
+
+#+begin_src latex
+  \newcommand{\QQ}{\mathbf{Q}}
+  \newcommand{\ZZ}{\mathbf{Z}}
+#+end_src
+* Uppgifterna
+** U1
+#+begin_src latex
+    Hur många lösningar har kongruensen
+  \begin{displaymath}
+    x^4 +2x +4 \equiv 0 \mod 625
+  \end{displaymath}
+#+end_src
+
+
+#+begin_src sage
+  :session 
+   p = 5
+   k = 4
+   q = p^k
+   R.<x> = ZZ[]
+   f = x^4 + 2*x + 3
+   [(s,f.roots(Integers(p^s),multiplicities=False)) for s in range(k+1)]
+   df = f.diff()
+   df, df(3)
+#+end_src
+
+#+RESULTS:
+: [(0, [0]), (1, [3]), (2, []), (3, []), (4, [])]
+: (4*x^3 + 2, 110)
+
+#+begin_src sage :session :results file :file "nu1.png"
+  H_L_tree(f,p,k).show(layout='tree')
+#+end_src
+
+#+RESULTS:
+[[file:nu1.png]]
+
+
+** U2
+
+
+#+begin_src latex
+  Hitta alla rationella punkter på kurvan
+  \begin{displaymath}
+    x^2 +2y^2 - 1 = 0
+  \end{displaymath}
+
+#+end_src
+
+Parametrisera med lutning, rationell punkt (-1,0)
+
+#+begin_src sage
+  :session
+  var('s,t,x,y')
+  kurva = x^2 + 2*y^2-1
+  linje = y - t*(x+1)
+  soln = solve([kurva,linje],[x,y])
+  soln
+  rp = soln[0]
+  rp
+  soln2 = solve([linje],[t])
+  soln2
+#+end_src
+
+#+RESULTS:
+: (s, t, x, y)
+: [[x == -(2*t^2 - 1)/(2*t^2 + 1), y == 2*t/(2*t^2 + 1)], [x == -1, y == 0]]
+: [x == -(2*t^2 - 1)/(2*t^2 + 1), y == 2*t/(2*t^2 + 1)]
+: [t == y/(x + 1)]
+
+#+begin_src sage :session :results file :file "nu2.png"
+epsi = 10^(-1)
+parametric_plot((rp[0].rhs(),rp[1].rhs()), (t,+epsi,1/sqrt(7) -epsi) )
+#+end_src
+
+#+RESULTS:
+[[file:nu2.png]]
+
+** U3
+#+begin_src sage :session
+  legendre_symbol(187,23)
+#+end_src
+
+#+RESULTS:
+: 1
+
+** U4 
+#+begin_src latex
+   Låt \(\tau(n)\) beteckna antalet positiva delare till heltalet \(n\).
+    Visa  att
+    \[
+      \sum_{k=1}^n \tau(k) = \sum_{k=1}^n  \left \lfloor \frac{n}{k} \right \rfloor
+    \]
+#+end_src
+
+#+begin_s
+rc sage :session
+n=20
+[(k,number_of_divisors(k),floor(n/k)) for k in range(1,n+1)]
+
+def u(n):
+    return sum(number_of_divisors(k) for k in range(1,n+1))
+
+def w(n):
+    return sum(floor(n/k) for k in range(1,n+1))
+
+[(k,u(k),w(k)) for k in range(1,20+1)]
+
+#+end_src
+
+#+RESULTS:
+
+#+begin_example
+[(1, 1, 20),
+ (2, 2, 10),
+ (3, 2, 6),
+ (4, 3, 5),
+ (5, 2, 4),
+ (6, 4, 3),
+ (7, 2, 2),
+ (8, 4, 2),
+ (9, 3, 2),
+ (10, 4, 2),
+ (11, 2, 1),
+ (12, 6, 1),
+ (13, 2, 1),
+ (14, 4, 1),
+ (15, 4, 1),
+ (16, 5, 1),
+ (17, 2, 1),
+ (18, 6, 1),
+ (19, 2, 1),
+ (20, 6, 1)]
+[(1, 1, 1),
+ (2, 3, 3),
+ (3, 5, 5),
+ (4, 8, 8),
+ (5, 10, 10),
+ (6, 14, 14),
+ (7, 16, 16),
+ (8, 20, 20),
+ (9, 23, 23),
+ (10, 27, 27),
+ (11, 29, 29),
+ (12, 35, 35),
+ (13, 37, 37),
+ (14, 41, 41),
+ (15, 45, 45),
+ (16, 50, 50),
+ (17, 52, 52),
+ (18, 58, 58),
+ (19, 60, 60),
+ (20, 66, 66)]
+#+end_example
+
+** U5
+
+
+#+begin_src latex
+
+#+end_src
+
+#+RESULTS:
+#+begin_export latex
+Hur många lösningar har ekvationen
+\(x^2 \equiv 187 \mod 23\)?
+Hur många har
+\(x^2 \equiv 23 \mod 187\)?
+#+end_export
+
+
+#+begin_src s
+age :session
+p = 23
+a = 11*17
+a
+legendre_symbol(a,p)
+
+R.<x> = ZZ[]
+f = x^2 -23
+f.roots(Integers(187),multiplicities=False)
+f.roots(Integers(11),multiplicities=False)
+f.roots(Integers(17),multiplicities=False)
+#+end_src
+
+#+RESULTS:
+: 187
+: 1
+: []
+: [10, 1]
+: []
+
+** U7
+#+begin_src latex
+Låt \(r=117/119\). Hitta positiva heltal \(a,b\) med \(b < 119\)
+så att \(|a/b - r| \le |c/d -r|\) för alla
+positiva heltal \(c,d\) med \(d < 119\).
+#+end_src
+
+#+RESULTS:
+#+begin_export latex
+Låt \(r=117/119\). Hitta positiva heltal \(a,b\) med \(b < 119\)
+så att \(|a/b - r| \le |c/d -r|\) för alla
+positiva heltal \(c,d\) med \(d < 119\).
+#+end_export
+
+#+begin_src sage :session
+p=117
+q= 119
+r=p/q
+cf = continued_fraction(r)
+cf
+best = cf.convergent(2)
+fel = abs(p/q - best)
+"kedjebråkapprox fel", fel
+
+for j in range(2,q):
+    under = floor(r*j)
+    over = under+1
+    funder = abs(under/j - r)
+    fover = abs(over/j - r)
+    print(j,funder.n(),fover.n(),fel.n())
+#+end_src
+
+#+RESULTS:
+#+begin_example
+[0; 1, 58, 2]
+('kedjebråkapprox fel', 1/7021)
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+#+end_example
+
+* Slut

TATA54/TENTOR/.#tenta-TATA54-20240822-skiss.org

+jansn@jansn19-thinkpad.130192:1724236765
\ No newline at end of file

TATA54/TENTOR/tenta-TATA54-20240822-losning.aux

+\relax 
+\providecommand\babel@aux[2]{}
+\@nameuse{bbl@beforestart}
+\catcode `"\active 
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+\gdef \@abspage@last{1}

TATA54/TENTOR/tenta-TATA54-20240822-losning.log

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TATA54/TENTOR/tenta-TATA54-20240822-losning.tex

+\documentclass[10pt,a4paper]{article}
+\usepackage{times}
+\usepackage{euler}
+\usepackage{a4wide}
+\usepackage{enumerate}
+\usepackage{amsmath,amssymb,amsthm}
+\usepackage[utf8]{inputenc}
+\usepackage[swedish]{babel}
+\newcommand{\sgd}{\mathrm{sgd}}
+\newcommand{\set}[1]{\left\{{#1}\right\}}
+\newcommand{\vek}[1]{\boldsymbol{#1}}
+\newcommand\setsuchas[2]{\left\{\,{#1}\,\vrule\,{#2}\,\right\}}
+\newcommand{\Z}{\mathbb{Z}}
+\newcommand{\QQ}{\mathbb{Q}}
+\newcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}
+\newcommand{\divides}[2]{{#1} \lvert  {#2}}
+\newcommand{\soln}{\textbf{Lösning: }}
+\pagestyle{empty}
+\begin{document}
+{
+  \noindent Lösningar till
+  Talteori 6hp, Kurskod TATA54, Provkod TEN1 \\
+22 aug 2024 kl 14-18 \\
+LINKÖPINGS UNIVERSITET\\
+Matematiska Institutionen\\
+Examinator: Jan Snellman}
+
+\bigskip
+
+
+\begin{enumerate}[1)]
+
+\item
+   Hur många lösningar har kongruensen
+  \begin{displaymath}
+    x^4 +2x +4 \equiv 0 \mod 625 ?
+  \end{displaymath}
+
+  \soln Kalla polynomet för \(f(x)\). Modulo 5 finns den enda lösningen
+  \(x \equiv 3 \mod 5\). Eftersom \(f'(3)=110 \equiv 0 \mod 5\)
+  så lyfter inga eller alla av \(3 + 5s\), \(s=0,1,2,3,4\),
+  till lösningar modulo 25. Eftersom \(f(3)=110 \not \equiv 0\)
+  så finns inga lösningar modulo 25, och alltså inga modulo 625.
+
+\item   Hitta alla rationella punkter på kurvan
+   \begin{displaymath}
+    x^2 +2y^2 - 1 = 0.
+  \end{displaymath}
+Annorlunda uttryckt, hitta alla rationella lösningar till ekvationen
+(inte bara heltalslösningar).
+
+\soln Vi har de triviala lösningarna \((1,0),(-1,0)\).
+Linjen \(y=t(x+1)\) skär ellipsen i \((-1,0)\)
+samt i \(\frac{1}{2t^2+1}(1-2t^2, 2t)\). Det ger en bijektion mellan
+rella linjen och kurvan minus
+punkten \((-1,0))\). Inversen ges av \((x,y) \mapsto \frac{y}{x+1}\).
+Båda dessa avbildningar respekterar rationalitet, så de rationella punkterna på kurvan
+är \((-1,0)\) samt \(\frac{1}{2t^2+1}(1-2t^2, 2t)\) för \(t \in \QQ\).
+
+
+
+
+\item
+  Hur många lösningar har kongruensen
+\(x^2 \equiv 187 \mod 23\)?
+
+\soln 23 är ett primtal, och
+\(\legendre{187}{23} = \legendre{3}{23} = - \legendre{23}{3} = - \legendre{2}{3}
+=1\). Alltså har 187 precis 2 kvadratrötter modulo 23.
+
+\item
+  Hur många  lösningar har kongruensen
+  \(x^2 \equiv 23 \mod 187\)?
+
+  \soln \(187=11*17\) och
+  \(\legendre{23}{17}=\legendre{6}{17} = \legendre{2}{17} \legendre{3}{17}\)
+
+ \item
+   Låt \(r=117/119\). Hitta positiva heltal \(a,b\) med \(b < 119\)
+   så att
+   \[
+   |a/b - r| \le |c/d -r|
+   \]
+   för alla
+   positiva heltal \(c,d\) med \(d < 119\).
+
+
+\item
+  Låt \(\tau(n)\) beteckna antalet positiva delare till det positiva heltalet \(n\),
+  och låt \(\lfloor r \rfloor\) beteckna heltalsdelen av \(r\).
+    Gäller det att
+     \[
+       \sum_{k=1}^n \tau(k) =
+       \sum_{k=1}^n  \left \lfloor \frac{n}{k} \right \rfloor
+     \]
+     för alla positiva heltal \(n\)? Ge bevis eller motexempel.
+
+     %https://math.stackexchange.com/questions/1749204/sum-of-number-of-divisors-function-equals-sum-j-1n-lfloor-n-j-rfloor
+\end{enumerate}
+
+
+
+
+\end{document}
+
+%%% Local Variables: 
+%%% mode: latex
+%%% TeX-master: t
+%%% End: 
+
+

TATA54/TENTOR/tenta-TATA54-20240822-losning.tex~

+\documentclass[12pt,a4paper]{article}
+%\usepackage{times}
+%\usepackage{euler}
+%\usepackage{a4wide}
+\usepackage{enumerate}
+\usepackage{amsmath,amssymb,amsthm}
+\usepackage[utf8]{inputenc}
+\usepackage[swedish]{babel}
+\newcommand{\sgd}{\mathrm{sgd}}
+\newcommand{\set}[1]{\left\{{#1}\right\}}
+\newcommand{\vek}[1]{\boldsymbol{#1}}
+\newcommand\setsuchas[2]{\left\{\,{#1}\,\vrule\,{#2}\,\right\}}
+\newcommand{\Z}{\mathbb{Z}}
+\newcommand{\QQ}{\mathbb{Q}}
+\newcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}
+\newcommand{\divides}[2]{{#1} \lvert  {#2}}
+\pagestyle{empty}
+\begin{document}
+{
+\noindent Talteori 6hp, Kurskod TATA54, Provkod TEN1 \\
+22 aug 2024 kl 14-18 \\
+LINKÖPINGS UNIVERSITET\\
+Matematiska Institutionen\\
+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
+
+\begin{enumerate}[1)]
+
+\item
+   Hur många lösningar har kongruensen
+  \begin{displaymath}
+    x^4 +2x +4 \equiv 0 \mod 625 ?
+  \end{displaymath}
+
+
+\item   Hitta alla rationella punkter på kurvan
+   \begin{displaymath}
+    x^2 +2y^2 - 1 = 0.
+  \end{displaymath}
+Annorlunda uttryckt, hitta alla rationella lösningar till ekvationen
+(inte bara heltalslösningar).
+
+
+\item
+  Hur många lösningar har kongruensen
+\(x^2 \equiv 187 \mod 23\)?
+
+\item
+  Hur många  lösningar har kongruensen
+ \(x^2 \equiv 23 \mod 187\)?
+
+ \item
+   Låt \(r=117/119\). Hitta positiva heltal \(a,b\) med \(b < 119\)
+   så att
+   \[
+   |a/b - r| \le |c/d -r|
+   \]
+   för alla
+   positiva heltal \(c,d\) med \(d < 119\).
+
+
+\item
+  Låt \(\tau(n)\) beteckna antalet positiva delare till det positiva heltalet \(n\),
+  och låt \(\lfloor r \rfloor\) beteckna heltalsdelen av \(r\).
+    Gäller det att
+     \[
+       \sum_{k=1}^n \tau(k) =
+       \sum_{k=1}^n  \left \lfloor \frac{n}{k} \right \rfloor
+     \]
+     för alla positiva heltal \(n\)? Ge bevis eller motexempel.
+
+     %https://math.stackexchange.com/questions/1749204/sum-of-number-of-divisors-function-equals-sum-j-1n-lfloor-n-j-rfloor
+\end{enumerate}
+
+
+
+
+\end{document}
+
+%%% Local Variables: 
+%%% mode: latex
+%%% TeX-master: t
+%%% End: 
+
+