A. Student

Math 124 Problem Set 5

2. We establish both identities by induction.
Claim: $[a_n,a_{n-1},\ldots,a_1,a_0]=\frac{p_n}{p_{n-1}}$.
Since $p_{-1}=1$ and $p_0=a_0$, $[a_0]=\frac{p_0}{p_{-1}}$. This establishes the base case. Now suppose that $[a_{n-1},\ldots,a_0]=\frac{p_{n-1}}{p_{n-2}}$. Then $[a_n,a_{n-1},\ldots,a_0]=a_n+1/[a_{n-1},\ldots,a_0]=a_n+\frac{1}{\frac{p_{n-1}}{p_{n-2}}} = a_n+\frac{p_{n-2}}{p_{n-1}}=\frac{a_np_{n-1}+p_{n-2}}{p_{n-1}}$. The numerator is just the definition of $p_n$; this proves the claim.
Claim: $[a_n,a_{n-1},\ldots,a_1]=\frac{q_n}{q_{n-1}}$.
Since $q_0=1$ and $q_1=a_1$, $[a_1]=\frac{q_1}{q_0}$. This establishes the base case. Now suppose that $[a_{n-1},\ldots,a_1]=\frac{q_{n-1}}{q_{n-2}}$. Then $[a_n,a_{n-1},\ldots,a_1]=a_n+1/[a_{n-1},\ldots,a_1]=a_n+\frac{1}{\frac{q_{n-1}}{q_{n-2}}} = a_n+\frac{q_{n-2}}{q_{n-1}}=\frac{a_nq_{n-1}+q_{n-2}}{q_{n-1}}$. The numerator is just the definition of $q_n$; this proves the claim.
3. If we compute $ellj(t)$ in PARI, where $t=-0.5+0.3281996289i$ the result is $61.7142856\ldots-6.2\ E-26I$. The imaginary part is effectively 0, so we wish to guess the rational number that gives the real part. The command $contfrac(61.7142856)$ gives $[61,1,2,2,178571]$, suggesting that a good guess for our rational number is given by the continued fraction $[61,1,2,2]$. This value is $\frac{432}{7}$.
4i. Let $\alpha=[\overline{2,3}]$. Then $\alpha=2+\frac{1}{3+\frac{1}{\alpha}}$, so $3\alpha^2-6\alpha-2=0$. Solving for $\alpha$ yields $1+\frac{\sqrt{15}}{3}$.
4ii. First we compute $\alpha=[\overline{1,2,1}]$. This gives $\alpha=1+(2+\frac{1}{(1+\alpha^{-1})})^{-1}$. Solving for $\alpha$ yields $3\alpha^2-2\alpha-3=0$, so $\alpha=\frac{1+\sqrt{10}}{3}$. Now $[2,\overline{1,2,1}]=2+\frac{1}{[\overline{1,2,1}]}$, so our final answer is $\frac{5+\sqrt{10}}{3}$.
4iii. This is $[\overline{1,2,3}]^{-1}$. As above, if $\alpha=[\overline{1,2,3}]$ then $\alpha=1+(2+\frac{1}{3+\frac{1}{\alpha}})^{-1}$. This simplifies to $\alpha=7\alpha^2+8\alpha-3=0$, so $\alpha=\frac{4+\sqrt{37}}{7}$. Therefore our desired answer is $\frac{-4+\sqrt{37}}{3}$.
5. For all three parts we use contfrac in PARI to find the continued fraction and prove that the answer is correct.
5i. We claim that $\sqrt{5}=[2,\overline{4}]$. Let $\alpha=[\overline{4}]$; then $\alpha=4+\frac{1}{\alpha}$, so $\alpha=2+\sqrt{5}$. Now $[2,\overline{4}]=2+\frac{1}{2+\sqrt{5}}=\sqrt{5},$ as desired.
5ii. We claim that $\frac{1+\sqrt{13}}{2}=[2,\overline{3}]$. Let $\alpha=[\overline{3}]$; then $\alpha=3+\frac{1}{\alpha}$, so $\alpha^2-3\alpha-1=0$. This gives $\alpha=\frac{3+\sqrt{13}}{2}$. Then $[2,\overline{3}]=2+\frac{2}{3+\sqrt{13}}=\frac{1+\sqrt{13}}{2}.$
5iii. We claim that $\frac{5+\sqrt{37}}{4}=[\overline{2,1,3}]$. Let $\alpha=[\overline{2,1,3}]$; then $\alpha=2+(1+(3+\frac{1}{\alpha})^{-1})^{-1}$, so $4\alpha^2-10\alpha-3=0$. Solving for $\alpha$ gives $\frac{5+\sqrt{37}}{4}$, as desired.
6i. First we compute $[\overline{2n}]$. Let $\alpha=\overline{[2n]}$; then $\alpha=2n+\frac{1}{\alpha}$. This gives $\alpha^2-2n\alpha-1=0$, so $\alpha=n+\sqrt{n^2+1}$. Now $[n,\overline{2n}]=n+\frac{1}{\alpha}=n-(n-\sqrt{n^2+1})=\sqrt{n^2+1},$ as desired.
6ii. Using the previous part, we know that $\sqrt{5}=[2,\overline{4}]$. We can try successive convergents until two agree up to four decimal places; once such convergent is $682/305$.
7. In PARI, use convergents(contfrac(Pi)) to obtain the convergents of the continued fraction of $\pi$. We can now test convergents for property described in the problem, noting that smaller denominators are more likely to work. One convergent that satisfies the property is $3/1$, since $\pi-3<.15$ and $\frac{1}{\sqrt{5}}>.44$. The next is $22/7$, since $22/7-\pi<.002$ and $\frac{1}{49\sqrt{5}}>.009$. A third is $355/113$, since $355/113-\pi<.0000003$ while $\frac{1}{113^2\sqrt{5}}>.00003$.
8. In PARI, the command contfrac(exp(2)) gives

$$[7,2,1,1,3,18,5,1,1,6,30,8,1,1,9,42,11,1,1,12,54,14,1,1,15,77,17,1,1,18,78,20,1,1,21,90,\ldots]$$

This suggests that after the initial 7, the $i$th group of 5 numbers is of the form

$$3(i-1)+2,1,1,3i,18+12(i-1).$$

9i. We are looking for natural number solutions to $n+1=x^2$, $\frac{n}{2}+1=y^2$. Isolating $n$ yields $n=x^2-1=2(y^2-1)$, implying that $x^2-2y^2=-1$. There are infinitely many solutions $(x,y)$ to this equation if the period of the continued fraction of $\sqrt{2}$ has odd order. Indeed, $\sqrt{2}=[1,\overline{2}]$. For each $(x,y)$ we can take $2(y^2-1)$ to find a unique (even) $n$; thus there are infinitely many $n$ satisfying the desired property.
9ii. We can use the convergents program introduced in class to produce a list of convergents for the continued fraction of $\sqrt{2}$. The odd terms are of interest, and since we want to find $n>389$, we want the denominator of the convergent to be at least 14. The first two such convergents are $\frac{41}{29}$ and $\frac{239}{169}$. They yield $n=2(29^2-1)=$ 1680 and $n=2(169^2-1)=$ 57120.
10. If $x$ and $y$ are consecutive integers, then we have $z^2=x^2+(x+1)^2=2x^2+2x+1$. Multiplying by 2, we have $2z^2=4x^2+4x+2=(2x+1)^2+1$. Put $u=2x+1$; then $u^2-2z^2=-1$. From the previous problem we know there are infinitely many solutions to this equation for $u$ and $z$. Clearly each solution $(u,z)$ gives a unique $(x,z)$. Lastly we verify that each of these solutions $(x,z)$ leads to a primitive Pythagorean triple. But this is trivial: clearly $x$ and $x+1$ are coprime, and if either shares a nontrivial divisor with $z$ then the third must also share this divisor, implying that $(x,x+1)>1$, which is impossible.