hcju/proofwikiqa huggingface dataset:
=== $(1)$ implies $(2)$ === Let $\alpha$ be the infimum of $S$. Then by definition, $\alpha$ is a lower bound for $S$. Let $\epsilon>0$. Because $\alpha+\epsilon>\alpha$, it is not a lower bound for $S$. Thus there exists $x\in S$ with $x < \alpha + \epsilon$. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $\alpha$ be a lower bound of $S$ such that $\forall \epsilon > 0$ there exists $x \in S$ with $x < \alpha + \epsilon$. Let $d \in \R$ be a lower bound of $S$. We have to prove that $d \le \alpha$. {{AimForCont}} $d > \alpha$. Let $\epsilon = d - \alpha > 0$. Then there exists $x \in S$ such that $x < \alpha + \left({d - \alpha}\right) = d$. But this contradicts our assumption that $d$ is a lower bound of $S$. {{qed}}
=== $(1)$ implies $(2)$ === Let $\omega$ be the supremum of $S$. Then by definition, $\omega$ is an upper bound for $S$. Let $\epsilon > 0$. Because $\omega - \epsilon < \omega$, it is not an upper bound for $S$. Thus there exists $x\in S$ with $x > \omega - \epsilon$. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $\omega$ be an upper bound of $S$ such that $\forall \epsilon > 0$ there exists $x \in S$ with $x > \omega - \epsilon$. Let $d \in \R$ be an upper bound of $S$. We have to prove that $d \ge \omega$. {{AimForCont}} $d < \omega$. Let $\epsilon = \omega - d > 0$. Then there exists $x \in S$ such that $x > \omega - \left({\omega - d}\right) = d$. But this contradicts our assumption that $d$ is an upper bound of $S$. {{qed}}
=== 1 implies 2 === Let $I\subset \R$ be a subset such that: :$\forall x, y \in I : \forall z \in \R : \paren {x \le z \le y \implies z \in I}$ Let: :$a = \map \inf I$ (or $a = -\infty$ if $I$ is not bounded below) :$b = \map \sup I$ (or $b = +\infty$ if $I$ is not bounded above). We will prove that: :$\openint a b \subseteq I \subseteq \closedint a b$ In order not to need to provide a long list of cases, the '''temporary''' convention is used: :A square bracket "$[$" or "$]$" next to $\pm \infty$ means the same as a round bracket "$($" or "$)$" :$\openint a b$ means $\O$ {{iff}} $a = b$. Let $z \in \openint a b$. Then $z \ge a$, so $\exists x \in I: x \le z$ by definition of $a$. Similarly $z \le b$, so $\exists y \in S: z \le y$ by definition of $b$. Hence by hypothesis $z \in I$. This proves that $\openint a b \subseteq I$. From the definitions of $a$ and $b$ it follows that $I \subseteq \closedint a b$. But if $\openint a b \subseteq I \subseteq \closedint a b$, then $I$ must be one of the various types of real interval. {{qed|lemma}} === 2 implies 1 === If $I$ is a real interval of any of the various kinds, then it clearly has this property by definition. {{qed}}
Suppose $a > 0$.  By Sequence Converges to Within Half Limit: :$\exists N \in \N: \forall n > N: x_n > \dfrac a 2 > 0$ Now suppose $a < 0$.  By Sequence Converges to Within Half Limit: :$\exists N \in \N: \forall n > N: x_n < \dfrac a 2 < 0$ This shows that if $a \ne 0$: :$\exists N \in \N: \forall n > N: x_n \ne 0$ {{qed}}
Let $\map \zeta s$ be the Riemann zeta function. Let $\displaystyle \map g z = \sum_{n \mathop = 1}^\infty z^n \map \zeta {2 n}$ be the  generating function of $\map \zeta {2 n}$ By Power Series Expansion for Cotangent Function, for $\size z < 1$: {{begin-eqn}} {{eqn | l = \pi \map \cot {\pi z}       | r = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n \pi^{2 n} 2^{2 n} B_{2 n} \, z^{2 n - 1} } {\paren {2 n}!} }} {{eqn | ll= \leadsto       | l = \pi z \map \cot {\pi z}       | r = 1 + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n \pi^{2 n} 2^{2 n} B_{2 n} \, z^{2 n} } {\paren {2 n}!} }} {{eqn | ll= \leadsto       | l = \frac {\pi z \map \cot {\pi z} - 1} {-2}       | r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} \pi^{2 n} 2^{2 n - 1} B_{2 n} \, z^{2 n} } {\paren {2 n}!} }} {{end-eqn}} By Riemann Zeta Function at Even Integers: :$\map \zeta {2 n} = \dfrac {\paren {-1}^{n + 1}  \pi^{2 n} 2^{2 n - 1} B_{2 n}  } {\paren {2 n}!}$ Thus: {{begin-eqn}} {{eqn | l = \dfrac {\pi z \map \cot {\pi z} - 1} {-2}       | r = \sum_{n \mathop = 1}^\infty \map \zeta {2 n} z^{2 n}  }} {{eqn | r = \map g {z^2} }} {{end-eqn}} By Analytic Continuation of Generating Function of Dirichlet Series and Uniqueness of Analytic Continuation:  :$\displaystyle \dfrac {\pi z \map \cot {\pi z} - 1} {-2} = \sum_{n \mathop = 1}^\infty \dfrac {z^2} {n^2 - z^2}$ for all of $\C$, as this is the overlap of their domains. Thus: {{begin-eqn}} {{eqn | l = \pi z \map \cot {\pi z} - 1       | r =  2 \sum_{n \mathop = 1}^\infty \frac {z^2} {z^2 - n^2} }} {{eqn | ll= \leadsto       | l = \pi z \map \cot {\pi z}       | r =  1 + 2 \sum_{n \mathop = 1}^\infty \frac {z^2} {z^2 - n^2} }} {{eqn | ll= \leadsto       | l = \pi \map \cot {\pi z}       | r = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \frac z {z^2 - n^2} }} {{end-eqn}} {{qed}}
Let $\mathcal L$ denote the logarithmic derivative. On the open set $\C \setminus \Z$ we have: {{begin-eqn}} {{eqn | l = \pi \cot \pi z        | r = \mathcal L \left({\sin \left({\pi z}\right)}\right)       | c = Primitive of Cotangent Function, or a complex version thereof }} {{eqn | r = \mathcal L \left({\pi z \prod_{n \mathop = 1}^\infty \left({1 - \frac {z^2} {n^2} }\right)}\right)       | c = Euler Formula for Sine Function }} {{eqn | r = \mathcal L \left({\pi z}\right) + \sum_{n \mathop = 1}^\infty \mathcal L \left({1 - \frac {z^2} {n^2} }\right)       | c = Logarithmic Derivative of Infinite Product of Analytic Functions }} {{eqn | r = \frac \pi {\pi z} + \sum_{n \mathop = 1}^\infty \frac 1 {1 - \frac {z^2} {n^2} } \cdot \frac {\mathrm d} {\mathrm d z} \left({1 - \frac {z^2} {n^2} }\right)       | c = {{Defof|Logarithmic Derivative of Meromorphic Function}} }} {{eqn | r = \frac 1 z - 2 \sum_{n \mathop = 1}^\infty \frac z {n^2 - z^2}       | c = Derivative of Power }} {{eqn | r = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \frac z {z^2 - n^2} }} {{end-eqn}} {{qed}}
From Mittag-Leffler Expansion for Cotangent Function: :$\displaystyle \pi \cot \pi z = \frac 1 z + 2 \sum_{k \mathop = 1}^\infty \frac z {z^2 - k^2}$ Factoring $-\dfrac 1 {k^2}$:  :$\displaystyle \pi \cot \pi z = \frac 1 z - 2 \sum_{k \mathop = 1}^\infty \frac z {k^2} \cdot \frac 1 {1 - \frac {z^2} {k^2} }$ Taking $\cmod z < 1$, and noting that $k \ge 1$, we have, by Sum of Infinite Geometric Sequence: :$\displaystyle \pi \cot \pi z = \frac 1 z - 2 \sum_{k \mathop = 1}^\infty \frac z {k^2} \cdot \sum_{n \mathop = 1}^\infty \paren {\frac {z^2} {k^2} }^{n - 1}$ from which: {{begin-eqn}} {{eqn | l = \pi \cot \pi z       | r = \frac 1 z - 2 \sum_{k \mathop = 1}^\infty \sum_{n \mathop = 1}^\infty \frac {z^{2 n - 2} \cdot z} {k^{2 n - 2} \cdot k^2} }} {{eqn | r = \frac 1 z - 2 \sum_{k \mathop = 1}^\infty \sum_{n \mathop = 1}^\infty \frac 1 {k^{2 n} } \cdot z^{2 n - 1} }} {{eqn | r = \frac 1 z - 2 \sum_{n \mathop = 1}^\infty \map \zeta {2 n} z^{2 n - 1}       | c = {{Defof|Riemann Zeta Function}} }} {{end-eqn}} {{qed}} Category:Laurent Series Expansion for Cotangent Function Category:Laurent Series Expansions Category:Cotangent Function tmigxwb63elg6ab0xcb4011tt72zbp1
=== 1 implies 2 === Let $\map P X \in R \sqbrk X$ be the polynomial: :$P = \ds \sum_{k \mathop = 0}^n a_k \cdot X^k$ where $\sum$ denotes indexed summation. We show that $\map P \iota = f$. {{finish}} === 2 implies 1 === Let $P \in R \sqbrk X$ be a polynomial and $f = \map P \iota \in R^S$. By Polynomial is Linear Combination of Monomials, there exist: :$n \in \N$ :$a_0, \ldots, a_n \in R$ such that: :$P = \ds \sum_{k \mathop = 0}^n a_k \cdot X^k$ where $\sum$ denotes indexed summation. Let $\operatorname{ev}_{\iota}$ denote the evaluation homomorphism at $\iota$. Then $\map {\operatorname{ev}_{\iota} } P = f$. We have: {{begin-eqn}} {{eqn | l = f       | r = \map {\operatorname{ev}_{\iota} } {\sum_{k \mathop = 0}^n a_k \cdot X^k} }} {{eqn | r = \sum_{k \mathop = 0}^n \map {\operatorname{ev}_{\iota} } {a_k \cdot X^k}       | c = Ring Homomorphism Preserves Indexed Summations }} {{eqn | r = \sum_{k \mathop = 0}^n a_k \cdot \map {\operatorname{ev}_{\iota} } {X^k}       | c = {{Defof|Polynomial Evaluation Homomorphism}} }} {{eqn | r = \sum_{k \mathop = 0}^n a_k \cdot \map {\operatorname{ev}_{\iota} } {X^k }       | c = {{Defof|Polynomial Evaluation Homomorphism}} }} {{end-eqn}} {{finish|but move the calculation to Explicit Formula for Polynomial Evaluation Homomorphism}} Category:Polynomial Theory a7c8wt4tngw3mr8y1r3e43mx6dcmw9e
Follows from: * Polynomial is Linear Combination of Monomials * Monomials of Polynomial Ring are Linearly Independent {{qed}} Category:Polynomial Rings Category:Monomials ibj1ycvcvta6451o5b2hkeptfdnrhkb
Let $X = \set {x_1, x_2, \ldots, x_v}$. Let $\mathcal B = \multiset {y_1, y_2,\ldots, y_b}$, where the notation denotes a multiset. Let $Y = \displaystyle \bigcup_{i \mathop = 1}^b y_i$. We shall show that $Y \subseteq X$ and $X \subseteq Y$. === $Y \subseteq X$: === By definition, $\mathcal B$ is a multiset of subsets of $X$. This means that each $y_i \in Y$ is a subset of $X$ for $i \in \closedint 1 b$. By Union of Subsets is Subset, $Y$ itself is a subset of $\mathcal B$. {{qed|lemma}} === $X \subseteq Y$: === Let $x_i \in X$ be arbitrary. Choose any $x_j \in X$ such that $i \ne j$. Such an $x_j$ necessarily exists because by hypothesis $\card X \ge 2$. Then $\set {x_i, x_j} \subseteq Y$ by the definition of blocks, as $\lambda > 0$ by hypothesis. But by the definition of subset, this implies that $x_i$ is an element in $Y$. Hence the result, as $x_i$ was arbitrary. {{qed}} === Proof of Lemma === By the definition of balanced incomplete block design, $\struct {X, \mathcal B}$ is itself a type of pairwise balanced design. Thus the lemma follows from the main result. {{qed}} Category:Design Theory tpi7f069keudxo9nh0u1i5giiqhl2bg
By: * Uniqueness of Polynomial Ring in One Variable * Homomorphism Preserves Indexed Products we may assume $R \sqbrk X$ is the ring of sequences of finite support over $R$, and $X$ is the sequence $\sequence {0, 1, 0, 0 \ldots}$. One verifies that, for $k \ge 0$, $X^k$ is the sequence with $\map {X^k} l = \delta_{k l}$, where $\delta$ is the Kronecker delta. {{theoremWanted|verify this somewhere else}} Thus $X^k \ne X^l$ if $k \ne l$. {{qed}}
Let $P' = P \cup Q$. We observe: :$P'$ is either equal to $P$ or finer than $P$ :$P'$ is either equal to $Q$ or finer than $Q$ We find: :$\map L P \le \map L {P'}$ by the definition of lower sum and $P'$ refining $P$ :$\map L {P'} \le \map U {P'}$ by Upper Sum Never Smaller than Lower Sum :$\map U {P'} \le \map U Q$ by the definition of upper sum and $P'$ refining $Q$ By combining these inequalities, we conclude: :$\map L P \le \map U Q$ {{qed}} Category:Real Analysis pl2sdzoid9vqslje2yplrl97iqur6q1
=== Necessary Condition === Suppose that $S$ is finite. By definition, this means there exists a bijection $f: \N_{< n} \to S$. Then $f$ is a fortiori also the sought surjection. {{qed|lemma}} === Sufficient Condition === Let $f: \N_{< n} \to S$ be a surjection. Define $g: S \to \N_{< n}$ by: :$g (s) := \min f^{-1} (s)$ where $f^{-1} (s)$ is the preimage of $s$ under $f$. Note that $f^{-1} (s)$ is not empty because $f$ is a surjection. By the Well-Ordering Principle, $f^{-1} (s) \subseteq \N$ has a smallest element. Hence $g$ is well-defined. Next we show that $g$ is injective. So suppose that $g(s) = g(s')$ for some $s, s' \in S$: {{begin-eqn}} {{eqn|l = g(s)      |r = g(s') }} {{eqn|ll= \implies      |l = f \left({ g(s) }\right)      |r = f \left({ g(s') }\right) }} {{eqn|ll= \implies      |l = s      |r = s'      |c = Definition of $g$ }} {{end-eqn}} Hence $g$ is injective. Then by Injection to Image is Bijection, $S$ is equivalent to a subset of $\N_{<n}$. By Subset of Finite Set is Finite, it follows that $S$ is finite. {{qed}}
=== Necessary Condition === Suppose that $S$ is finite. By definition, this means there exists a bijection $f: S \to \N_{< n}$. Then $f$ is a fortiori also the sought injection. {{qed|lemma}} === Sufficient Condition === Let $f: S \to \N_{< n}$ be an injection. Then by Injection to Image is Bijection, $S$ is equivalent to a subset of $\N_{<n}$. By Subset of Finite Set is Finite, it follows that $S$ is finite. {{qed}}
Let $p$ be such an integer. Apart from the first digit, each of its digits forms the second digit of a $2$-digit prime number. Thus, apart from the first digit, $p$ cannot contain $2$, $4$, $5$, $6$, $8$ or $0$. $0$ cannot of course be the first digit either. Thus, apart from the first pair of its digits, the $2$-digit prime numbers contained in $p$ can only ever be in the set: :$S = \left\{ {11, 13, 17, 19, 31, 37, 71, 73, 79, 97}\right\}$ Let it be assumed that the largest such $p$ contains all of the above. There are $10$ elements of $S$. Each of the elements of $S$ apart from one shares both digits with one other element of $S$, apart from the final one in $p$. Thus the elements of $S$ contribute $11$ digits of $p$ in total. The first digit of the first pair contributes an extra $1$ digit to $p$. Thus $p$ contains $12$ digits. Only $1$ element of $S$ begins with $9$, but $2$ of them end with $9$. Therefore $9$ must be the last digit of $p$. Only $3$ elements of $S$ end in $1$, but $4$ of them begin with $1$. Thus the first pair of digits of $p$ must be a prime number which does not appear in $S$ whose $2$nd digit is $1$. Thus the first $2$ digits of $p$ can only be $41$ or $61$. Let us construct $p$ by starting from the left and use the greedy algorithm consisting of select the largest digit available. :$61$ is larger than $41$, so $p$ can be assumed to start with $61$. Using this algorithm, we select the elements of $S$ in the following order: :$61$, $19$, $97$ (which is forced), $73$, $37$, $71$ The next element of $S$ cannot be $17$, because that would force $79$ to follow. But as $79$ is the only remaining element of $S$ ending in $9$, $79$ must be at the end of $p$. So $71$ must be followed by $11$ or $13$. Continuing to use the greedy algorithm: :$61$, $19$, $97$ $73$, $37$, $71$, $13$, $31$ (forced) For the same reason as above, we cannot then select $17$, as this will be followed by $79$ which cannot be followed. So we continue with the greedy algorithm: :$61$, $19$, $97$ $73$, $37$, $71$, $13$, $31$, $11$, $17$, $79$ and $p$ is complete: :$619 \, 737 \, 131 \, 179$ As a bonus, $p$ is itself a prime number. {{qed}}
From Upper Sum Never Smaller than Lower Sum for any Pair of Subdivisions, $\map L S$ is a lower bound for the real set: :$T = \leftset {\map U P: P}$ is a finite subdivision of $\rightset {\closedint a b}$ Since $\inf_P \map U P$ is the infumum of $T$: :$\inf_P \map U P \ge \map L S$ Hence the result. {{qed}} Category:Real Analysis jq9jbzo0wiugc4xg797a485pbx5zxaq
From Upper Sum Never Smaller than Lower Sum for any Pair of Subdivisions, $\map U S$ is an upper bound for the real set: :$T = \leftset {\map L P: P}$ is a finite subdivision of $\rightset {\closedint a b}$ Since $\sup_P \map L P$ is the supremum of $T$: :$\sup_P \map L P \le \map U S$ Hence the result. {{qed}} Category:Real Analysis 1eh713jwmo7suop90lls86p3n0c3inu
Let $C$ denote the sequence in question. We have that $758 \, 083 \, 947 \, 856 \, 951$ is prime. First note that: :$\dfrac {758 \, 083 \, 947 \, 856 \, 951 + 1} 2 = 379 \, 041 \, 973 \, 928 \, 476 = 2^2 \times 94 \, 760 \, 493 \, 482 \, 119$ and so is not prime. Thus $758 \, 083 \, 947 \, 856 \, 951$ fulfils the requirement for $C$ to be a Cunningham chain of the second kind. Then: {{begin-eqn}} {{eqn | n = 1       | l = 2 \times 758 \, 083 \, 947 \, 856 \, 951 - 1       | r = 1 \, 516 \, 167 \, 895 \, 713 \, 901       | c = which is prime }} {{eqn | n = 2       | l = 2 \times 1 \, 516 \, 167 \, 895 \, 713 \, 901 - 1       | r = 3 \, 032 \, 335 \, 791 \, 427 \, 801       | c = which is prime }} {{eqn | n = 3       | l = 2 \times 3 \, 032 \, 335 \, 791 \, 427 \, 801 - 1       | r = 6 \, 064 \, 671 \, 582 \, 855 \, 601       | c = which is prime }} {{eqn | n = 4       | l = 2 \times 6 \, 064 \, 671 \, 582 \, 855 \, 601 - 1       | r = 12 \, 129 \, 343 \, 165 \, 711 \, 201       | c = which is prime }} {{eqn | n = 5       | l = 2 \times 12 \, 129 \, 343 \, 165 \, 711 \, 201 - 1       | r = 24 \, 258 \, 686 \, 331 \, 422 \, 401       | c = which is prime }} {{eqn | n = 6       | l = 2 \times 24 \, 258 \, 686 \, 331 \, 422 \, 401 - 1       | r = 48 \, 517 \, 372 \, 662 \, 844 \, 801       | c = which is prime }} {{eqn | n = 7       | l = 2 \times 48 \, 517 \, 372 \, 662 \, 844 \, 801 - 1       | r = 97 \, 034 \, 745 \, 325 \, 689 \, 601       | c = which is prime }} {{eqn | n = 8       | l = 2 \times 97 \, 034 \, 745 \, 325 \, 689 \, 601 - 1       | r = 194 \, 069 \, 490 \, 651 \, 379 \, 201       | c = which is prime }} {{eqn | n = 9       | l = 2 \times 194 \, 069 \, 490 \, 651 \, 379 \, 201 - 1       | r = 388 \, 138 \, 981 \, 302 \, 758 \, 401       | c = which is prime }} {{eqn | n = 10       | l = 2 \times 388 \, 138 \, 981 \, 302 \, 758 \, 401 - 1       | r = 776 \, 277 \, 962 \, 605 \, 516 \, 801       | c = which is prime }} {{eqn | n = 11       | l = 2 \times 776 \, 277 \, 962 \, 605 \, 516 \, 801 - 1       | r = 1 \, 552 \, 555 \, 925 \, 211 \, 033 \, 601       | c = which is prime }} {{eqn | n = 12       | l = 2 \times 1 \, 552 \, 555 \, 925 \, 211 \, 033 \, 601 - 1       | r = 3 \, 105 \, 111 \, 850 \, 422 \, 067 \, 201       | c = which is prime }} {{eqn | n = 13       | l = 2 \times 3 \, 105 \, 111 \, 850 \, 422 \, 067 \, 201 - 1       | r = 6 \, 210 \, 223 \, 700 \, 844 \, 134 \, 401       | c = which is $41 \times 691 \times 739 \times 4957 \times 59 \, 838 \, 677$ and so not prime }} {{end-eqn}} Establishing that this is indeed the smallest such Cunningham chain of the second kind of length $13$ can be done by a computer search. {{qed}}
$6 \, 505 \, 941 \, 701 \, 960 \, 039$ is a prime number. $6 \, 505 \, 941 \, 701 \, 960 \, 903$ is a prime number. It can be checked that all numbers between these two are composite. It can also be checked by computer that there is no prime gap of $864$ between smaller prime numbers. {{qed}}
Suppose $T$ were finite. Then by Set Finite iff Injection to Initial Segment of Natural Numbers, there is an injection $f: T \to \N_{<n}$ for some $n \in \N$. But then by Restriction of Injection is Injection, also the restriction of $f$ to $S$: :$f \restriction_{S}: S \to \N_{<n}$ is an injection. Again by Set Finite iff Injection to Initial Segment of Natural Numbers, this contradicts the assumption that $S$ is infinite. Hence $T$ is infinite. {{qed}}
From Aurifeuillian Factorization of 2 Mod 4th Power of Two plus 1, we have: :$2^{4 n + 2} + 1 = \left({2^{2 n + 1} - 2^{n + 1} + 1}\right) \left({2^{2 n + 1} + 2^{n + 1} + 1}\right)$ Setting $n = 14$: {{begin-eqn}} {{eqn | l = 2^{58} + 1       | r = \left({2^{29} - 2^{15} + 1}\right) \left({2^{29} + 2^{15} + 1}\right)       | c =  }} {{eqn | r = \left({536 \, 870 \, 912 - 32 \, 768 + 1}\right) \left({536 \, 870 \, 912 + 32 \, 768 + 1}\right)       | c =  }} {{eqn | r = 536 \, 838 \, 145 \times 536 \, 903 \, 681       | c =  }} {{eqn | r = 5 \times 107 \, 367 \, 629 \times 536 \, 903 \, 681       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac {10^n - 1} 9       | r = \dfrac {10^n - 1} {10 - 1}       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^{n - 1} 10^k       | c = Sum of Geometric Sequence }} {{eqn | r = 1 + 10 + 100 + \ldots + 10^{n - 2} + 10^{n - 1}       | c =  }} {{end-eqn}} The result follows from the Basis Representation Theorem. {{qed}}
Let $m$ be a repunit with $r$ digits such that $r > 1$. By definition, $m$ is odd. Thus from Square Modulo 4, if $m$ were square it would be of the form: :$m \equiv 1 \pmod 4$. $m$ is of the form $\displaystyle \sum_{k \mathop = 0}^{r - 1} 10^k$ where $r$ is the number of digits. Thus for $r \ge 2$: {{begin-eqn}} {{eqn | l = m       | r = 11 + 100 s       | c = for some $s \in \Z$ }} {{eqn | r = \paren {2 \times 4} + 3 + 4 \times \paren {25 s}       | c =  }} {{eqn | r = 3 + 4 t       | c = for some $t \in \Z$ }} {{end-eqn}} Hence: :$m \equiv 3 \pmod 4$ and so cannot be square. {{qed}}
Define $T^*$ to be the Kleene closure of $T$, i.e.: :$T^* = \displaystyle \bigcup_{i = 1}^\infty T^i$ Note how we, for convenience, exclude the empty sequence from $T^*$. Now define a mapping $\mathcal G: T^* \to T^*$ by: :$\mathcal G \left({t_1, \ldots, t_n}\right) = \left({t_1, \ldots, t_n, G_n \left({t_1, \ldots, t_n}\right)}\right)$ that is, extending each finite sequence $(t_1, \ldots, t_n)$ with the element $G_n \left({t_1, \ldots, t_n}\right) \in T$. By the Principle of Recursive Definition applied to $\mathcal G$ and the finite sequence $( a )$, we obtain a unique mapping: :$\mathcal F: \N \to T^*, \mathcal F(x) = \begin{cases}(a) & : x = 0 \\ \mathcal G \left({ \mathcal F(n) }\right) & : x = n + 1\end{cases}$ Next define $f: \N \to T$ by: :$f(n) = \text{the last element of $\mathcal F(n)$}$ We claim that this $f$ has the sought properties, which will be proven by the Principle of Mathematical Induction. We prove the following assertions by induction: :$\mathcal F(n) = \left({f(0), f(1), \ldots, f(n-1), G_n \left({f(0), \ldots, f(n-1)}\right)}\right)$ :$f(n) = G_n \left({f(0), \ldots, f(n-1)}\right)$ For $n = 0$, these statements do not make sense, however it is immediate that $f(0) = \operatorname{last}((a)) = a$. For the base case, $n = 1$, we have: {{begin-eqn}} {{eqn|l = \mathcal F(1)      |r = \mathcal G((a)) }} {{eqn|r = (a, G_1(a)) }} {{eqn|ll= \implies      |l = f(1)      |r = G_1(a) }} {{end-eqn}} Now assume that we have that: :$\mathcal F(n) = \left({f(0), f(1), \ldots, f(n-1), G_n \left({f(0), \ldots, f(n-1)}\right)}\right)$ :$f(n) = G_n \left({f(0), \ldots, f(n-1)}\right)$. Then: {{begin-eqn}} {{eqn|l = \mathcal F(n+1)      |r = \mathcal G( \mathcal F(n) ) }} {{eqn|r = \mathcal G \left({ (f(0),\ldots, f(n-1), G_{n-1} \left({f(0),\ldots,f(n-1)}\right)}\right)      |c = Induction hypothesis on $\mathcal F$ }} {{eqn|r = \mathcal G \left({ (f(0),\ldots, f(n-1), f(n)}\right)      |c = Induction hypothesis on $f$ }} {{eqn|r = \left({f(0), \ldots, f(n), G_n \left({f(0), \ldots, f(n)}\right)}\right)      |c = Definition of $\mathcal G$ }} {{eqn|ll = \implies      |l = f(n+1)      |r = \operatorname{last}( \mathcal F(n+1) ) }} {{eqn|r = G_n \left({f(0), \ldots, f(n)}\right) }} {{end-eqn}} The result follows by the Principle of Mathematical Induction. {{qed}}
=== Necessary Condition === Let $R_p$ and $R_q$ be coprime. {{AimForCont}} $p$ and $q$ are not coprime. Let $p = d m, q = d n$ for some $d > 1$. Then by Divisors of Repunit with Composite Index: :$R_d \divides R_p$ and: :$R_d \divides R_q$ where $\divides$ denotes divisibility. That is, $R_d$ is a common divisor of $R_p$ and $R_q$. This contradicts our hypothesis that $R_p$ and $R_q$ are coprime. Thus by Proof by Contradiction, $p$ and $q$ are coprime. {{qed|lemma}} === Sufficient Condition === Let $p$ and $q$ be coprime. Let $R_p$ and $R_q$ be repunits in base $b$. By Integer Combination of Coprime Integers, there exists integers $m, n$ such that $m p + n q = 1$. One of $m, n$ is strictly positive while the other is negative. {{WLOG}}, suppose $m > 0$ and $n \le 0$. Then from $m p = 1 - n q$: We have $R_{m p} - b R_{- n q} = 1$. By Divisors of Repunit with Composite Index, $R_p$ is a divisor of $R_{m p}$, while $R_q$ is a divisor of $R_{- n q}$. By definition of divisibility, there exists integers $c, d$ such that $R_{m p} = c R_p, R_{- n q} = d R_q$. Therefore $c R_p - b d R_q = 1$. Note that both $c$ and $- b d$ are integers. By Integer Combination of Coprime Integers, $R_p$ and $R_q$ are coprime. {{qed}}