hcju/proofwikiqa huggingface dataset:
These are the primes which coincide with the upper end of a prime gap greater than $6$. These can be found at: :$89$ to $97$: prime gap of $8$ :$113$ to $127$: prime gap of $14$ :$139$ to $149$: prime gap of $10$ and so on. We have that: {{begin-eqn}} {{eqn | l = 98       | r = 19 + 79 }} {{eqn | l = 122       | r = 13 + 109 }} {{eqn | l = 124       | r = 11 + 113 }} {{eqn | l = 126       | r = 13 + 113 }} {{eqn | l = 128       | r = 19 + 109 }} {{end-eqn}} and so on. {{qed}}
Performing the calculation using long division: <pre>    0.0101...   ---------- 99)1.0000000      99      --       100        99       ---       ... </pre> {{qed}}
We note as examples: {{begin-eqn}} {{eqn | l = 9^2       | r = 81 }} {{eqn | l = 8 + 1       | r = 9 }} {{eqn | l = 99^2       | r = 9801 }} {{eqn | l = 98 + 01       | r = 99 }} {{eqn | l = 999^2       | r = 998001 }} {{eqn | l = 998 + 001       | r = 999 }} {{end-eqn}} Now we consider: {{begin-eqn}} {{eqn | l = \paren {\sum_{k \mathop = 0}^n 9 \times 10^k}^2       | r = \paren {10^{n + 1} - 1}^2       | c = Sum of Geometric Sequence }} {{eqn | r = 10^{n + 1} \paren {10^{n + 1} - 2} + 1 }} {{end-eqn}} So $\paren {10^{n + 1} - 1}^2$ can be split as: :$\paren {10^{n + 1} - 2} + 1 = 10^{n + 1} - 1 = \displaystyle \sum_{k \mathop = 0}^n 9 \times 10^k$ Thus $\displaystyle \sum_{k \mathop = 0}^n 9 \times 10^k$ is a Kaprekar number. {{qed}}
It should be noted that for any two ordinals, $x \lt y \iff x \in y$. Let $P \left({x}\right)$ be a property that satisfies the above conditions. Aiming for contradiction, let $y$ be an ordinal such that $\neg P \left({y}\right)$. By the Rule of Transposition, it follows that if $y$ is an ordinal such that $\neg P \left({y}\right)$, then there exists an ordinal $x \lt y \iff x \in y$ such that $\neg P \left({x}\right)$. Let $\alpha$ be the smallest ordinal in $y$ such that $\neg P \left({\alpha}\right)$. We can choose such an element because an ordinal is strictly well-ordered by definition. Since $\neg P \left({\alpha}\right)$, let $\beta$ be the smallest ordinal in $\alpha$ such that $\neg P \left({\beta}\right)$. Since $y$ is transitive by definition: :$\beta \in \alpha \land \alpha \in y \implies \beta \in y$ And so $\beta$ is an ordinal in $y$ such that $\neg P \left({\beta}\right)$. But since $\beta \lt \alpha$, this contradicts the fact that $\alpha$ is the smallest ordinal of $y$ such that $\neg P \left({\alpha}\right)$. Therefore $P \left({x}\right)$ holds for all ordinals. Hence the result. {{qed}}
It should be noted that for any two ordinals, $x \lt y \iff x \in y$. Let $P \left({x}\right)$ be a property that satisfies the above conditions. Aiming for contradiction, let $y$ be an ordinal such that $\neg P \left({y}\right)$. Let $S$ be the set defined as: :$S = \left\{{x \in y^+: \neg P \left({x}\right)}\right\}$ It is seen that this set is non-empty because by Set is Element of Successor: :$y \in y^+$ $y^+$ is an ordinal because of Successor Set of Ordinal is Ordinal. By Element of Ordinal is Ordinal, $S$ is a set of ordinals. By the Rule of Transposition, it follows that for every $x \in S$, there exists an ordinal $\alpha \lt x \iff \alpha \in x$ such that $\neg P \left({\alpha}\right)$. But by transitivity: :$\alpha \in x \land x \in y^+ \implies \alpha \in y^+$ So the epsilon restriction $\in_S$ is a right-total relation on $S$. Therefore by the Axiom of Dependent Choice there exists a sequence $\left\langle{x_n}\right\rangle_{n \in \N}$ in $S$ such that: :$\forall n \in \N: x_{n+1} \in x_n$ But this contradicts No Infinitely Descending Membership Chains. And thus $P \left({x}\right)$ holds for all ordinals. Hence the result. {{qed}}
It should be noted that for any two ordinals, $x \lt y \iff x \in y$. Let $\phi \left({x}\right)$ be a property that satisfies the above conditions. Aiming for contradiction, let $y$ be an ordinal such that $\neg \phi \left({y}\right)$. It is noted that $y \ne \varnothing$. Therefore $y$ must be either a successor ordinal or a limit ordinal. If $y$ is a successor ordinal, then let $x$ be the ordinal such that $x^+ = y$. By the Rule of Transposition, it is seen that $\neg \phi \left({x}\right)$. By Set is Element of Successor, it follows that $x \in y$, and so $x \lt y$. If $y$ is a limit ordinal, then by the Rule of Transposition there exists an ordinal $x \lt y \iff x \in y$ such that $\neg \phi \left({x}\right)$. It has been shown that if $y$ is an ordinal such that $\neg \phi \left({y}\right)$, then there exists an ordinal $x \in y$ such that $\neg \phi \left({x}\right)$. The rest follows from the proof of Schema 1. {{qed}}
Define the class: :$A := \left\{{x \in \operatorname{On}: \phi \left({x}\right) = \mathrm T}\right\}$. Then $\phi \left({x}\right) = \mathrm T$ is equivalent to the statement: :that $x \in A$ The three conditions in the hypothesis become: :$(1a): \quad \varnothing \in A$ :$(2a): \quad x \in A \implies x^+ \in A$ :$(3a): \quad \left({\forall x < y: x \in A}\right) \implies y \in A$ These are precisely the conditions for the class $A$ in the second principle of transfinite induction. Therefore, $\operatorname{On} \subseteq A$. Thus, $\phi \left({x}\right)$ holds for all $x \in \operatorname{On}$. {{qed}}
Performing the calculation using long division: <pre>     0.0097087378640776699029126213592233009...   -------------------------------------------- 103)1.0000000000000000000000000000000000000...       927     618     927     103    309       ---     ---     ---     ---    ---        730     420      300    370     1000        721     412      206    309      927        ---     ---      ---    ---     ----          900     800     940    610    ....          824     721     927    515          ---     ---     ---    ---           760     790     130    950           721     721     103    927           ---     ---     ---    ---            390     690     270    230            309     618     206    206            ---     ---     ---    ---             810     720     640    240             721     618     618    206             ---     ---     ---    ---              890    1020     220    340              824     927     206    309              ---    ----     ---    ---               660     930     140    310               618     927     103    309 </pre> {{ProofWanted|Still to be demonstrated that it is the smallest.}}
First it is demonstrated that $105$ itself satisfies this property. Let $d \in \Z_{> 1}$ be odd and coprime to $105$. Then $d$ does not have $3$, $5$ or $7$ as a prime factor. Thus $d$ must have at least one odd prime as a divisor which is $11$ or greater. The smallest such composite number is $11^2$. But $11^2 = 121 > 105$. Thus $d$ must be prime. Thus it has been demonstrated that all odd integers greater than $1$ and smaller than $105$ which are coprime to $105$ are prime. {{qed|lemma}} Using an argument similar to the above, we see that for an integer to have this property,  if it is greater than $p^2$ for some odd prime $p$, then it must be divisible by $p$. If not, it will be coprime to $p^2$, a composite number. Let $p_n$ denote the $n$th prime. Suppose $N$ has this property. By the argument above, if $p_{n + 1}^2 \ge N > p_n^2$, we must have $p_2 p_3 \cdots p_n \divides N$. By Absolute Value of Integer is not less than Divisors, we have $p_2 p_3 \cdots p_n \le N$. Bertrand-Chebyshev Theorem asserts that there is a prime between $p_n$ and $2 p_n$. Thus we have $2 p_n > p_{n + 1}$. Hence for $n \ge 5$: {{begin-eqn}} {{eqn | l = N       | o = \ge       | r = p_2 p_3 \cdots p_n }} {{eqn | r = 3 \times 5 p_4 \cdots p_n }} {{eqn | o = >       | r = 8 p_{n - 1} p_n }} {{eqn | o = >       | r = 4 p_n^2       | c = Bertrand-Chebyshev Theorem }} {{eqn | o = >       | r = p_{n + 1}^2       | c = Bertrand-Chebyshev Theorem }} {{eqn | o = \ge       | r = N       | c = From assumption }} {{end-eqn}} This is a contradiction.  Hence we must have $N \le p_5^2 = 121$. From the argument above we also have: :$3 \divides N$ for $9 < N \le 25$ :$3, 5 \divides N$ for $25 < N \le 49$ :$3, 5, 7 \divides N$ for $49 < N \le 121$ So we end up with the list $N = 1, 3, 5, 7, 9, 15, 21, 45, 105$. {{qed}}
{{begin-eqn}} {{eqn | l = \map \phi {105}       | r = 48       | c = {{EulerPhiLink|105}} }} {{eqn | l = \map \phi {106}       | r = 52       | c = {{EulerPhiLink|106}} }} {{eqn | l = \map \phi {107}       | r = 106       | c = Euler Phi Function of Prime: $107$ is prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map \phi {165}       | r = 80       | c = {{EulerPhiLink|165}} }} {{eqn | l = \map \phi {166}       | r = 82       | c = {{EulerPhiLink|166}} }} {{eqn | l = \map \phi {167}       | r = 166       | c = Euler Phi Function of Prime: $167$ is prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map \phi {315}       | r = 144       | c = {{EulerPhiLink|315}} }} {{eqn | l = \map \phi {316}       | r = 156       | c = {{EulerPhiLink|316}} }} {{eqn | l = \map \phi {317}       | r = 316       | c = Euler Phi Function of Prime: $317$ is prime }} {{end-eqn}} {{qed}}
We have: {{begin-eqn}} {{eqn | l = \phi \left({35}\right)       | r = 24       | c = Euler Phi Function of 35 }} {{eqn | l = \tau \left({35}\right)       | r = 4       | c = Tau Function of 35 }} {{eqn | ll= \leadsto       | l = \phi \left({35}\right) \times \tau \left({35}\right)       | r = \dfrac {24 \times 4} 2       | c =  }} {{eqn | r = 48       | c =  }} {{eqn | r = \sigma \left({35}\right)       | c = Sigma Function of 35 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \phi \left({105}\right)       | r = 48       | c = Euler Phi Function of 105 }} {{eqn | l = \tau \left({105}\right)       | r = 8       | c = Tau Function of 105 }} {{eqn | ll= \leadsto       | l = \phi \left({105}\right) \times \tau \left({105}\right)       | r = \dfrac {48 \times 8} 2       | c =  }} {{eqn | r = 192       | c =  }} {{eqn | r = \sigma \left({105}\right)       | c = Sigma Function of 105 }} {{end-eqn}} {{qed}}
From 35 Hexominoes and by inspection, none of the $35$ hexominoes has a hole. From Number of Heptominoes and by inspection, exactly one of the $108$ heptominoes has a hole. This is the smallest polyomino with a hole. {{qed}}
The smallest prime magic square (including $1$) is: {{:Prime Magic Square/Examples/Order 3/Smallest}} As can be seen by inspection, the sums of the elements in the rows, columns and diagonals are $111$: {{begin-eqn}} {{eqn | l = 67 + 1 + 43       | r = 111 }} {{eqn | l = 13 + 37 + 61       | r = 111 }} {{eqn | l = 31 + 73 + 7       | r = 111 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 67 + 13 + 31       | r = 111 }} {{eqn | l = 1 + 37 + 73       | r = 111 }} {{eqn | l = 43 + 61 + 7       | r = 111 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 67 + 37 + 7       | r = 111 }} {{eqn | l = 43 + 37 + 31       | r = 111 }} {{end-eqn}} {{qed}}
{{:Definition:Lucky Number/Sequence}} Of these, the palindromic ones can be picked out by inspection. {{qed}}
{{:Definition:Lucky Number/Sequence}} Of these, the square ones can be picked out by inspection. {{qed}} Category:Lucky Numbers Category:Square Numbers 926u5u5kr0ql3e1xd83oaoyatuxjcbw
Let $c \in \R$ be conjugate to $a$, such that $c < a$. By assumption: :$c \notin \closedint a b$ Hence, there exists a set $\closedint c b$ such that: :$\closedint c b = \closedint c a \cup \closedint a b$ where $\size {c - a} > 0$. By there exists a real point between two real points: :$\exists \epsilon: \size {c - a} > \epsilon > 0$ Therefore, there exists $\epsilon > 0$ such that: :$\gamma$ can be extended onto the whole interval $\closedint {a - \epsilon} b$, where extension means definition of some mapping in $\hointr {a - \epsilon} a$ {{explain|the author provides this as a condition, but does a definability affect existence ? }} :the interval $\closedint {a - \epsilon} b$ contains no points conjugate to $a$. Consider a family of extremals leaving the point $\tuple {a - \epsilon, \map {\mathbf y} {a - \epsilon} }$. There are no points conjugate to $a - \epsilon$ in $\closedint {a - \epsilon} b$. Hence, for $x \in \closedint a b$ no two extremals in this family which are sufficiently close to the original extremal $\gamma$ can intersect. {{explain|Follows from the definition, but a picture or a better explanation would help}} Since all the functions are extremals, they satisfy same differential equations. Lack of intersection implies different boundary conditions. Denote these conditions collectively as $\map {\boldsymbol \psi} {x, \mathbf y} = \map {\mathbf y'} x$. Thus, in some region $R$ containing $\gamma$ extremals sufficiently close to $\gamma$ define a central field in which $\gamma$ is embedded. By Central Field is Field of Functional, $\gamma$ can be embedded in the field of functional. {{qed}}
Let $x^2 - y^2 = R_n$ for some $n$, where $R_n$ denotes the $n$-digit repunit. From Integer as Difference between Two Squares: :$R_n$ has at least two distinct divisors of the same parity. Then from Difference of Two Squares: :$x = \dfrac {a + b} 2$ :$y = \dfrac {a - b} 2$ where: :$R_n = a b$ for all $a, b$ where: :$a b = R_n$ :$a$ and $b$ are of the same parity. Here we have that $R_n$ is odd. So both $a$ and $b$ are always odd and therefore always of the same parity. It remains to perform the calculations and evaluate the examples.
:$113$ is prime. :$131$ is prime. :$311$ is prime. Consider the $3$-digit primes smaller than $113$: :$101, 103, 107, 109$ They all contain a zero. Thus, for each of these, at least one permutation ends in a zero. Hence it is divisible by $10$ and so is not prime. {{qed}}
There are two cases to consider: either $\struct {G, \circ}$ has an infinite cyclic subgroup, or it does not. === Case 1 === Suppose that $\struct {G, \circ}$ has an infinite cyclic subgroup denoted as $H$. Let $a \in G$ be the element of $G$ such that $\gen a = H$. Then by Distinct Subgroups of Infinite Cyclic Group, $\sequence {\gen {a^k} }_{k \mathop \in \N^*}$ is a sequence of distinct subgroups of $H$. So, by Subgroup of Subgroup is Subgroup, it follows that $\struct {G, \circ}$ has an infinite number of subgroups. === Case 2 === Suppose that $\struct {G, \circ}$ does ''not'' have an infinite cyclic subgroup. Then for every $x \in G$, we have that $\gen x$ is a finite group. Let $a, b \in G$. Let $\sim$ be the equivalence relation on $G$ defined as: :$a \sim b \iff \gen a = \gen b$ Let $\eqclass a \sim$ denote the equivalence class of $a$ under $\sim$. Let $x \sim a$. Then: :$\gen x = \gen a$ Since $x \in \gen x$, it follows that: :$x \in \gen a$ It has been shown that: :$\eqclass a \sim \subseteq \gen a$ By Subset of Finite Set is Finite, every equivalence class under $\sim$ is finite. By Equivalence Class holds Equivalent Elements: :$\eqclass a \sim \ne \eqclass b \sim \implies a \not \sim b \implies \gen a \ne \gen b$ So every equivalence class generates a unique subgroup of $\struct {G, \circ}$. By Union of Equivalence Classes is Whole Set, the union of all equivalence classes under $\sim$ must equal $G$. From Finite Union of Finite Sets is Finite, it follows that this must be a (countably) infinite union. Thus there must exist an infinite number of equivalence classes, and hence an infinite number of subgroups. The result then follows from Proof by Cases. {{qed}} Category:Infinite Groups 897cv70vtox46af0sbs3np9gwkuncrc
=== Necessary Condition === Suppose that $\struct {G, \circ}$ is a finite group. Let $\struct {H, \circ}$ be a subgroup of $\struct {G, \circ}$. $H \subseteq G$ by definition. Therefore: : $H \in \powerset G$ where $\powerset G$ denotes the power set of $G$. By Power Set of Finite Set is Finite, $\powerset G$ is finite. It is seen that the set of all subgroups form a subset of $\powerset G$. The result then follows from Subset of Finite Set is Finite. {{qed|lemma}} === Sufficient Condition === Suppose that $\struct {G, \circ}$ is a group with only a finite number of subgroups. It is noted that an Infinite Group has Infinite Number of Subgroups. The result then follows from the Rule of Transposition. {{qed}}
It is confirmed that: :$113, 131, 311$ are all prime :$199, 919, 991$ are all prime :$337, 373, 733$ are all prime. From Digits of Permutable Prime, all permutable primes contain digits in the set: :$\left\{ {1, 3, 7, 9}\right\}$ The sum of a $3$-digit repdigit number is divisible by $3$. By Divisibility by 3 it follows that all $3$-digit repdigit numbers are divisible by $3$ and therefore composite. Hence the possibly permutable primes are: :$113, 117, 119, 133, 137, 139, 177, 179, 199, 337, 339, 377, 379, 399, 779, 799$ and their anagrams. We eliminate $113, 199, 337$ from this list, as it has been established that they are permutable primes. Of those remaining, the following are composite: {{begin-eqn}} {{eqn | l = 117       | r = 3^2 \times 13 }} {{eqn | l = 119       | r = 7 \times 17 }} {{eqn | l = 133       | r = 7 \times 19 }} {{eqn | l = 177       | r = 3 \times 59 }} {{eqn | l = 339       | r = 3 \times 113 }} {{eqn | l = 377       | r = 13 \times 29 }} {{eqn | l = 779       | r = 19 \times 41 }} {{eqn | l = 799       | r = 17 \times 47 }} {{end-eqn}} It remains to demonstrate that at least one anagram of the remaining numbers: :$137, 139, 179, 379$ is composite. We find that: {{begin-eqn}} {{eqn | l = 371       | r = 7 \times 53 }} {{eqn | l = 319       | r = 11 \times 29 }} {{eqn | l = 791       | r = 7 \times 113 }} {{eqn | l = 793       | r = 13 \times 61 }} {{end-eqn}} All contenders are eliminated except for the established permutable primes $113, 119, 337$ and their anagrams. {{qed}}
First note that from 3-Digit Permutable Primes, both $337$ and $199$ are permutable primes. Hence it follows that all the elements of $\left\{ {1, 3, 7, 9}\right\}$ appear in at least one permutable prime. Let $p$ contain an even digit $d$. Then at least one anagram $p'$ of $p$ has $d$ at the end. From Divisibility by 2, it follows that $p'$ is even. As $p$ has more than $1$ digit it follows that $p'$ is composite. Let $p$ contain the digit $5$. Then at least one anagram $p'$ of $p$ has $5$ at the end. From Divisibility by 5, it follows that $p'$ is divisible by $5$. Except for $5$ itself, all integers which are divisible by $5$ are composite. As $p$ has more than $1$ digit it follows that $p'$ is composite. The result follows by definition of permutable prime. {{qed}} Category:Permutable Primes fg1nktn31ejo7yn7ruv3oli7m6hih4k
Let $N$ be divisible by $5$. Then: {{begin-eqn}} {{eqn | l = N       | o = \equiv       | r = 0 \pmod 5 }} {{eqn | ll= \leadstoandfrom       | l = \sum_{k \mathop = 0}^n a_k 10^k       | o = \equiv       | r = 0 \pmod 5 }} {{eqn | ll= \leadstoandfrom       | l = a_0 + 10 \sum_{k \mathop = 1}^n a_k 10^{k - 1}       | o = \equiv       | r = 0 \pmod 5 }} {{eqn | ll= \leadstoandfrom       | l = a_0       | o = \equiv       | r = 0 \pmod 5       | c = as $10 \equiv 0 \pmod 5$ }} {{end-eqn}} {{qed}} Category:Divisibility Tests Category:5 3tvnw0m20osbfmwu5igoai5vzdssl35
Let $n$ be a repdigit number with $k$ digits. Then by the Basis Representation Theorem: :$\displaystyle n = \sum_{j \mathop = 0}^k m b^j$ for some $m$ such that $1 \le m < b$. Let $m \ge 2$. Then: :$\displaystyle n = m \sum_{j \mathop = 0}^k b^j$ and so has $m$ as a divisor. Hence $n$ is not prime. The result follows by the Rule of Transposition. {{qed}} Category:Repdigit Numbers Category:Repunits Category:Repunit Primes 35tgqldbngld9g08gyfadp7qrakygnz
It is confirmed that: :$13$ and $31$ are both prime :$17$ and $71$ are both prime :$37$ and $73$ are both prime. :$79$ and $97$ are both prime. From Digits of Permutable Prime, all permutable primes contain digits in the set: :$\left\{ {1, 3, 7, 9}\right\}$ By Prime Repdigit Number is Repunit, all $2$-digit repdigit numbers are composite except for $11$ which is prime. From Repunit Prime is Permutable Prime, it follows that $11$ is a permutable prime. Apart from $11$ whose status is known, the possibly permutable primes are: :$13, 17, 19, 37, 39, 79$ and their anagrams. We eliminate $13, 17, 37, 79$ from this list, as it has been established that they are permutable primes. Of those remaining, the following is composite: {{begin-eqn}} {{eqn | l = 39       | r = 3 \times 13 }} {{end-eqn}} It remains to demonstrate that the anagram of $19$, that is, $91$, composite. We find that: {{begin-eqn}} {{eqn | l = 91       | r = 7 \times 13 }} {{end-eqn}} All contenders are eliminated except for the established permutable primes $11, 13, 17, 37, 79$ and their anagrams. {{qed}}
{{questionable|See below.}} It is demonstrated that the result is true for $n = 118$: :$\dfrac {4 \times 118} 3 = 157 \cdotp \dot 3$ The primes between $118$ and $157$ are: {{begin-eqn}} {{eqn | l = 127       | r = 4 \times 32 - 1 }} {{eqn | l =        | r = 6 \times 21 + 1 }} {{eqn | l = 131       | r = 4 \times 33 - 1 }} {{eqn | l =        | r = 6 \times 22 - 1 }} {{eqn | l = 137       | r = 4 \times 34 + 1 }} {{eqn | l =        | r = 6 \times 23 - 1 }} {{eqn | l = 139       | r = 4 \times 35 - 1 }} {{eqn | l =        | r = 6 \times 23 + 1 }} {{eqn | l = 149       | r = 4 \times 37 + 1 }} {{eqn | l =        | r = 6 \times 25 - 1 }} {{eqn | l = 151       | r = 4 \times 38 - 1 }} {{eqn | l =        | r = 6 \times 25 + 1 }} {{eqn | l = 157       | r = 4 \times 39 + 1 }} {{eqn | l =        | r = 6 \times 26 + 1 }} {{end-eqn}} So it can be seen that: :the primes of the form $4 m - 1$ are: ::$127, 131, 139, 151$ :the primes of the form $4 m + 1$ are: ::$137, 139, 157$ :the primes of the form $6 m - 1$ are: ::$131, 137, 149$ :the primes of the form $6 m + 1$ are: ::$127, 139, 151, 157$ and so the conditions of the result are fulfilled for $118$. But then note we have: :$\dfrac {4 \times 117} 3 = 156$ and so the primes between $117$ and $157$ are the same ones as between $118$ and $157$, excluding $157$ itself. Thus the conditions also hold for $117$. It is puzzling as to why the condition $n \ge 118$ has been applied to the initial statement.
{{begin-eqn}} {{eqn | l = 1^3 + 2^3 + 3^3 + 4^3 + 5^3       | r = 1 + 8 + 27 + 64 + 125       | c =  }} {{eqn | r = 225       | c =  }} {{eqn | r = 15^2       | c = also see Sum of Sequence of Cubes }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 96^3 + 97^3 + 98^3 + 99^3 + 100^3       | r = 884 \, 736 + 912 \, 673 + 941 \, 192 + 970 \, 299 + 1 \, 000 \, 000       | c =  }} {{eqn | r = 4 \, 708 \, 900       | c =  }} {{eqn | r = 2170^2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 118^3 + 119^3 + 120^3 + 121^3 + 122^3       | r = 1 \, 643 \, 032 + 1 \, 685 \, 159 + 1 \, 728 \, 000 + 1 \, 771 \, 561 + 1 \, 815 \, 848       | c =  }} {{eqn | r = 8 \, 643 \, 600       | c =  }} {{eqn | r = 2940^2       | c =  }} {{end-eqn}} {{ProofWanted|It remains to be shown that these are the only ones.}}
From Tau Function of 120: :$\map \tau {120} = 16$ The result is a specific instance of Smallest Number with $2^n$ Divisors: :$120 = 2 \times 3 \times 4 \times 5$ {{qed}}
The aliquot sum of an integer $n$ is the integer sum  of the aliquot parts of $n$. That is, the aliquot sum of $360$ is the $\sigma$ function of $360$ minus $360$. Thus: {{begin-eqn}} {{eqn | l = \map \sigma {360} - 360       | r = 1170 - 360       | c = {{SigmaLink|360}} }} {{eqn | r = 810       | c =  }} {{eqn | ll= \leadsto       | l = 4 \times \paren {\map \sigma {360} - 360}       | r = 3240       | c =  }} {{eqn | r = 9 \times 360       | c =  }} {{end-eqn}} {{qed}}
Consider $11_b$. By the Basis Representation Theorem: :$11_b = b + 1$ Thus: {{begin-eqn}} {{eqn | l = {11_b}^2       | r = \paren {b + 1}^2       | c =  }} {{eqn | r = b^2 + 2 b + 1       | c = Square of Sum }} {{eqn | r = 121_b       | c =  }} {{end-eqn}} Thus: :$121_b = {11_b}^2$ and so is a square number. {{qed}}
From Square of 11 is Palindromic: :$11^2 = 121$ Thus: <pre>  121 x 11 -----  121 1210 ----- 1331 </pre>{{qed}}
From Square of 11 is Palindromic: :$11^2 = 121$ Then from Square of 121 is Palindromic: :$121^2 = 14641 = \paren {11^2}^2 = 11^4$ {{qed}}
:$121 = 1 + 3 + 9 + 27 + 81$ {{ProofWanted|It remains to be shown that this is the only such square number.}}