hcju/proofwikiqa huggingface dataset:
{{ProofWanted|Can easily be proved that $n$ must itself be prime for the expression to be prime. Then it's a matter of checking them all.}} Note that if $4 p + 1$ is prime for prime $p$, then $\dfrac {p^p - 1} {p - 1}$ is divisible by $4 p + 1$: Let $q = 4 p + 1$ be prime. By First Supplement to Law of Quadratic Reciprocity: :$\paren {\dfrac {-1} q} = 1$ that is, there exists some integer $I$ such that $I^2 \equiv -1 \pmod q$. Then: {{begin-eqn}} {{eqn | l = \paren {1 + I}^4       | r = \paren {1 + 2 I + I^2}^2       | c = Square of Sum }} {{eqn | o = \equiv       | r = \paren {2 I}^2       | rr= \pmod q       | c = Congruence of Powers }} {{eqn | o = \equiv       | r = -4       | rr= \pmod q }} {{end-eqn}} and thus: {{begin-eqn}} {{eqn | l = p^p       | o = \equiv       | r = p^p \paren {1 + I}^{q - 1}       | rr= \pmod q       | c = Fermat's Little Theorem }} {{eqn | o = \equiv       | r = p^p \paren {1 + I}^{4 p}       | rr= \pmod q }} {{eqn | o = \equiv       | r = p^p \paren {-4}^p       | rr= \pmod q }} {{eqn | o = \equiv       | r = \paren {-4 p}^p       | rr= \pmod q }} {{eqn | o = \equiv       | r = 1^p       | rr= \pmod q }} {{eqn | o = \equiv       | r = 1       | rr= \pmod q }} {{end-eqn}} Hence $q \divides \paren {p^p - 1}$. Obviously $q > p - 1$. Therefore $q \divides \dfrac {p^p - 1} {p - 1}$.
=== Case $1$: $\mathtt {xy}$ is odd === Consider the last $3$ digits of $n$: $\mathtt {yxy}$. By Odd Square Modulo 8: :$n \equiv \mathtt {yxy} \equiv 1 \pmod 8$ By Square Modulo 5: :$n \equiv \mathtt {yxy} \equiv 0, 1, 4 \pmod 5$ We have: {{begin-eqn}} {{eqn | l = \mathtt {yxy}       | o = \equiv       | r = 25, 1, 9       | rr = \pmod {40}       | c = Chinese Remainder Theorem }} {{eqn | o = \equiv       | r = 1, 9, 25, 41, 49, 65, 81, 89, 105, 121, 129, 145, 161, 169, 185       | rr = \pmod {200} }} {{end-eqn}} Since $y$ is odd, we have $\mathtt {yxy} \equiv 1 \mathtt {xy} \pmod {200}$. Hence $\mathtt {xy}$ can only be $05, 21, 29, 45, 61, 69, 85$. Suppose $n$ ends in $5$ in decimal representation. By Divisibility by 5, $5 \divides n$. Since $5$ is a prime, $5 \divides \sqrt n$. By Odd Square Modulo 8 and Chinese Remainder Theorem: :$n \equiv 25 \pmod {200}$ for which none of $\mathtt {xy} = 05, 45, 85$ satisfy. This shows that $\mathtt {xy}$ can only be $21, 29, 61, 69$. {{qed|lemma}} === Case $2$: $\mathtt {xy}$ is even === Consider the last $2$ digits of $n$: $\mathtt {xy}$. We omit the trivial case $\mathtt {xy} = 00$. This case occurs whenever $10 \divides \sqrt n$. Since $\sqrt n$ is even: :$n \equiv \mathtt {xy} \equiv 0 \pmod 4$ Since $5 \nmid \sqrt n$, by Square Modulo 5: :$n \equiv \mathtt {xy} \equiv 1, 4 \pmod 5$ We have: {{begin-eqn}} {{eqn | l = \mathtt {xy}       | o = \equiv       | r = 16, 4       | rr = \pmod {20}       | c = Chinese Remainder Theorem }} {{eqn | o = \equiv       | r = 4, 16, 24, 36, 44, 56, 64, 76, 84, 96       | rr = \pmod {100} }} {{end-eqn}} In the list above: :$4, 24, 36, 44, 56, 76, 84$ are divisible by $2^2$ but not $2^4$ :$16, 96$ are divisible by $2^4$ but not $2^6$ :$64$ is divisible by $2^6$ Write $n = 10^{10} k + 101010101 \times \mathtt {xy}$. We require $\dfrac n {2^{2 d}}$ to be a square number as well, where $d$ is the largest integer where $2^{2d} \divides \mathtt {xy}$ (shown above). We have: {{begin-eqn}} {{eqn | l = \frac n {2^{2 d} }       | r = 5^{10} \times 2^{10 - 2 d} \times k + 101010101 \times \frac {\mathtt {xy} } {2^{2 d} } }} {{eqn | o = \equiv       | r = \frac {5 \paren {\mathtt {xy} } } {2^{2 d} }       | rr = \pmod 8       | c = as $10 - 2 d > 3$ }} {{eqn | o = \equiv       | r = 0, 1, 4       | rr = \pmod 8       | c = Square Modulo 8 }} {{end-eqn}} and: :$\mathtt {xy} = 04, 16, 24, 36, 44, 56, 64, 76, 84, 96$ corresponds to: {{begin-eqn}} {{eqn | l = \frac {3 \paren {\mathtt {xy} } } {2^{2 d} }       | r = 5, 5, 30, 45, 55, 70, 5, 85, 105, 30 }} {{eqn | o = \equiv       | r = 5, 5, 6, 5, 7, 6, 5, 5, 1, 6       | rr = \pmod 8 }} {{end-eqn}} Thus the only valid $\mathtt {xy}$ is $84$. {{qed|lemma}} We have considered all possibilities, and the only possible $\mathtt {xy}$ are: :$21, 29, 61, 69, 84$ and by the examples shown, all these numbers do indeed produce squares. {{qed}}
By definition of subset: :$\mathit{Ids}\left({L}\right) \subseteq \mathcal P\left({S}\right)$ Then :$\mathord\precsim = \mathord\precsim' \cap \left({\mathit{Ids}\left({L}\right) \times \mathit{Ids}\left({L}\right)}\right)$ Hence $I$ is ordered subset of $P$. === Infima Inheriting === Let $A$ be a subset of $\mathit{Ids}\left({L}\right)$ such that :$A$ admits an infimum in $P$. By the proof of Power Set is Complete Lattice: :$\inf_P A = \bigcap A$ By Intersection of Semilattice Ideals is Ideal/Set of Sets: :$\inf_P A \in \mathit{Ids}\left({L}\right)$ Thus by Infimum in Ordered Subset: :$A$ admits an infimum in $I$ and $\inf_I A = \inf_P A$ Hence $I$ inherits infima. {{qed|lemma}} === Directed Suprema Inheriting === Let $D$ be a directed subset of $\mathit{Ids}\left({L}\right)$ such that :$D$ admits a supremum in $P$. By the proof of Power Set is Complete Lattice: :$\sup_P D = \bigcup D$ We will prove that :$\bigcup D$ is an ideal in $L$. ==== Directed ==== Let $x, y \in \bigcup D$. By definition of union: :$\exists I_1 \in D: x \in I_1$ and :$\exists I_2 \in D: y \in I_2$ By definition of directed: :$\exists I \in D: I_1 \precsim I \land I_2 \precsim I$ By definition of $\precsim$: :$I_1 \subseteq I$ and $I_2 \subseteq I$ By definition of subset: :$x, y \in I$ By definition of directed: :$\exists z \in I: x \preceq z \land y \preceq z$ Thus by definition of union: :$\exists z \in \bigcup D: x \preceq z \land y \preceq z$ {{qed|lemma}} ==== Lower Set ==== Let $x \in \bigcup D$, $y \in S$ such that :$y \preceq x$ By definition of union: :$\exists I \in D: x \in I$ By definition of lower set: :$y \in I$ Thus by definition of union: :$y \in \bigcup D$ {{qed|lemma}} ==== Non-Empty Set ==== By definition of directed: :$D$ is non-empty and $\forall I \in D: I$ is non-empty. Thus by definitions of non-empty set and union: :$\bigcup D$ is non-empty. {{qed|lemma}} By definition of $\mathit{Ids}$: :$\bigcup D \in \mathit{Ids}\left({L}\right)$ Hence $I$ inherits directed suprema. {{qed}}
By definition, $ \Delta J \sqbrk y$ can be expressed as: :$\Delta J \sqbrk {y; h} = \delta J \sqbrk {y; h} + \delta^2 J \sqbrk {y; h} + \epsilon \size h^2$ By assumption: :$\delta J \sqbrk {\hat y; h} = 0$ Hence: :$\Delta J \sqbrk {\hat y; h} = \delta^2 J \sqbrk {\hat y; h} + \epsilon \size h^2$ Therefore, for sufficiently small $\size h$ both $\Delta J \sqbrk {\hat y; h}$ and $\delta^2 J \sqbrk {\hat y; h}$ will have the same sign.   {{qed|lemma}} {{AimForCont}} there exists $h = h_0$ such that: :$\delta^2 J \sqbrk {\hat y; h_0} < 0$  Then, for any $\alpha \ne 0$: {{begin-eqn}} {{eqn | l = \delta^2 J \sqbrk {\hat y; \alpha h_0}       | r = \alpha^2 \delta^2 J \sqbrk {\hat y; h_0} }} {{eqn | o = <       | r = 0 }} {{end-eqn}} Therefore, $\Delta J \sqbrk {\hat y; h}$ can be made negative for arbitrary small $\size h$. However, by assumption $\Delta J \sqbrk {\hat y; h}$ is a minimum of $\Delta J \sqbrk {y; h}$ for all sufficiently small $\size h$. This is a contradiction. Thus, a function $h_0: \delta^2 J \sqbrk {\hat y; h_0} < 0$ does not exist. In other words: :$\delta^2 J \sqbrk {\hat y; h} \ge 0$  for all $h$. {{qed}}
First we note that: :$1! = 1$ which has $1$ digit. Then from Examples of Factorials: :$2! = 2$ which has $1$ digit. Multiplying $n$ by a $1$-digit number increases the number of digits of $n$ by no more than $1$. Thus from $n = 3$ to $n = 9$, $n!$ has no more than $n - 1$ digits. From $21$ Factorial: :$21! = 51 \, 090 \, 942 \, 171 \, 709 \, 440 \, 000$ and so $21!$ has $20$ digits. Multiplying $n$ by a $2$-digit number increases the number of digits of $n$ by at least $1$. Thus from $n = 10$ to $n = 20$, $n!$ has no more than $n - 1$ digits. It is noted that: :From $22$ Factorial: ::$22! = 1 \, 124 \, 000 \, 727 \, 777 \, 607 \, 680 \, 000$ :which has $22$ digits. :From $23$ Factorial: ::$23! = 25 \, 852 \, 016 \, 738 \, 884 \, 976 \, 640 \, 000$ :which has $23$ digits. :From $24$ Factorial: ::$24! = 620 \, 448 \, 401 \, 733 \, 239 \, 439 \, 360 \, 000$ :which has $24$ digits. But from $25$ Factorial: :$25! = 15 \, 511 \, 210 \, 043 \, 330 \, 985 \, 984 \, 000 \, 000$ which has $26$ digits. As above, multiplying $n$ by a number with $2$ digits or more increases the number of digits of $n$ by at least $1$. So if $n!$ has more than $n$ digits, $\paren {n + 1}!$ has more than $n + 1$ digits. It follows by induction that for all $n \ge 25$, $n!$ has more than $n$ digits. Hence the result. {{qed}}
Let $\displaystyle n = \sum_{k \mathop = 0}^m a_k 10^k$ be a number satisfying the conditions above. Then: {{begin-eqn}} {{eqn | n = 1       | l = \sum_{k \mathop = 0}^m a_k^2       | o = <       | r = 10 }} {{eqn | n = 2       | l = a_k       | r = a_{m - k}       | rr = \forall k: 0 \le k \le m }} {{end-eqn}} Consider $\displaystyle n^2 = \paren {\sum_{k \mathop = 0}^m a_k 10^k}^2$. From definition of Multiplication of Polynomials, the coefficient of $10^l$ in the product is: {{begin-eqn}} {{eqn | o =       | r = \sum_{\substack {j \mathop + k \mathop = l \\ j, k \mathop \in \Z} } a_j a_k }} {{eqn | o = \le       | r = \sqrt {\paren {\sum_{j \mathop = 0}^l a_j^2} \paren {\sum_{k \mathop = 0}^l a_k^2} }       | c = Cauchy's Inequality }} {{eqn | r = \sum_{j \mathop = 0}^l a_j^2 }} {{eqn | o = \le       | r = \sum_{j \mathop = 0}^m a_j^2 }} {{eqn | o = <       | r = 10       | c = From (1) }} {{end-eqn}} so no carries occur in the multiplication, and this form satisfies Basis Representation Theorem. Moreover: {{begin-eqn}} {{eqn | l = \sum_{\substack {j \mathop + k \mathop = 2 m \mathop - l \\ j, k \mathop \in \Z} } a_j a_k       | r = \sum_{\substack {m \mathop - j \mathop + m \mathop - k \mathop = 2 m \mathop - l \\ m \mathop - j, m \mathop - k \mathop \in \Z} } a_{m - j} a_{m - k} }} {{eqn | r = \sum_{\substack {j \mathop + k \mathop = l \\ j, k \mathop \in \Z} } a_{m - j} a_{m - k} }} {{eqn | r = \sum_{\substack {j \mathop + k \mathop = l \\ j, k \mathop \in \Z} } a_j a_k       | c = From (2) }} {{end-eqn}} so the coefficient of $10^{2 m - l}$ is equal to the coefficient of $10^l$ in the expansion of $n^2$. This shows that $n^2$ is palindromic. {{qed}}
By assumption, $J$ has an extremum for $y = \hat y$: :$\delta J \sqbrk {\hat y; h} = 0$ The increment is expressible then as: :$\Delta J \sqbrk {\hat y; h} = \delta^2 J \sqbrk {\hat y; h} + \epsilon \size h^2$ where $\epsilon \to 0$ as $\size h \to 0$. By assumption, the second variation is strongly positive: :$\delta^2 J \sqbrk {\hat y; h} \ge k \size h^2, \quad k \in \R_{>0}$ Hence: :$\Delta J \sqbrk {\hat y; h} \ge \paren {k + \epsilon} \size h^2$ What remains to be shown is that there exists a set of $h$ such that $\epsilon$ is small enough so that {{RHS}} is always positive. Since $\epsilon \to 0$ as $\size h \to 0$, there exist $c \in \R_{>0}$, such that: :$\size h < c \implies \size \epsilon < \dfrac 1 2 k$ Choose $h$ such that this inequality holds. Then {{begin-eqn}} {{eqn | l = \frac 1 2 k       | o = >       | r = \epsilon > -\frac 1 2 k       | c = $\big \vert + k$, by Membership is Left Compatible with Ordinal Addition }} {{eqn | l = \frac 3 2 k       | o = >       | r = k + \epsilon > \frac 1 2 k       | c = $\big \vert \cdot \size h^2$, by Membership is Left Compatible with Ordinal Multiplication }} {{eqn | l = \frac 3 2 k \size h^2       | o = >       | r = \paren {k + \epsilon} \size h^2 > \frac 1 2 k \size h^2 }} {{end-eqn}} {{Explain|What does this mean? : $\big \vert + k$ and $\big \vert \cdot \size h^2$}} Therefore: :$\Delta J \sqbrk {\hat y; h} \ge \paren {k + \epsilon} \size h^2 > \dfrac 1 2 k \size h^2 $ For $k \in \R_{>0}$ and $\size h \ne 0$ {{RHS}} is always positive. Thus, there exists a neighbourhood around $y = \hat y$ where the increment is always positive:  :$\exists c \in \R_{>0}: \size h < c \implies \Delta J \sqbrk {\hat y; h} > 0$  and $J$ has a minimum for $y = \hat y$. {{qed}}
Let $\sigma \left({m}\right) = k^2$. Let $\sigma \left({n}\right) = l^2$. Thus: {{begin-eqn}} {{eqn | l = \sigma \left({m n}\right)       | r = \sigma \left({m}\right) \sigma \left({n}\right)       | c = Sigma Function is Multiplicative }} {{eqn | r = k^2 l^2       | c = from above }} {{eqn | r = \left({k l}\right)^2       | c = from above }} {{end-eqn}} {{qed}} Category:Numbers whose Sigma is Square Category:Coprime Integers cn1rc6qys6nf6095jaspy1iqih8vqou
Let $P' \left({n}\right)$ denote the $n$th second pentagonal number. We have: {{begin-eqn}} {{eqn | l = P' \left({n}\right)       | r = \dfrac {-n \left({-3 n - 1}\right)} 2       | c = Definition of Second Pentagonal Number: Definition 1 }} {{eqn | r = \dfrac {n \left({3 n + 1}\right)} 2       | c = simplifying }} {{end-eqn}} But: :$P' \left({n}\right) = \dfrac {n \left({3 n + 1}\right)} 2$ is the definition of a second pentagonal Number by definition 2. Hence the result. {{qed}} Category:Second Pentagonal Numbers i6rcybg0oq1gfgo3o816oq275t0gxo2
Recall the definition of the generalised pentagonal numbers $GP_n$ for $n = 0, 1, 2, \ldots$ :$GP_n = \begin{cases} \dfrac {m \paren {3 m + 1} } 2 & : n = 2 m \\ \dfrac {m \paren {3 m - 1} } 2 & : n = 2 m - 1 \end{cases}$ for $n = 0, 1, 2, \ldots$ Hence: {{begin-eqn}} {{eqn | l = \Delta_{2 n - 1}       | r = GP_{2 n} - GP_{2 n - 1}       | c =  }} {{eqn | r = \dfrac {n \paren {3 n + 1} } 2 - \dfrac {n \paren {3 n - 1} } 2       | c =  }} {{eqn | r = \dfrac {3 n^2 + n - 3 n^2 + n} 2       | c =  }} {{eqn | r = \dfrac {2 n} 2       | c =  }} {{eqn | r = n       | c =  }} {{end-eqn}} which defines the sequence of natural numbers. Then: {{begin-eqn}} {{eqn | l = \Delta_{2 n}       | r = GP_{2 n + 1} - GP_{2 n}       | c =  }} {{eqn | r = GP_{2 \paren {n + 1} - 1} - GP_{2 n}       | c =  }} {{eqn | r = \dfrac {\paren {n + 1} \paren {3 \paren {n + 1} - 1} } 2 - \dfrac {n \paren {3 n + 1} } 2       | c =  }} {{eqn | r = \dfrac {\paren {3 n^2 + 6 n + 3} - \paren {n + 1} - 3 n^2 - n} 2       | c =  }} {{eqn | r = \dfrac {4 n + 2} 2       | c =  }} {{eqn | r = 2 n + 1       | c =  }} {{end-eqn}} which defines the sequence of odd numbers. {{qed}}
It is sufficient to verify that the sequence $GP_n$ defined in Definition 2 contains both sequences of pentagonal numbers and second pentagonal numbers, and is in ascending order. From Closed Form for Pentagonal Numbers: :$P_n = \dfrac {n \paren {3 n - 1} } 2$ which are the odd terms (and the zeroth term) of $GP_n$. From definition of second pentagonal numbers: :$P'_n = \dfrac {n \paren {3 n + 1} } 2$ which are the even terms of $GP_n$. Finally it is shown that $GP_n$ is increasing. We have: :$GP_{2 m - 1} = \dfrac {m \paren {3 m - 1} } 2 < \dfrac {m \paren {3 m + 1} } 2 = GP_{2 m}$ and: :$GP_{2 m} = \dfrac {m \paren {3 m + 1} } 2 < \dfrac {\paren {m + 1} \paren {3 m + 2} } 2 = GP_{2 m + 1}$ Hence the result. {{qed}} Category:Generalized Pentagonal Numbers a3hl8mward6gbf7tn8cabvxrwaovxoj
=== Non-Empty Set === By Bottom in Ideal: :$\forall I \in \II: \bot \in I$ where $\bot$ denotes the smallest element in $S$. By definition of intersection: :$\bot \in \bigcap \II$ Hence $\bigcap \II$ is non-empty. {{qed|lemma}} === Lower Set === Let $x \in \bigcap \II$, $y \in S$ such that :$y \preceq x$ We will prove that: :$\forall I \in \II: y \in I$ Let $I \in \II$. By definition of intersection: :$x \in I$ Thus by definition of lower set: :$y \in I$ {{qed|lemma}} Thus by definition of intersection: :$y \in \bigcap \II$ {{qed|lemma}} === Directed Subset === Let $x, y \in \bigcap \II$. We will prove that: :$\forall I \in \II: x \vee y \in I$ Let $I \in \II$. By definition of intersection: :$x, y \in I$ By definition of directed: :$\exists z \in I: x \preceq z \land y \preceq z$ By definition of join: :$x \vee y = \sup \set {x, y}$ By definition: :$z$ is an upper bound for $\set {x, y}$. By definition of supremum: :$x \vee y \preceq z$ Thus by definition of lower set: :$x \vee y \in I$ {{qed|lemma}} Thus by definition of intersection: :$x \vee y \in \bigcap \II$ Hence by Directed in Join Semilattice: :$\bigcap \II$ is directed. {{qed}}
=== Lemma 1 === Let $y = \map y x$ be a real function, such that: :$\map y a = A, \quad \map y b = B$ Let $J \sqbrk y$ be a functional, such that: :$\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$ where: :$F \in C^2 \closedint a b$  {{WRT}} all its variables. Then: :$\ds \delta^2 J \sqbrk {y; h} = \int_a^b \paren {\map P {x, \map y x } h'^2 + \map Q {x, \map y x} h^2} \rd x$ where: :$\map P {x, \map y x} = \dfrac 1 2 F_{y' y'}, \quad \map Q {x, \map y x} = \dfrac 1 2 \paren {F_{yy} - \dfrac \d {\d x} F_{yy'} }$ === Lemma 2 === Let $h$ be a real function such that: :$h \in C^1 \openint a b, \quad \map h a = 0, \quad \map h b = 0$ Let: :$\ds \delta^2 J \sqbrk {y; h} = \int_a^b \paren {\map P {x, \map y x} h'^2 + \map Q {x, \map y x} h^2} \rd x$ where $P \in C^0 \closedint a b$.  Then a necessary condition for: :$\delta^2 J \sqbrk {y; h} \ge 0$  is: :$\map P {x, \map y x} \ge 0 \quad \forall x \in \closedint a b$ {{qed}} {{Namedfor|Adrien-Marie Legendre|cat = Legendre}}
{{ProofWanted|Follows somehow from Euler's Pentagonal Numbers Theorem, but at this time of night I have not a clue how.}}
Performing the calculation using long division: <pre>    0.043478260869565217391304...   ----------------------------- 23)1.00000000000000000000000000      92     184     23      --     ---     --       80     160    170       69     138    161       --     ---    ---       110     220     90        92     207     69       ---     ---     --        180     130    210        161     115    207        ---     ---    ---         190     150     30         184     138     23         ---     ---     --           60     120     70           46     115     69           --     ---     --           140      50     100           138      46      92           ---      --     ---             200     40    ...             184     23 </pre> {{qed}}
The first few perfect powers are: :$1, 4, 8, 9, 16, 25, 27, 32, \dots$ First we show that $23$ cannot be expressed as the sum of distinct perfect powers. Only $1, 4, 8, 9, 16$ are perfect powers less than $23$. Suppose $23$ can be so expressed. Since $1 + 4 + 8 + 9 = 22 < 23$, $16$ must be used in the sum. However $23 - 16 = 7$ cannot be expressed as a sum of $1$ and $4$. Thus $23$ cannot be expressed as the sum of distinct perfect powers. {{qed|lemma}} Now we show that all numbers greater than $23$ can be so expressed. By Richert's Theorem, we just need to show: :For any $23 < n \le 23 + 32$, $n$ can be expressed as a sum of distinct elements in $\set {1, 4, 8, 9, 16, 25, 27}$ :$s_{i + 1} \le 2 s_i$ for every $i \ge 7$, where $s_i$ is the $i$th perfect power Verification of the first statement is included in the bottom of this proof. To verify the second statement: Let $i \ge 7$. Let $m$ be the integer satisfying: :$2^{m + 1} > s_i \ge 2^m$ We have that $m \ge 4$. Hence $2^{m + 1}$ is also a perfect power. There must be a perfect power greater than $s_i$ but not greater than $2^{m + 1}$. Thus: :$2 s_i \ge 2^{m + 1} \ge s_{i + 1}$ Therefore $23$ is the largest integer that cannot be expressed as the sum of distinct perfect powers.. {{qed|lemma}} Here is $23 < n \le 55$ expressed as a sum of distinct elements in $\set {1, 4, 8, 9, 16, 25, 27}$: {{begin-eqn}} {{eqn | l = 24       | r = 16 + 8 }} {{eqn | l = 25       | r = 25 }} {{eqn | l = 26       | r = 25 + 1 }} {{eqn | l = 27       | r = 27 }} {{eqn | l = 28       | r = 27 + 1 }} {{eqn | l = 29       | r = 25 + 4 }} {{eqn | l = 30       | r = 25 + 4 + 1 }} {{eqn | l = 31       | r = 27 + 4 }} {{eqn | l = 32       | r = 27 + 4 + 1 }} {{eqn | l = 33       | r = 25 + 8 }} {{eqn | l = 34       | r = 25 + 9 }} {{eqn | l = 35       | r = 25 + 9 + 1 }} {{eqn | l = 36       | r = 27 + 9 }} {{eqn | l = 37       | r = 27 + 9 + 1 }} {{eqn | l = 38       | r = 25 + 9 + 4 }} {{eqn | l = 39       | r = 25 + 9 + 4 + 1 }} {{eqn | l = 40       | r = 27 + 9 + 4 }} {{eqn | l = 41       | r = 27 + 9 + 4 + 1 }} {{eqn | l = 42       | r = 25 + 9 + 8 }} {{eqn | l = 43       | r = 25 + 9 + 8 + 1 }} {{eqn | l = 44       | r = 27 + 9 + 8 }} {{eqn | l = 45       | r = 27 + 9 + 8 + 1 }} {{eqn | l = 46       | r = 25 + 9 + 8 + 4 }} {{eqn | l = 47       | r = 25 + 9 + 8 + 4 + 1 }} {{eqn | l = 48       | r = 27 + 9 + 8 + 4 }} {{eqn | l = 49       | r = 27 + 9 + 8 + 4 + 1 }} {{eqn | l = 50       | r = 25 + 16 + 9 }} {{eqn | l = 51       | r = 25 + 16 + 9 + 1 }} {{eqn | l = 52       | r = 27 + 25 }} {{eqn | l = 53       | r = 27 + 25 + 1 }} {{eqn | l = 54       | r = 25 + 16 + 9 + 4 }} {{eqn | l = 55       | r = 25 + 16 + 9 + 4 + 1 }} {{end-eqn}} {{qed}}
Recall the Ulam numbers: {{:Definition:Ulam Number/Sequence}} We have: {{begin-eqn}} {{eqn | l = 2       | r = 1 + 1 }} {{eqn | l = 3       | r = 2 + 1 }} {{eqn | l = 4       | r = 3 + 1 }} {{eqn | r = 2 + 2 }} {{eqn | l = 5       | r = 4 + 1 }} {{eqn | r = 3 + 2 }} {{eqn | l = 6       | r = 4 + 2 }} {{eqn | r = 3 + 3 }} {{eqn | l = 7       | r = 6 + 1 }} {{eqn | r = 4 + 3 }} {{eqn | l = 8       | r = 6 + 2 }} {{eqn | r = 4 + 4 }} {{eqn | l = 9       | r = 8 + 1 }} {{eqn | r = 6 + 3 }} {{eqn | l = 10       | r = 8 + 2 }} {{eqn | r = 6 + 4 }} {{eqn | l = 11       | r = 8 + 3 }} {{eqn | l = 12       | r = 11 + 1 }} {{eqn | r = 8 + 4 }} {{eqn | r = 6 + 6 }} {{eqn | l = 13       | r = 11 + 2 }} {{eqn | l = 14       | r = 13 + 1 }} {{eqn | r = 11 + 3 }} {{eqn | r = 8 + 6 }} {{eqn | l = 15       | r = 13 + 2 }} {{eqn | r = 11 + 4 }} {{eqn | l = 16       | r = 13 + 3 }} {{eqn | r = 8 + 8 }} {{eqn | l = 17       | r = 16 + 1 }} {{eqn | r = 13 + 4 }} {{eqn | r = 11 + 6 }} {{eqn | l = 18       | r = 16 + 2 }} {{eqn | l = 19       | r = 18 + 1 }} {{eqn | r = 16 + 3 }} {{eqn | r = 13 + 6 }} {{eqn | r = 11 + 8 }} {{eqn | l = 20       | r = 18 + 2 }} {{eqn | r = 16 + 4 }} {{eqn | l = 21       | r = 18 + 3 }} {{eqn | r = 13 + 8 }} {{eqn | l = 22       | r = 18 + 4 }} {{eqn | r = 16 + 6 }} {{eqn | r = 11 + 11 }} {{end-eqn}} Now consider the the difference between $23$ and successive Ulam numbers: {{begin-eqn}} {{eqn | l = 23 - 18       | r = 5       | c = not a Ulam number }} {{eqn | l = 23 - 16       | r = 7       | c = not a Ulam number }} {{eqn | l = 23 - 13       | r = 10       | c = not a Ulam number }} {{eqn | l = 23 - 11       | r = 12       | c = not a Ulam number }} {{end-eqn}} and it is not necessary to go further back. Hence the result. {{qed}}