hcju/proofwikiqa huggingface dataset:
We have that: :$720 = 2^4 \times 3^2 \times 5$ It remains to inspect the divisibility of $2$, $3$ and $5$ in $720!$ Thus: === Multiplicity of $2$ in $720!$ === {{:De Polignac's Formula/Examples/2 in 720 Factorial}} === Multiplicity of $3$ in $720!$ === {{:De Polignac's Formula/Examples/3 in 720 Factorial}} === Multiplicity of $5$ in $720!$ === {{:De Polignac's Formula/Examples/5 in 720 Factorial}} We calculate the multiplicity of the powers of $2$ and $3$ in $720!$ thus: {{begin-eqn}} {{eqn | l = 716       | r = 4 \times 179       | c =  }} {{eqn | ll= \leadsto       | l = \paren {2^4}^{179}       | o = \divides       | r = 720!       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 356       | r = 2 \times 178       | c =  }} {{eqn | ll= \leadsto       | l = \paren {3^2}^{178}       | o = \divides       | r = 720!       | c =  }} {{end-eqn}} Thus it is seen that the smallest power of the prime powers that are divisors of $720$ that divide $720!$ is that of $3^2$, which is $178$. Hence: :$720^{178} \divides 720!$ but: :$720^{179} \nmid 720!$ {{qed}}
=== Sufficient Condition === Assume that :$f$ is continuous. By Continuous iff Directed Suprema Preserving: :$f$ is preserves directed suprema. Thus by Directed Suprema Preserving Mapping at Element is Supremum: :$\forall x \in S: f\left({x}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$ {{qed|lemma}} === Necessary Condition === Assume that :$\forall x \in S: f\left({x}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$ By Mapping at Element is Supremum implies Way Below iff There Exists Element that Way Below and Way Below: :$\forall x \in S, y \in T: y \ll f\left({x}\right) \iff \exists w \in S: w \ll x \land y \ll f\left({w}\right)$ Thus by Continuous iff Way Below iff There Exists Element that Way Below and Way Below: :$f$ is continuous. {{qed}}
=== 1 implies 2 === Let $U, V \in \mathcal B$. Let $x \in U \cap V$. By hypothesis: :$\displaystyle \exists \mathcal A \subseteq \mathcal B: U \cap V = \bigcup \mathcal A$ By definition of union, $\exists W \in\mathcal A : x \in W$. By Set is Subset of Union: General Result, $W \subset U \cap V$. Therefore: :$\displaystyle \forall x \in A \cap B: \exists W \in \mathcal A \subseteq \mathcal B: x \in W \subseteq U \cap V$ {{qed|lemma}} === 2 implies 1 === Let $U, V \in \mathcal B$. Define the set: :$\displaystyle \mathcal A = \left\{{W \in \mathcal B: W \subseteq U \cap V}\right\} \subseteq \mathcal B$ By Union is Smallest Superset: General Result: :$\displaystyle \bigcup \mathcal A \subseteq U \cap V$ By hypothesis: :$\displaystyle \forall x \in U \cap V: \exists W \in \mathcal A: x \in W$ Thus $\displaystyle U \cap V \subseteq \bigcup \mathcal A$ By definition of set equality: :$\displaystyle U \cap V = \bigcup \mathcal A$ {{qed}} Category:Topological Bases m15b6r1d0o3c8wzdll9jdc3352t5fbm
{{ProofWanted|compare Equivalence of Definitions of Derivative}} Category:Differentiable Real Functions 34ueqe29g1aqkgizzbncuvs99ypu7tq
By Algebraic iff Continuous and For Every Way Below Exists Compact Between: :$L$ and $R$ are continuous. === Sufficient Condition === Assume that@ :$f$ is continuous. By Continuous iff Mapping at Element is Supremum: :$\forall x \in S: f\left({x}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$ Let $x \in S$. By definitions of image of set and compact closure: :$\left\{ {f\left({w}\right): w \in S \land w \preceq_1 x \land w}\right.$ is compact$\left.{}\right\} = f\left[{x^{\mathrm{compact} } }\right]$ By Compact Closure is Directed: :$D := x^{\mathrm{compact} }$ is directed. By Continuous iff Directed Suprema Preserving: :$f$ preserves directed suprema. By definition of mapping preserves directed suprema: :$f$ preserves the supremum of $D$. By definition of complete lattice: :$D$ admits a supremum. By definition of algebraic: :$L$ satisfies axiom of K-approximation. Thus {{begin-eqn}} {{eqn | l = f\left({x}\right) | r = f\left({\sup D}\right) | c = axiom of K-approximation }} {{eqn | r = \sup \left({f\left[{D}\right]}\right) | c = definition of mapping preserves the supremum }} {{eqn | r = \sup \left\{ {f\left({w}\right): w \in S \land w \preceq_1 x \land w \mathrm{\ is\ compact} }\right\} }} {{end-eqn}} {{qed|lemma}} === Necessary Condition === Assume that :$\forall x \in S: f \left({x}\right) = \sup \left\{ {f \left({w}\right): w \in S \land w \preceq_1 x \land w}\right.$ is compact$\left.{}\right\}$ By Mapping at Element is Supremum of Compact Elements implies Mapping at Element is Supremum that Way Below: :$\forall x \in S: f \left({x}\right) = \sup \left\{ {f \left({w}\right): w \in S \land w \ll x}\right\}$ Thus by Continuous iff Mapping at Element is Supremum: :$f$ is continuous. {{qed}}
{{begin-eqn}} {{eqn | l = 9^3       | r = 729       | c =  }} {{eqn | r = 1 + 27 + 64 + 125 + 512       | c =  }} {{eqn | r = 1^3 + 3^3 + 4^3 + 5^3 + 8^3       | c =  }} {{end-eqn}} {{expand|Add the proof based on $9^3 {{=}} 1^3 + 6^3 + 8^3$ and $6^3 {{=}} 3^3 + 4^3 + 5^3$ from Cubes which are Sum of Three Cubes}}
Performing the calculation using long division: <pre>     0.00137174211248285322359396433470507544581618655692729766803840877914951989026063100    --------------------------------------------------------------------------------------- 729)1.000000000000000000000000000000000000000000000000000000000000000000000000000000000000       729    729   3645   2187   5103     5832   3645   5103     5832   3645     4374       ---    ---   ----   ----   ----     ----   ----   ----     ----   ----     ----       2710    910   2350   7030    3700    1180   5050   4870     5680   1450     2260       2187    729   2187   6561    3645     729   4374   4374     5103    729     2187       ----    ---   ----   ----    ----    ----   ----   ----     ----   ----     ----        5230   1810   1630   4690     5500   4510   6760   4960     5770   7210      730        5103   1458   1458   4374     5103   4374   6561   4374     5103   6561      729        ----   ----   ----   ----     ----   ----   ----   ----     ----   ----      ---         1270   3520   1720   3160     3970   1360   1990   5860     6670   6490       1000          729   2916   1458   2916     3645    729   1458   5832     6561   5832        729         ----   ----   ----   ----     ----   ----   ----   ----     ----   ----       ----          5410   6040   2620   2440     3250   6310   5320    2800    1090   6580       ...          5103   5832   2187   2187     2916   5832   5103    2187     729   6561          ----   ----   ----   ----     ----   ----   ----    ----    ----   ----           3070   2080   4330   2530     3340   4780   2170    6130    3610    1900           2916   1458   3645   2187     2916   4374   1458    5832    2916    1458           ----   ----   ----   ----     ----   ----   ----    ----    ----    ----            1540   6220   6850   3430     4240   4060   7120    2980    6940    4420            1458   5832   6561   2916     3645   3645   6561    2916    6561    4374            ----   ----   ----   ----     ----   ----   ----    ----    ----    ----              820   3880   2890   5140     5950   4150   5590     6400   3790     4600              729   3645   2187   5103     5832   3645   5103     5832   3645     4374 </pre> {{qed}}
For every $t\in N$, the mapping: :$f_t : M \to M : x \mapsto f(x,t)$ is a contraction. By the Banach Fixed-Point Theorem, there exists a unique $g(t) \in M$ such that $f_t(g(t)) = g(t)$. We show that $g$ is Lipschitz continuous. Let $K<1$ be a uniform Lipschitz constant for $f$. Let $L$ be a Lipschitz constant for $f$. Let $s,t\in N$. Then {{begin-eqn}} {{eqn | l = d(g(s), g(t))       | r = d(f(g(s), s), f(g(t), t))       | c = Definition of $g$ }} {{eqn | o = \leq       | r = d(f(g(s), s), f(g(t), s)) + d(f(g(t), s), f(g(t), t))        | c = {{defof|Metric}} }} {{eqn | o = \leq       | r = K \cdot d(g(s), g(t)) + d(f(g(t), s), f(g(t), t))       | c = $f$ is a uniform contraction }} {{end-eqn}} and thus: {{begin-eqn}} {{eqn | l = d(g(s), g(t))       | o = \leq       | r = \dfrac1{1-K}d(f(g(t), s), f(g(t), t)) }} {{eqn | o = \leq       | r = \dfrac L{1-K} d(s,t)       | c = $f$ is lipschitz continuous }} {{end-eqn}} Thus $g$ is lipschitz continuous. {{qed}} Category:Implicit Functions 66rdm27e18ivrvk98jlfgzalfq9xt22
For every $t\in N$, the mapping: :$f_t : M \to M : x \mapsto f(x,t)$ is a contraction. By the Banach Fixed-Point Theorem, there exists a unique $g(t) \in M$ such that $f_t(g(t)) = g(t)$. Let $f$ be Lipschitz continuous at $(g(t),t)$. We show that $g$ is Lipschitz continuous at $t$. Let $K<1$ be a uniform Lipschitz constant for $f$. Let $L$ be a Lipschitz constant for $f$ at $a$. Let $s\in N$. Then {{begin-eqn}} {{eqn | l = d(g(s), g(t))       | r = d(f(g(s), s), f(g(t), t))       | c = Definition of $g$ }} {{eqn | o = \leq       | r = d(f(g(s), s), f(g(t), s)) + d(f(g(t), s), f(g(t), t))        | c = {{defof|Metric}} }} {{eqn | o = \leq       | r = K \cdot d(g(s), g(t)) + d(f(g(t), s), f(g(t), t))       | c = $f$ is a uniform contraction }} {{end-eqn}} and thus: {{begin-eqn}} {{eqn | l = d(g(s), g(t))       | o = \leq       | r = \dfrac1{1-K}d(f(g(t), s), f(g(t), t)) }} {{eqn | o = \leq       | r = \dfrac L{1-K} d(s,t)       | c = $f$ is lipschitz continuous at $(g(t),t)$ }} {{end-eqn}} Thus $g$ is lipschitz continuous at $t$. {{qed}}
{{begin-eqn}} {{eqn | l = 836^2       | r = 698 \, 896       | c = $6$ digits }} {{eqn | l = 798 \, 644^2       | r = 637 \, 832 \, 238 \, 736       | c = $12$ digits }} {{eqn | l = 64 \, 030 \, 648^2       | r = 4 \, 099 \, 923 \, 883 \, 299 \, 904       | c = $16$ digits }} {{eqn | l = 83 \, 163 \, 115 \, 486^2       | r = 6 \, 916 \, 103 \, 777 \, 337 \, 773 \, 016 \, 196       | c = $22$ digits }} {{eqn | l = 6 \, 360 \, 832 \, 925 \, 898^2       | r = 40 \, 460 \, 195 \, 511 \, 188 \, 111 \, 559 \, 106 \, 404       | c = $26$ digits }} {{end-eqn}}
A primitive semiperfect number is a semiperfect number which has no aliquot parts which are themselves semiperfect. Thus by definition a primitive semiperfect number is either perfect or abundant. A primitive abundant number is an abundant number whose aliquot parts are all deficient. Thus the perfect numbers: :$6, 28, 496, \ldots$ are not primitive abundant. However, by Divisor of Perfect Number is Deficient, the perfect numbers are all by definition primitive semiperfect. Hence the presence of the perfect numbers in this sequence. Next, consider the sequence of weird numbers: :$70, 836, \ldots$ These are numbers which are abundant but not semiperfect. Thus an abundant number whose aliquot parts are all deficient except for one or more weird numbers is a primitive semiperfect number which is not primitive abundant. However, such a number is not primitive abundant because it has an aliquot part which is abundant, though weird. So: :while $70$ is primitive abundant it is not primitive semiperfect number :while $350$, $490$, $710$, $910$, $1190$, and so on, are primitive semiperfect, they are not primitive abundant. Hence the result. {{qed}} Category:Primitive Semiperfect Numbers Category:Primitive Abundant Numbers 8aalkwm0o7fcvh38xcqvh31x3muet2s
All three-digit primes end in $1, 3, 7, 9$. Suppose $1$ is used as the units digit of a prime. Since the digit $1$ cannot be used again, the sum of the primes is at least: :$221 + 333 + 447 = 1001$ so $1$ cannot be used as a units digit . The units digits of the primes are $3, 7, 9$. To minimise the sum, the hundreds digits must be $1, 2, 4$. This leaves $5, 6, 8$ be the tens digits. The primes satisfying these conditions are: :$157, 163, 167$ :$257, 263, 269, 283$ :$457, 463, 467, 487$ Only $269$ contain $9$, so we must choose it. Only $157$ does not contain $6$, so must choose it next. But then all primes beginning with $4$ have some digit coinciding with the above primes. Hence the next minimal sum of the primes (if they exist) is: :$10^2 \paren {1 + 2 + 5} + 10 \paren {4 + 6 + 8} + \paren {3 + 7 + 9} = 999$ and we have shown that these primes do exist. {{qed}}
Let $N$ be expressed as: :$N = \displaystyle \sum_{k \mathop = 0}^n a_k 1000^k = a_0 + a_1 1000 + a_2 1000^2 + \cdots + a_n 1000^n$ where $n$ is the number of groups of $3$ digits. We have that: :$1000 \equiv -1 \pmod {1001}$ Hence from Congruence of Powers: :$1000^r \equiv \paren {-1}^r \pmod {1001}$ Thus: :$N \equiv a_0 + \paren {-1} a_1 + \paren {-1}^2 a_2 + \cdots + \paren {-1}^n a_n \pmod {1001}$ from the definition of Modulo Addition. Then we note that: :$1001 = 7 \times 11 \times 13$ and the result follows. {{qed}}
Let k represent the number of $9$s in the middle of the number. For $k > 0$ We can rewrite the number as follows: {{begin-eqn}} {{eqn | l = \sqbrk {10 (9) 89}_{10}       | r = 10 \times 10^{k + 2 } + 900 \sum_{i \mathop = 0}^{k - 1} 10^i + 89       | c = {{Defof|Geometric Series}} }} {{end-eqn}} Taking numbers of this form and multiplying by $9$ produces: {{begin-eqn}} {{eqn | l = 9 \times \paren {10 \times 10^{k + 2 } + 900 \sum_{i \mathop = 0}^{k - 1} 10^i + 89 }       | r = 90 \times 10^{k + 2 } + 8100 \sum_{i \mathop = 0}^{k - 1} 10^i + 801 }} {{end-eqn}} The first part is composed of $k + 4$ digits. The first digit will be $9$ followed by $k + 3$ digits of $0$ {{begin-eqn}} {{eqn | l = 90 \times 10^{k + 2 }        | r = 9 \times 10^{k + 3 }  }} {{end-eqn}} The sum in the middle is composed of $k + 3$ digits. The first digit will be $8$ followed by $k - 1$ digits of $9$ and then the remaining three digits at the end are $100$ {{begin-eqn}} {{eqn | l = 8100 \sum_{i \mathop = 0}^{k - 1} 10^i       | r = 899 \cdots 99100 }} {{end-eqn}} Summing the three pieces, the final answer will have $k + 4$ digits. The first digit is $9$  followed by $8$ which is the first digit of the middle part followed by $k$ digits of $9$ where the last $9$ is the sum of the $1$ from the middle part and the $8$ of the last part and then ending in $01$: {{begin-eqn}} {{eqn | l = \sqbrk {10 (9) 89}_{10} \times 9       | r = \sqbrk {98 (9) 01}_{10}       | c = }} {{end-eqn}} {{qed}}
Performing the calculation using long division: <pre>      0.00091827364554637281910009...     -------------------------------------------- 1089)1.00000000000000000000000000...        9801    5445    1089        ----    ----    ----         1990    5950    9910         1089    5445    9801         ----    ----    ----          9010    5050    1090          8712    4356    1089          ----    ----    ----           2980    6940      10000           2178    6534       9801           ----    ----       ----            8020    4060      ....            7623    3267            ----    ----             3970    7930             3267    7623             ----    ----              7030    3070              6534    2178              ----    ----               4960    8920               4356    8712               ----    ----                6040    2080                5445    1089         </pre>{{qed}}
{{begin-eqn}} {{eqn | l = 33^2       | r = 1089       | c =  }} {{eqn | l = 65^2 - 56^2       | r = 4225 - 3136       | c =  }} {{eqn | r = 1089       | c =  }} {{end-eqn}} Let $\sqbrk {xy}$ be a $2$-digit number such that $x \le y$ and $\sqbrk {xy}^2 - \sqbrk {yx}^2$ is a square of a $2$-digit number. The case $x = y$ gives the solution $\sqbrk {xy}^2 - \sqbrk {yx}^2 = 0$, which is not a square of a $2$-digit number. For $x \ne y$: {{begin-eqn}} {{eqn | l = \sqbrk {xy}^2 - \sqbrk {yx}^2       | r = \paren {10 x + y}^2 - \paren {10 y + x}^2 }} {{eqn | r = 100 x^2 + 20 x y + y^2 - 100 y^2 - 20 y x - x^2 }} {{eqn | r = 99 \paren {x^2 - y^2} }} {{eqn | r = 3^2 \times 11 \paren {x - y} \paren {x + y} }} {{eqn | ll = \leadsto       | l = \paren {x - y} \paren {x + y}       | r = 11 n^2       | c = for some integer $n$ }} {{end-eqn}} By Euclid's Lemma for Prime Divisors, one of $x - y, x + y$ must be divisible by $11$. Hence from Absolute Value of Integer is not less than Divisors, either $x - y$ or $x + y$ must be greater than or equal to $11$. Since $x - y < x < 9$, $x - y$ cannot be a multiple of $11$.  From $1 = 1 + 0 \le x + y < 9 + 9 = 18$, we have that $x + y = 11$. This implies that $x - y$ is a square number. $x + y = 11$ gives $\tuple {x,y} = \tuple {6,5}, \tuple {7,4}, \tuple {8,3}, \tuple {9,2}$ as the possible solutions. Among these solutions, only $\tuple {6,5}$ has a difference of a square number: $1$. Therefore the only square of a $2$-digit number expressible as a difference between a square and the square of its reversal is: :$65^2 - 56^2 = 33^2$. {{qed}}
{{begin-eqn}} {{eqn | l = n       | r = a^2 - b^2       | c =  }} {{eqn | r = \left({a + b}\right) \left({a - b}\right)       | c = Difference of Two Squares }} {{end-eqn}} Thus $n = p q$ where: {{begin-eqn}} {{eqn | n = 1       | l = p       | r = \left({a + b}\right)       | c =  }} {{eqn | n = 2       | l = q       | r = \left({a - b}\right)       | c =  }} {{eqn | ll= \leadsto       | l = p + q       | r = 2 a       | c = $(1)  + (2)$ }} {{eqn | l = p - q       | r = 2 b       | c = $(1)  - (2)$ }} {{eqn | ll= \leadsto       | l = a       | r = \dfrac {p + q} 2       | c =  }} {{eqn | l = b       | r = \dfrac {p - q} 2       | c =  }} {{end-eqn}} Thus for $a$ and $b$ to be integers, both $a$ and $b$ must be: :distinct, otherwise $p = q$ and so $b = 0$ :either both even or both odd, otherwise both $\dfrac {p + q} 2$ and $\dfrac {p - q} 2$ will be odd. Hence the result. {{qed}} Category:Square Numbers mxz53irrff1a78q6brxyd4g53y7yiax
From Difference between Two Squares equal to Repunit, $R_{2 n} = x^2 - y^2$ exactly when $R_{2 n} = a b$ where $x = \dfrac {a + b} 2$ and $y = \dfrac {a - b} 2$. By the Basis Representation Theorem   {{begin-eqn}} {{eqn | l = R_{2n}       | r = \sum_{k \mathop = 0}^{2 n - 1} 10^k       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^{n - 1} 10^k + \sum_{k \mathop = n}^{2 n - 1} 10^k       | c = splitting the summation into two }} {{eqn | r = \sum_{k \mathop = 0}^{n - 1} 10^k + 10^n \sum_{k \mathop = 0}^{n - 1} 10^k       | c = factoring $10^n$ out of the second part }} {{eqn | r = \paren {10^n + 1} \sum_{k \mathop = 0}^{n - 1} 10^k       | c =  }} {{eqn | r = \underbrace {100 \ldots 01}_{n - 1 \ 0 \text{'s} } \times \underbrace {111 \ldots 1}_{n \ 1 \text{'s} }       | c =  }} {{end-eqn}}   Thus, let: :$a = \paren {10^n + 1}$ :$b = \displaystyle \sum_{k \mathop = 0}^{n - 1}$ So: {{begin-eqn}} {{eqn | l = a + b       | r = \sum_{k \mathop = 0}^n 10^k + 1       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^n 10^k + 2       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {a + b} 2       | r = \sum_{k \mathop = 1}^n \frac {10^k} 2 + 1       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^n 5 \times 10^{k - 1} + 1       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^{n - 1} 5 \times 10^k + 1       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^{n - 1} 5 \times 10^k + 6       | c =  }} {{eqn | r = \underbrace {55 \ldots 50}_{n - 1 \ 5 \text{'s} } + 6       | c =  }} {{eqn | r = \underbrace {55 \ldots 56}_{n - 1 \ 5 \text{'s} }       | c =  }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = a - b       | r = \paren {10^n + 1} - \sum_{k \mathop = 0}^{n - 1} 10^k       | c =  }} {{eqn | r = \paren {\sum_{k \mathop = 0}^{n - 1} 9 \times 10^k + 1} + 1 - \sum_{k \mathop = 0}^{n - 1} 10^k       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^{n - 1} 8 \times 10^k + 2       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {a - b} 2       | r = \sum_{k \mathop = 0}^{n - 1} 4 \times 10^k + 1       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^{n - 1} 4 \times 10^k + 5       | c =  }} {{eqn | r = \underbrace {44 \ldots 45}_{n - 1 \ 4 \text{'s} }       | c =  }} {{end-eqn}} Hence the result. {{qed}}
From Difference between Two Squares equal to Repunit, $R_{2n} = x^2 - y^2$ exactly when $R_{2n} = a b$ where $x = \dfrac {a + b} 2$ and $y = \dfrac {a - b} 2$. By the Basis Representation Theorem   {{begin-eqn}} {{eqn | l = R_{2n}       | r = \sum_{0 \mathop \le k \mathop < 2 n} 10^k       | c =  }} {{eqn | r = \sum_{\substack {0 \mathop \le k \mathop < 2 n \\ k \text { even} } } 10^k + \sum_{\substack {0 \mathop \le k \mathop < 2 n \\ k \text { odd} } } 10^k       | c =  }} {{eqn | r = \sum_{\substack {0 \mathop \le k \mathop < 2 n \\ k \text { even} } } 10^k + 10 \times \sum_{\substack {0 \mathop \le k \mathop < 2 n \\ k \text { even} } } 10^k       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^{n - 1} 10^{2 k} + 10 \times \sum_{k \mathop = 0}^n 10^{2 k}       | c = change of indices }} {{eqn | r = \sum_{k \mathop = 0}^{n - 1} 11 \times 10^{2 k}        | c =  }} {{eqn | r = 11 \sum_{k \mathop = 0}^{n - 1} 10^{2 k}       | c =  }} {{eqn | r = 11 \times \underbrace {10101 \ldots 01}_{n \ 1 \text{'s} }       | c =  }} {{end-eqn}} Thus, let: :$a = \displaystyle \sum_{k \mathop = 0}^{n - 1} \times 10^{2 k}$ :$b = 11$ So: {{begin-eqn}} {{eqn | l = a + b       | r = \sum_{k \mathop = 0}^{n - 1} 10^{2 k} + 11       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^{n - 1} 10^{2 k} + 12       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {a + b} 2       | r = \sum_{k \mathop = 1}^{n - 1} \frac {10^{2 k} } 2 + 6       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^{n - 1} 5 \times 10^{2 k - 1} + 6       | c =  }} {{eqn | r = \underbrace {5050 \ldots 50}_{n - 1 \ 5 \text{'s} } + 6       | c =  }} {{eqn | r = \underbrace{5050 \ldots 56}_{n - 1 \ 5 \text{'s} }       | c =  }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = a - b       | r = \sum_{k \mathop = 0}^{n - 1} \times 10^{2 k} - 11       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^{n - 1} 10^{2 k} - 10       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {a - b} 2       | r = \sum_{k \mathop = 1}^{n - 1} \frac {10^{2 k} } 2 - 5       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^{n - 1} 5 \times 10^{2 k - 1} - 5       | c =  }} {{eqn | r = \underbrace {5050 \ldots 50}_{n - 1 \ 5 \text{'s} } - 5       | c =  }} {{eqn | r = \underbrace{5050 \ldots 45}_{n - 1 \ 5 \text{'s} }       | c =  }} {{end-eqn}} Hence the result. {{qed}} Category:Difference between Two Squares equal to Repunit 6a93mucf5ecaddxmcog2ma01dprm4w5
We have that: {{begin-eqn}} {{eqn | l = 1225       | r = 35^2       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^5 \paren {2 k - 1}^3 = 1^3 + 3^3 + 5^3 + 7^3 + 9^3       | c =  }} {{eqn | l = 1 \, 413 \, 721       | r = 1189^2       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^{29} \paren {2 k - 1}^3 = 1^3 + 3^3 + 5^3 + \dotsb + 55^3 + 57^3       | c =  }} {{end-eqn}} From Sum of Sequence of Odd Cubes we have: :$\displaystyle \sum_{j \mathop = 1}^n \paren {2 j - 1}^3 = 1^3 + 3^3 + 5^3 + \dotsb + \paren {2 n − 1}^3 = n^2 \paren {2 n^2 − 1}$ Thus we need to find all $n$ such that $2 n^2 − 1$ is square. {{finish|see if there's a proof that can be offered up}}
By definition of retract: :there exists a continuous retraction $r: S \to Z$ of $T$. Let $\mathcal Y = \left({Y, \sigma}\right)$ be a topological space. Let $f: Y \to Z$ be a continuous mapping. Let $\mathcal X = \left({X, \sigma'}\right)$ such that :$\mathcal Y$ is topological subspace of $\mathcal X$. By Inclusion Mapping is Continuous: :$i_Z$ is continuous $\left({R \to T}\right)$ where $i_Z$ denotes the inclusion mapping from $Z$ in $S$. By Composite of Continuous Mappings is Continuous: :$i_Z \circ f: Y \to S$ is continuous. By definition of injective space: :there exists a continuous mapping $g:X \to S: g \restriction Y = i_Z \circ f$ Define $h := r \circ g$ By Composite of Continuous Mappings is Continuous: :$h$ is continuous. We will prove that :$h \restriction Y = f$ By definition of topological subspace: :$Y \subseteq X$ and $Z \subseteq S$ Thus by definitions of composition of mappings and restriction of mapping: :$h \restriction Y: Y \to Z$ and $f: Y \to Z$ Let $y \in Y$. By Restriction of Composition is Composition of Restriction: :$h \restriction Y = r \circ \left({g \restriction Y}\right)$ By definition of mapping: :$f\left({y}\right) \in Z$ Thus {{begin-eqn}} {{eqn | l = \left({h \restriction Y}\right)\left({y}\right) | r = r\left({\left({i_Z \circ f}\right)\left({y}\right)}\right) | c = definition of composition of mappings }} {{eqn | r = r\left({i_Z\left({f\left({y}\right)}\right)}\right) | c = definition of composition of mappings }} {{eqn | r = r\left({f\left({y}\right)}\right) | c = definition of inclusion mapping }} {{eqn | r = f\left({y}\right) | c = definition of retraction }} {{end-eqn}} {{qed}}