hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^\infty \frac 1 {n^2}       | r = \map \zeta 2       | c = {{Defof|Riemann Zeta Function}} }} {{eqn | r = \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n}^2} + \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2} }} {{eqn | r = \frac 1 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} + \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2} }} {{eqn | r = \frac 1 4 \map \zeta 2 + \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2} }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}       | r = \frac 3 4 \map \zeta 2 }}  {{eqn | r = \frac 3 4 \cdot \frac{\pi^2} 6       | c = Basel Problem }} {{eqn | r = \frac {\pi^2} 8 }} {{end-eqn}} {{qed}}
Let $n$ be a positive integer. {{begin-eqn}} {{eqn | l = \int_0^1 \frac {x^{2 n + 1} } {\sqrt{1 - x^2} } \mathrm d x        | r = \int_0^{\pi / 2} \sin^{2 n + 1} x \ \mathrm d x       | c = substituting $x \to \sin x$ }} {{eqn | n = 1       | r = \frac {2^{2 n} \left({n!}\right)^2} {\left({2 n + 1}\right)!}        | c = Reduction Formula for Definite Integral of Power of Sine }} {{end-eqn}} We have: {{begin-eqn}} {{eqn | l = \int_0^1 \frac {\arcsin x} {\sqrt {1 - x^2} } \ \mathrm d x       | r = \int_0^1 \sum_{n \mathop = 0}^\infty \frac {\left({2 n}\right)!} {2^{2 n} \left({n!}\right)^2 \left({2 n + 1}\right)} \frac {x^{2 n + 1} } {\sqrt {1 - x^2} } \mathrm d x       | c = Power Series Expansion for Real Arcsine Function }} {{eqn | r =  \sum_{n \mathop = 0}^\infty \frac {\left({2 n}\right)!} {2^{2 n} \left({n!}\right)^2 \left({2 n + 1}\right)} \int_0^1 \frac {x^{2 n + 1} } {\sqrt {1 - x^2} } \mathrm d x       | c = Interchange of sum and integral is valid by Tonelli's Theorem  }} {{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\left({2 n}\right)!} {2^{2 n} \left({n!}\right)^2 \left({2 n + 1}\right)} \frac {2^{2 n} \left({n!}\right)^2} {\left({2 n + 1}\right)!}        | c = by $(1)$ }} {{eqn | r = \sum_{n \mathop = 0}^\infty \frac 1 {\left({2 n + 1}\right)^2}       | c =  }} {{end-eqn}} We also have: {{begin-eqn}} {{eqn | l = \int_0^1 \frac {\arcsin x} {\sqrt {1 - x^2} } \ \mathrm d x       | r = \left[{\frac 1 2 \left({\arcsin x}\right)^2}\right]_0^1       | c = Fundamental Theorem of Calculus }} {{eqn | r = \frac {\pi^2} 8       | c =  }} {{end-eqn}} So we can deduce: :$\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {\left({2 n + 1}\right)^2} = \frac {\pi^2} 8$  {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n - 1}\right)^2}       | c = This "proof" leads to a dead end. Its value lies in showing that this approach does not work. }} {{eqn | r = \int_0^1 \int_0^1 \frac 1 {1 - x^2 y^2} \, \mathrm d x \, \mathrm d y       | c = Sum of Reciprocals of Squares of Odd Integers as Double Integral }} {{eqn | r = \int_0^1 \int_0^1 \frac 1 {\left({1 + x y}\right) \left({1 - x y}\right)} \, \mathrm d x \, \mathrm d y }} {{eqn | r = \int_0^1 \int_0^1 \frac 1 {2 \left({1 + x y}\right)} \, \mathrm d x \, \mathrm d y - \int_0^1 \int_0^1 \frac 1 {2 \left({1 - x y}\right)} \, \mathrm d x \, \mathrm d y       | c = Primitive of Function of Constant Multiple, partial fraction expansion }} {{eqn | r = \frac 1 2 \int_0^1 \left[\frac{\ln \left( {1+xy} \right) } x\right]_0^1 \mathrm d x - \frac 1 2\int_0^1 \left[\frac{\ln \left({1 - x y}\right)} x\right]_0^1 \, \mathrm d x       | c = Primitive of Reciprocal  }} {{eqn | r = \frac 1 2 \int_0^1 \left[{\frac {\ln \left( {1 + x} \right)} x - \frac{\ln 1} x}\right] \mathrm d x - \frac 1 2 \int_0^1 \left[{\frac {\ln \left({1 - x}\right)} x - \frac {\ln 1} x}\right] \, \mathrm d x }} {{eqn | r = \frac 1 2 \int_0^1 \frac 1 x \ln \left({\frac {1 + x} {1 - x} }\right) \, \mathrm d x       | c = Sum of Logarithms }} {{eqn | r = \int_0^1 \frac{\tanh^{-1} x} x \, \mathrm d x       | c = {{Defof|Inverse Hyperbolic Tangent}} }} {{eqn | r = \int_0^1 \frac 1 x \paren {x + \frac {x^3} 3 + \frac {x^5} 5 + \frac {x^7} 7 + \cdots} \mathrm d x       | c = Power Series Expansion for Real Inverse Hyperbolic Tangent }} {{eqn | r = \int_0^1 \paren {1 + \frac {x^2} 3 + \frac {x^4} 5 + \frac {x^6} 7 + \cdots} \mathrm d x       | c =  }} {{eqn | r = \left[x + \frac {x^3} 9 + \frac {x^5} {25} + \frac {x^7} {49} + \cdots \right]_0^1       | c =  }} {{eqn | r = 1 + \frac 1 9 + \frac 1 {25} + \frac 1 {49} + \cdots       | c = And we have reached our dead-end.  }} {{end-eqn}} {{finish|Unsure how to proceed}}
{{begin-eqn}} {{eqn	| l = \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2n - 1}^2} 	| r = \int_0^1 \int_0^1 \frac 1 {1 - x^2y^2} \rd x \rd y 	| c = Sum of Reciprocals of Squares of Odd Integers as Double Integral }} {{end-eqn}} Applying the substitution: :$\tuple {x, y} = \tuple {\dfrac {\map \sin u} {\map \cos v}, \dfrac {\map \sin v} {\map \cos u} }$ the Jacobian determinant is: {{begin-eqn}} {{eqn | l = \size {\frac {\partial \tuple {x, y} } {\partial \tuple {u, v} } }       | r = \frac \partial {\partial u} \frac {\map \sin u} {\map \cos v} \frac \partial {\partial v} \frac {\map \sin v} {\map \cos u} - \frac \partial {\partial v} \frac {\map \sin u} {\map \cos v} \frac \partial {\partial u} \frac {\map \sin v} {\map \cos u} }} {{eqn | r = \frac {\map \cos u \, \map \cos v} {\map \cos v \, \map \cos u} - \frac {\map {\sin^2} u \, \map {\sin^2} v} {\map {\cos^2} v \, \map {\cos^2} u}       | c = Derivative of Secant Function, Derivative of Sine Function }} {{eqn | r = 1 - \map {\tan^2} u \map {\tan^2} v }} {{end-eqn}} Under this substitution, the image of the region $\closedint 0 1^2$, that is the unit square, is an isosceles triangle, $\bigtriangleup$ with base/height $\dfrac \pi 2$. (This may be observed by sketching the region $\set {\tuple {u, v}: u, v \ge 0 \land \map \cos u \ge \map \sin v \land \map \cos v \ge \map \sin u}$.) Hence,: {{begin-eqn}} {{eqn | l = \int_0^1 \int_0^1 \frac 1 {1 - x^2y^2} \rd x \rd y       | r = \iint_{\bigtriangleup} \frac {1 - \map {\tan^2} u \, \map {\tan^2} v} {1 - \paren {\map \tan u \, \map \tan v}^2} \rd u \rd v }} {{eqn | r = \iint_{\bigtriangleup} \rd u \rd v }} {{eqn | r = \frac 1 2 \int_0^{\frac \pi 2} \int_0^{\frac \pi 2} \rd u \rd v       | c = Area of Triangle in Terms of Side and Altitude }} {{eqn | r = \frac 1 2 \paren {\frac \pi 2}^2       | c = Primitive of Constant }} {{eqn | r = \frac {\pi^2} 8 }} {{end-eqn}} {{qed}}
By Fourier Series of Absolute Value of x, for $x \in \closedint {-\pi} \pi$: :$\displaystyle \size x = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {\paren {2 n - 1} x} } {\paren {2 n - 1}^2}$ Setting $x = \pi$: {{begin-eqn}} {{eqn | l = \size \pi       | r = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 \pi n - \pi} } {\paren {2 n - 1}^2} }} {{eqn | r = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {-\pi} } {\paren {2 n - 1}^2}       | c = Sine and Cosine are Periodic on Reals }} {{eqn | r = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\cos \pi} {\paren {2 n - 1}^2}       | c = Cosine Function is Even }} {{eqn | r = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {-1} {\paren {2 n - 1}^2}       | c = Cosine of Multiple of Pi }} {{eqn | ll= \leadsto       | l = \frac \pi 2       | r = \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2n - 1}^2}       | c = {{Defof|Absolute Value}} and rearranging }} {{eqn | ll= \leadsto       | l = \frac {\pi^2} 8       | r = \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2n - 1}^2}       | c = multiplying through by $\dfrac \pi 4$ }} {{end-eqn}} {{qed}}
Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as: :$\map f x = \begin{cases} -\pi & : 0 < x \le \pi \\ x - \pi & : \pi < x < 2 \pi \end{cases}$ By Fourier Series: $-\pi$ over $\openint 0 \pi$, $x - \pi$ over $\openint \pi {2 \pi}$, its Fourier series can be expressed as: :$\map f x \sim \map S x = \displaystyle -\dfrac \pi 4 + \frac 2 \pi \sum_{r \mathop = 0}^\infty \dfrac {\cos \paren {2 r + 1} x} {\paren {2 r + 1}^2} - \sum_{n \mathop = 1}^\infty \dfrac {2 - \paren {-1}^n \sin n x} n$ Consider the point $x = \pi$. By Fourier's Theorem: {{begin-eqn}} {{eqn | l = \map S \pi       | r = \frac 1 2 \paren {\lim_{x \mathop \to \pi^+} \map f x + \lim_{x \mathop \to \pi^-} \map f x}       | c =  }} {{eqn | r = \dfrac {-\pi + \paren {\pi - \pi} } 2       | c =  }} {{eqn | r = -\dfrac \pi 2       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = -\dfrac \pi 2       | r = -\dfrac \pi 4 + \frac 2 \pi \sum_{r \mathop = 0}^\infty \dfrac {\cos \paren {2 r + 1} \pi} {\paren {2 r + 1}^2} - \sum_{n \mathop = 1}^\infty \dfrac {2 - \paren {-1}^n \sin n \pi} n       | c =  }} {{eqn | ll= \leadsto       | l = -\dfrac \pi 4       | r = \frac 2 \pi \sum_{r \mathop = 0}^\infty \dfrac {\cos \paren {2 r + 1} \pi} {\paren {2 r + 1}^2}       | c = Sine of Multiple of Pi and simplifying }} {{eqn | r = \frac 2 \pi \sum_{r \mathop = 0}^\infty \dfrac {-1} {\paren {2 r + 1}^2}       | c = Cosine of Multiple of Pi and simplifying }} {{eqn | ll= \leadsto       | l = \dfrac {\pi^2} 8       | r = \sum_{r \mathop = 1}^\infty \frac 1 {\paren {2 r - 1}^2}       | c = multiplying both sides by $-\dfrac \pi 2$ and adjusting indices }} {{end-eqn}} {{qed}}
By Half-Range Fourier Cosine Series for Identity Function over $\openint 0 \pi$: :$\displaystyle x = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 n - 1} x} {\paren {2 n - 1}^2}$ for $x \in \openint 0 \pi$. We have that: :$\map f \pi = \map f {\pi - 2 \pi} = \map f {-\pi} = \pi$ and so: :$\map f {\pi^-} = \map f {\pi^+}$ Hence we can set $x = \pi$: {{begin-eqn}} {{eqn | l = \pi       | r = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 n - 1} \pi} {\paren {2 n - 1}^2} }} {{eqn | r = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {-1} {\paren {2 n - 1}^2}       | c = Cosine of Multiple of Pi }} {{eqn | ll= \leadsto       | l = \frac {\pi^2} 8       | r = \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}       | c = simplification }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}       | r = \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1} \paren {2 n - 1} } }} {{eqn | r = \sum_{n \mathop = 1}^\infty \int_0^1 x^{2 n - 2} \rd x \int_0^1 y^{2 n - 2} \rd y }} {{eqn | r = \sum_{n \mathop = 1}^\infty \int_0^1 \int_0^1 \paren {x^2 y^2}^{n - 1} \rd x \rd y }} {{eqn | r = \int_0^1 \int_0^1 \sum_{n \mathop = 0}^\infty \paren {x^2 y^2}^n \rd x \rd y       | c = Fubini's Theorem }} {{eqn | r = \int_0^1 \int_0^1 \frac 1 {1 - x^2 y^2} \rd x \rd y       | c = Sum of Infinite Geometric Sequence }} {{end-eqn}} {{qed}} {{MissingLinks|The jump from the $1$st to $2$nd lines needs to be justified. So does the $2$nd to the $3$rd, now I come to think about it.}}
{{begin-eqn}} {{eqn | l = \int_0^1 \int_0^1 \frac 1 {1 - x^2 y^2} \rd x \rd y       | r = \int_0^1 \int_0^1 \paren {1 + x^2 y^2 + x^4 y^4 + x^6 y^6 + \cdots} \rd x \rd y       | c = Sum of Infinite Geometric Sequence }} {{eqn | r = \int_0^1 \intlimits {y + \frac {x^2y^3} 3 + \frac {x^4y^5} 5 + \frac {x^6y^7} 7 + \cdots } {y \mathop = 0} {y \mathop = 1} \rd x }} {{eqn | r = \int_0^1 \paren {1 + \frac 1 3 x^2 + \frac 1 5 x^4 + \frac 1 7 x^6 + \cdots} \rd x }} {{eqn | r = \intlimits {x + \frac {x^3} 9 + \frac {x^5} {25} + \frac {x^7} {49} + \cdots } {x \mathop = 0} {x \mathop = 1} }} {{eqn | r = \paren {1 + \frac 1 9 + \frac 1 {25} + \frac 1 {49} + \cdots } }} {{eqn | r = \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}       | c =  }} {{end-eqn}} {{qed}}
From Sum of Reciprocals of Powers as Euler Product: :$\displaystyle \sum_{n \mathop \ge 1} \dfrac 1 {n^z} = \prod_p \frac 1 {1 - p^{-z} }$ for $z \in \C$ such that $\map \Re z > 1$. Putting $z = 2$: {{begin-eqn}} {{eqn | l = \sum_{n \mathop \ge 1} \dfrac 1 {n^2}       | r = \prod_p \frac 1 {1 - p^{-2} }       | c =  }} {{eqn | r = \prod_p \frac {p^2} {p^2 - 1}       | c = multiplying top and bottom by $p^2$ }} {{end-eqn}} The result follows from Riemann Zeta Function of $2$. {{qed}}
The result follows by Convergence of Taylor Series of Function Analytic on Disk for the case $R = \infty$. {{qed}}
We have that $F$ is analytic everywhere except at its singularities. Also, the distance from $x_0$ to the closest singularity of $F$ is $R$. Therefore: :$F$ is analytic at every point $z \in \C$ satisfying $\size {z - x_0} < R$ The result follows by Convergence of Taylor Series of Function Analytic on Disk. {{qed}}
Let $10^n$ have $m$ digits when expressed in binary notation. By the Basis Representation Theorem and its implications, a positive integer $x$ has $m$ digits {{iff}}: :$2^{m - 1} \le x < 2^m$ Thus: {{begin-eqn}} {{eqn | l = 2^{m - 1}       | o = \le       | m = 10^n       | mo= <       | r = 2^m       | c =  }} {{eqn | ll= \leadsto       | l = m - 1       | o = \le       | m = \map {\log_2} {10^n}       | mo= <       | r = m       | c =  }} {{eqn | ll= \leadsto       | l = m - 1       | o = \le       | m = n \log_2 10       | mo= <       | r = m       | c =  }} {{end-eqn}} Because a power of $10$ cannot equal a power of $2$, it will always be the case that: :$m - 1 < n \log_2 10 < m$ and so: :$m - 1 < n \log_2 10 \le m$ Hence from Integer equals Ceiling iff Number between Integer and One Less: :$m = \ceiling {n \log_2 10}$ {{qed}}
Let $m$ be the number of digits in $1000$. From Number of Binary Digits in Power of 10: :$m = \ceiling {3 \log_2 10}$ From Logarithm Base 2 of 10: :$\log_2 10 \approx 3 \cdotp 32192 \, 8 \ldots$ and so: :$m \approx 9 \cdotp 96$ Hence the result. The actual number is: :$1000_{10} = 1 \, 111 \, 101 \, 100_2$ {{qed}}
=== Necessary Condition === We have that $3! = 6$ and that $4$ does not divide $6$. So in order for $n$ to divide $n - 1$ it is necessary that $n \ne 4$. {{qed|lemma}} === Sufficient Condition === Let $n \ne 4$ be composite. Let $n = r s$ where: :$r, s \in \Z_{> 1}$ :$r \ne s$ :$r, s < n$ This is always possible unless $n = p^2$ for some prime number $p$. {{WLOG}}, let $r < s$. Then by definition of factorial: :$\paren {n - 1}! = 1 \times 2 \times \ldots \times r \times \ldots \times s \times \ldots \times \paren {n - 2} \times \paren {n - 1}$ and so: :$n = r s \divides \paren {n - 1}!$ Let $n = p^2$. As $n \ne 4$, we have that $p \ne 2$. Hence $p > 2$. Hence: :$2 p < p^2$ and so: :$2 p \divides \paren {n - 1}!$ By definition of factorial: :$\paren {n - 1}! = 1 \times 2 \times \ldots \times p \times \ldots \times 2 p \times \ldots \times \paren {n - 2} \times \paren {n - 1}$ and so: :$n = p^2 \divides \paren {n - 1}!$ Hence the result. {{qed}}
Let $N$ be divisible by $4$. Then: {{begin-eqn}} {{eqn | l = N       | o = \equiv       | r = 0 \pmod 4 }} {{eqn | ll= \leadstoandfrom       | l = \sum_{k \mathop = 0}^n a_k 10^k       | o = \equiv       | r = 0 \pmod 4 }} {{eqn | ll= \leadstoandfrom       | l = 10 a_1 + a_0 + 10^2 \sum_{k \mathop = 2}^n a_k 10^{k - 2}       | o = \equiv       | r = 0 \pmod 4 }} {{eqn | ll= \leadstoandfrom       | l = 10 a_1 + a_0       | o = \equiv       | r = 0 \pmod 4       | c = as $10^2 \equiv 0 \pmod 4$ }} {{end-eqn}} {{qed}}
=== Existence === $8712 = 4 \times 2178$ $9801 = 9 \times 1089$ === Uniqueness === Write $n = \sqbrk {a \dots b}$. Then $m = \sqbrk {b \dots a} = k n$. Comparing the leftmost digit, we must have $k a \le b < k \paren {a + 1}$. Hence $k a < 10$. Comparing the rightmost digit, we must have $a \equiv k b \pmod {10}$. We only consider $2 \le k < 10$ since the case $k = 1$ is trivially satisfied by palindromes. For $k \ge 5$, we must have $a = 1$. $1 \equiv k b \pmod {10}$ has solutions only for $k = 7$ or $9$. For $k = 7$, we must have $b = 3$. However $3 < 7 \times 1$. Hence $k = 9$. {{qed|lemma}} For $k = 3$, $a \le 3$. From $a \equiv 3 b \pmod {10}$ we have: :$\tuple {a, b} = \tuple {1, 7}, \tuple {2, 4}, \tuple {3, 1}$ However none of the above satisfy the condition: :$3 a \le b < 3 \paren {a + 1}$ {{qed|lemma}} For $k = 2$, $a \le 4$. From $a \equiv 2 b \pmod {10}$, both sides must be even. Hence $a = 2$ or $4$, giving $b = 1$ or $6$ or $2$ or $7$. However none of the above satisfy the condition: :$2 a \le b < 2 \paren {a + 1}$ {{qed|lemma}} Hence the only possible values of $k$ are $4$ and $9$. {{qed}}
From Solutions of Pythagorean Equation, all Pythagorean triangles, the set of all primitive Pythagorean triples is generated by: :$\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ where: :$m, n \in \Z_{>0}$ are (strictly) positive integers :$m \perp n$, that is, $m$ and $n$ are coprime :$m$ and $n$ are of opposite parity :$m > n$. The smallest two (strictly) positive integers which satisfy the above criteria are: :$n = 1$ :$m = 2$ Hence: :$2 m n = 2 \times 2 \times 1 = 4$ :$m^2 - n^2 = 2^2 - 1^2 = 3$ :$m^2 + n^2 = 2^2 + 1^2 = 5$ and to confirm: :$3^2 + 4^2 = 9 + 16 = 25 = 5^2$ {{qed}}
Let $a, b, c$ be the lengths of the sides of a Pythagorean triangle such that $a < b < c$. Let $a, b, c$ be in arithmetic sequence: :$b - a = c - b$ Let $a, b, c$ form a primitive Pythagorean triple: :$a \perp b$ By definition of primitive Pythagorean triple, $a, b, c$ are in the form: :$2 m n, m^2 - n^2, m^2 + n^2$ We have that $m^2 + n^2$ is always the hypotenuse. There are two cases: :$(1): \quad 2 m n > m^2 - n^2$, as in, for example, $3, 4, 5$, where $m = 2, n = 1$. :$(2): \quad 2 m n < m^2 - n^2$, as in, for example, $8-15-17$, where $m = 4, n = 1$. First, let $2 m n > m^2 - n^2$: :$a = m^2 - n^2$ :$b = 2 m n$ :$c = m^2 + n^2$ Then: {{begin-eqn}} {{eqn | l = 2 m n - \paren {m^2 - n^2}       | r = m^2 + n^2 - 2 m n       | c = {{Defof|Arithmetic Sequence}} }} {{eqn | ll= \leadsto       | l = 4 m n       | r = 2 m^2       | c = adding $2 m n + m^2 - n^2$ to both sides }} {{eqn | ll= \leadsto       | l = 2 n       | r = m       | c = dividing both sides by $2 m$ }} {{end-eqn}} From Solutions of Pythagorean Equation: Primitive, $m$ and $n$ must be coprime. Hence $n = 1$ and $m = 2$ are the only $m$ and $n$ which fulfil the requirements. This leads to the $3-4-5$ triangle. Now let $2 m n < m^2 - n^2$: :$a = 2 m n$ :$b = m^2 - n^2$ :$c = m^2 + n^2$ Then: {{begin-eqn}} {{eqn | l = m^2 - n^2 - 2 m n       | r = m^2 + n^2 - \paren {m^2 - n^2}       | c = {{Defof|Arithmetic Sequence}} }} {{eqn | ll= \leadsto       | l = m^2 - n^2 - 2 m n       | r = 2 n^2       | c = simplifying {{RHS}} }} {{eqn | ll= \leadsto       | l = m^2 - 2 m n - 3 n^2       | r = 0       | c = subtracting $2 n^2$ from both sides }} {{end-eqn}} In order for $a, b, c$ to form a primitive Pythagorean triple, then $m$ and $n$ must be of opposite parity. If $m$ is even, then $m^2 - 2 m n$ is even. But then $3 n^2$ is even, which makes $n$ even. Otherwise, if $m$ is odd, then $m^2 - 2 m n$ is odd. But then $3 n^2$ is odd, which makes $n$ odd. So when $2 m n < m^2 - n^2$, $a, b, c$ cannot be both in arithmetic sequence and be a primitive Pythagorean triple. Hence follows the result. {{qed}}
Let $a, b, c$ be the lengths of the sides of a Pythagorean triangle $T$. Thus $a, b, c$ form a Pythagorean triple. By definition of Pythagorean triple, $a, b, c$ are in the form: :$2 m n, m^2 - n^2, m^2 + n^2$ We have that $m^2 + n^2$ is always the hypotenuse. Thus the area of $T$ is given by: :$\AA = m n \paren {m^2 - n^2}$ The perimeter of $T$ is given by: :$\PP = m^2 - n^2 + 2 m n + m^2 + n^2 = 2 m^2 + 2 m n$ We need to find all $m$ and $n$ such that $\PP = 2 \AA$.  Thus: {{begin-eqn}} {{eqn | l = 2 m^2 + 2 m n       | r = 2 m n \paren {m^2 - n^2}       | c =  }} {{eqn | ll= \leadsto       | l = m + n       | r = n \paren {m + n} \paren {m - n}       | c =  }} {{eqn | ll= \leadsto       | l = n \paren {m - n}       | r = 1       | c =  }} {{end-eqn}} As $m$ and $n$ are both (strictly) positive integers, it follows immediately that: :$n = 1$ :$m - n = 1$ and so: :$m = 2, n = 1$ and the result follows. {{qed}}
Note that the following proof, as written, only holds for real $\theta$. Define: :$\map x \theta = e^{i \theta}$ :$\map y \theta = \cos \theta + i \sin \theta$ Consider first $\theta \ge 0$. Taking Laplace transforms: {{begin-eqn}} {{eqn | l = \map {\laptrans {\map x \theta} } s       | r = \map {\laptrans {e^{i \theta} } } s }} {{eqn| r = \frac 1 {s - i}      | c = Laplace Transform of Exponential }} {{eqn | l = \map {\laptrans {\map y \theta} } s       | r = \map {\laptrans {\cos \theta + i \sin \theta} } s }} {{eqn | r = \map {\laptrans {\cos \theta} } s + i \, \map {\laptrans {\sin \theta} } s       | c = Linear Combination of Laplace Transforms }} {{eqn | r = \frac s {s^2 + 1} + \frac i {s^2 + 1}       | c = Laplace Transform of Cosine, Laplace Transform of Sine }} {{eqn | r = \frac {s + i} {\paren {s + i} \paren {s - i} } }} {{eqn | r = \frac 1 {s - i} }} {{end-eqn}} So $x$ and $y$ have the same Laplace transform for $\theta \ge 0$. Now define $\tau = -\theta, \sigma = -s$, and consider $\theta < 0$ so that $\tau > 0$. Taking Laplace transforms of $\map x \tau$ and $\map y \tau$:  {{begin-eqn}} {{eqn | l = \map {\laptrans {\map x \tau} } \sigma       | r = \frac 1 {\sigma - i}       | c = from above }} {{eqn | l = \map {\laptrans {\map y \tau} } \sigma       | r = \frac 1 {\sigma - i}       | c = from above }} {{end-eqn}} So $\map x \theta$ and $\map y \theta$ have the same Laplace transforms for $\theta < 0$. The result follows from Injectivity of Laplace Transform. {{qed}}
The absence of $2$ and $3$ from the digit that can end a square follows directly from Square Modulo 5. As $7 \equiv 2 \pmod 5$ and $8 \equiv 3 \pmod 5$, the result for $7$ and $8$ follows directly. {{qed}}
To demonstrate that there are prime numbers of either form, note: :$5 = 6 \times 1 - 1$ :$7 = 6 \times 1 + 1$ The only other possibilities for $p$ are: :$p = 6 n$, in which case $6 \divides p$ and so $p$ is not prime :$p = 6 n + 2$, in which case $2 \divides p$ and so $p$ is not prime :$p = 6 n + 3$, in which case $3 \divides p$ and so $p$ is not prime :$p = 6 n + 4$, in which case $2 \divides p$ and so $p$ is not prime :$p = 6 n + 5$, which is the same as $p = 6 \paren {n + 1} - 1$, which is covered above. Hence the result. {{qed}}
We have for any $m, n \in \Z$: {{begin-eqn}} {{eqn | l = \paren {6 m + n}^3       | o = \equiv       | r = n^3       | rr = \pmod 6       | c = Congruence of Powers }} {{eqn | o = \equiv       | r = n       | rr = \pmod 6       | c = Euler's Theorem: $\map \phi 6 = 2$ }} {{end-eqn}} By definition of modulo arithmetic: :$\exists k \in \Z: \paren {6 m + n}^3 = n + 6 k$ We also have: {{begin-eqn}} {{eqn | l = \paren {k + 1}^3 + \paren {k - 1}^3 - k^3 - k^3        | r = \paren {k^3 + 3 k^2 + 3 k + 1} + \paren {k^3 - 3 k^2 + 3 k - 1} - k^3 - k^3       | c = Cube of Sum }} {{eqn | r = 6 k }} {{end-eqn}} Thus $n = \paren {6 m + n}^3 + k^3 + k^3 - \paren {k + 1}^3 - \paren {k - 1}^3$ is an expression of $n$ as a sum of $5$ cubes. In the equation above, $m$ is arbitrary and $k$ depends on both $m$ and $n$. As there is an infinite number of choices for $m$, there is an infinite number of such expressions. {{qed}}
{{ProofWanted|Needs a considerable amount of background work to be completed first.}} === Sequence of Volumes of Unit Hyperspheres === {{:Volume of Unit Hypersphere/Sequence}}