Assigning Parts

Let $a = \sqrt{3!\sqrt{3!\sqrt{3!\sqrt{3!\sqrt{...}}}}}$ , $b = \cfrac{3}{1^2} + \cfrac{3}{2^2} + \cfrac{3}{3^2} + \cfrac{3}{4^2} + \cfrac{3}{5^2} ...$ , and $c = (1 \times 10 + 9)(9 \times 10 + 1) + 1 - 9^3 + 12^3 - 10^3$ . Then $x = \cfrac{a \cdot 1729 \cdot b}{\pi^2 \cdot c}$ .

Solving for $a$

Since $a = \sqrt{3!\sqrt{3!\sqrt{3!\sqrt{3!\sqrt{...}}}}}$ , that means $a = \sqrt{3!a}$ . Squaring both sides gives $a^2 = 3!a$ , and since $a \neq 0$ , $a = 3! = 6$ .

Solving for $b$

Writing $b = \cfrac{3}{1^2} + \cfrac{3}{2^2} + \cfrac{3}{3^2} + \cfrac{3}{4^2} + \cfrac{3}{5^2} ...$ as a sum gives $b = \displaystyle \sum _{n=1}^{\infty }{\frac {3}{n^2}} = \displaystyle 3\sum _{n=1}^{\infty }{\frac {1}{n^2}}$ . Since the sum of the reciprocals of the squares of the natural numbers is a Basel problem that has been shown to be $\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^2}} = \frac{\pi^2}{6}$ , this can be substituted in to show that $b = 3 \cdot \cfrac{\pi^2}{6} = \cfrac{\pi^2}{2}$ .

Solving for $c$

By direct calculation, $c = (1 \times 10 + 9)(9 \times 10 + 1) + 1 - 9^3 + 12^3 - 10^3 = 1729$ .

(Bonus: $1729$ is the smallest number that can be expressed as the sum of two positive cubes in two different ways, and is called a taxicab number .)

Solving for $x$

Substituting $a = 6$ , $b = \cfrac{\pi^2}{2}$ , and $c = 1729$ into $x = \cfrac{a \cdot 1729 \cdot b}{\pi^2 \cdot c}$ gives $x = \cfrac{6 \cdot 1729 \cdot \frac{\pi^2}{2}}{\pi^2 \cdot 1729} = 3$ .

Solving for $p$

Since $u$ is a positive integer, it is either in the form of $4n$ , $4n + 1$ , $4n + 2$ , or $4n + 3$ for some integer $n$ .

That means there are four cases to consider for $p = u^2 + (\sqrt{2})^4 = u^2 + 4$ :

Case 1: If $u = 4n$ , then $p = (4n)^2 + 4 = 16n^2 + 4 \equiv 0 \pmod{4}$

Case 2: If $u = 4n + 1$ , then $p = (4n + 1)^2 + 4 = 16n^2 + 8n + 5 \equiv 1 \pmod{4}$

Case 3: If $u = 4n + 1$ , then $p = (4n + 2)^2 + 4 = 16n^2 + 16n + 4 \equiv 0 \pmod{4}$

Case 4: If $u = 4n + 1$ , then $p = (4n + 3)^2 + 4 = 16n^2 + 24n + 9 \equiv 1 \pmod{4}$

Since there are no cases where $p \equiv 3 \pmod{4}$ , $p$ does not exist .

Analyzing $\tan (x)$

According to an extension of Niven's Theorem , the only rational values of tangent are $0$ and $\pm 1$ , which would occur at $\tan(\pi n)$ and $\tan(\pi n \pm \cfrac{\pi}{4})$ for some integer $n$ , or put another way, would occur at $\tan(\pi k)$ for some rational number $k$ . However, since $x = 3 = \pi \cdot \cfrac{3}{\pi}$ , the $k$ value would be $k = \cfrac{3}{\pi}$ , which is not rational. Therefore, $\tan(3)$ must be irrational .

In the graph above, rational values of $\tan(x)$ occur at the intersection of blue and black lines, when $\tan (x) = 0$ or $\tan (x) = \pm 1$ . As proved above, the green line $x = 3$ will not give $\tan(x)$ a rational value because it does not pass through the blue and black intersection points. Therefore, $\tan(3)$ is irrational.

Conclusion

Therefore, Chetan would say that $p$ does not exist and that $\tan(x)$ is irrational .

$\colorbox{#333333}{\color{#FFFFFF}{\text{Solving First Part}}}$

• Square Root Part

$k_1 = \sqrt{3!\sqrt{3!\sqrt{3!\sqrt{.....}}}}$ $k_1 = \sqrt{3! × k_1}$ $k_1^2 = 6k_1$ $k_1(k_1 - 6) = 0$ $\text{As } k_1 \neq 0, \boxed{k_1 = 6}$

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• Basel Problem Part

$\text{Here is a }$ $\text{link}$ $\text{to solve it. It's long so I can't tell here}$ $\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} \ldots = \frac{\pi^2}{6}$ $k_2 = \frac{3}{1^2} + \frac{3}{2^2} + \frac{3}{3^2} \ldots = 3 × \frac{\pi^2}{6}$ $\boxed{k_2 = \frac{\pi^2}{2}}$

$\color{grey}{\text{Basel Problem Solution}}$

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• Denominator part

$k_3 = \pi^2 (1 × 10 + 9)(9 × 10 + 1) + 1 - 9^3 + 12^3 - 10^3$ $\boxed{k_3 = \pi^2 × 1729}$

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• Combining everything

$x = \frac{k_1 × 1729 × k_2}{k_3}$ $x = \frac{6 × 1729 × \frac{\pi^2}{2}}{\pi^2 × 1729}$ $\color{#3D99F6}{\boxed{x = 3}}$

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$\colorbox{#333333}{\color{#FFFFFF}{\text{Solving Second Part}}}$

$\text{As } p \equiv 3 \pmod{4}, \text{ Let } p = 4k + 3 \text{ where } k \in I$ $\text{Putting } p = 4k + 3 \text{ in } p = u^2 + (\sqrt{2})^4$ $4k + 3 = u^2 + 4$ $k = \frac{u^2 + 1}{4}$

$\text{Now, we know that all squares can be expressed of form } 4m \text{ or } 4m + 1 \text{ where } m \in I \text{ Here is a proof if you like}$ $\text{link}$

$\color{grey}{\text{Proof that squares are of form 4m and 4m + 1}}$

$k = \frac{(4m) + 1}{4} \text{ or } k = \frac{(4m + 1) + 1}{4} = \frac{4m + 2}{4}$

$\text{Now, R.H.S. in neither is an integer whereas L.H.S. in both is an integer}$

$\color{#3D99F6}{\boxed{p\text{ is not existent}}}$

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$\colorbox{#333333}{\color{#FFFFFF}{\text{Solving Whole Question}}}$

$\text{Now, as } p \text{ is not existent, we'll take second Statement }$

$\tan x = \tan{3} \text{ which is irrational}$

$\text{So, Answer is }\color{#3D99F6}{\boxed{\text{Option 1}}}$

@Siddharth Chakravarty ,i am not as great as leonardo eulur,this question was a big trouble for me, i know that the continued fractions will converge at pi/6 but,i have no proof till. Please remove me from leader board

peace

The question has been made to look unnecessarily complex it would be better to split this problem into various sub problems.Also this has nothing special to itself it just a collection of problems combining them in this way doesn't makes it better neither does it adds extra thinking.No offence.

After all questions are supposed to teach us how to think and learn not to just 'look' hard.

also @Siddharth Chakravarty you can kindly remove me from the leader board I am no longer participating.

This question can be broken down into pieces:

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$\huge Finding P$
Since $p\equiv 3 \pmod{4}$ and $p=u^2+4$ , $\color{#D61F06} u^2\equiv 3\pmod 4$ . We know immediately that this isn’t possible for integer $u$ as for integer $u$ , $u^2\equiv 0,1\pmod{4}$ . So $p$ doesn’t exist.

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$\huge Breaking$
$\frac{\color{#D61F06}(\sqrt{3!\sqrt {3!\sqrt{...}}})\color{#3D99F6}(1729)(\frac{3}{1}+\frac{3}{4}...)}{\color{#20A900}\pi ^2(19\times 91+1-9^3+12^3-10^3)}$

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$\huge \color{#D61F06} RED$
Call this bit $x$ . Since $3!=6$ , $x=\sqrt{6x}$ $x^2=6x\Rightarrow x=0,\underline{6}$ So $x=6$ .

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$\huge \color{#3D99F6} BLUE$
Study $\sum_{i=1}^n\dfrac{3}{i^2}$ . This is actually three times the answer to the Basel problem .

Proof and graph-to-theorem

So this bit $=\dfrac{1729\pi ^2}{2}$ .

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$\huge \color{#20A900} GREEN$
This requires some simple maths. You get $1729\pi ^2$ immediately.

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$\huge END$
Put these together to get $\frac{(6)(1729)\frac{\pi ^2}{2}}{1729\pi^2}=3.$ Now find out whether $\tan 3(\text{rad})$ is rational.

We see this isn’t $0$ or $\pm 1$ so it’s irrational. So $p$ doesn’t exist, and $\tan 3$ is irrational. Done, @Siddharth Chakravarty . Pls give me marks now :)

$\text{The value of } x \text{, solving each term separately}$

$\text{First part}$

$\sqrt{3!{\sqrt{\sqrt{3!\cdots}}}}=y \text{(Say)}$ $\implies \sqrt{3!y}=y$

$\text{Squaring both sides of the equation, and taking positive square root since } y>0$ $\implies y^2 =3!y$ $\implies y \cancel{y} =3!\cancel{y}$ $\implies y=3!=3\times 2\times 1 =6 \text{ (By definition of factorial function)}$ $\text{Link to factorial function}$

$\text{Second part}$

$\dfrac{3}{1^2}+\dfrac{3}{2^2}+\dfrac{3}{3^2}+\dfrac{3}{4^2} \cdots$ $3\times (\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2} \cdots )$ $\text{By the famous Basel Problem, solved by the great Leonhard Euler,}$

$\text{Link 1 to Basel Problem}$ $\text{Link 2 to Basel Problem(Wonderful explanation!)}$ $\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2} \cdots= \dfrac{{\pi}^2}{6}$ $3\times (\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2} \cdots )= 3\times\dfrac{{\pi}^2}{6}=\dfrac{{\pi}^2}{2}$

$\text{Denominator}$

$\text{We leave } \pi^2 \text{ just now, and solve the rest}$ $(1\times 10 + 9)(9\times 10 +1)+1+9^3-12^3+10^3=1729 \text{(Can be solved using algebraic identities, and someone may use a calculator also.)}$

$\text{Combining everything}$

$\text{The expression becomes:}$ $x= \dfrac{6 \times (1729) \times \dfrac{\pi^2}{2}}{\pi^2 \times (1729)} =\dfrac{3 \times (1729) \times \pi^2}{\pi^2 \times (1729)}=\dfrac{3 \times \cancel{(1729)} \times \cancel{\pi^2}}{\cancel{\pi^2} \times \cancel{(1729)}}$ $\Large{\color{#D61F06}{\implies \boxed{x=3}}}$

$\text{There is no value of } p$

$\text{Given,}$ $p=u^2+({\sqrt{2}})^4=u^2+4 \implies p \equiv u^2 \pmod{4}$ $\text{Now, according to "Square Modulo 4"}$ $\text{Link to Square Modulo 4}$

$\text{All perfect squares are equivalent to either 0 or 1} \pmod{4}$

$\text{But, it is also given that } p \equiv 3 \pmod{4}$ , $\text{Hence, we reach a contradiction, and}$

$\Large{\color{#D61F06} p\text{ does not exist.}}$

$\text{Is } \tan{(x)} \text{ rational}$

$\text{According to my search on WolframAlpha,}$ $\text{Link to WolframAlpha}$

$\text{A short proof}$

$\text{According to Niven's Theorem,}$ $\text{Link to Niven's Theorem}$

$\text{Link to Proof of Niven's Theorem}$

$\text{Now, since we know that}$ $3 \text{radians} = 3 \times \left(\dfrac{180}{\pi}\right)^{\circ} = \left(\dfrac{540}{\pi}\right)^{\circ}$ $\text{Since it is irrational, and not a multiple of some special values, which make it 0 or }\pm1\text{, it needs to be irrational.}$

$\text{Final Answer!!!!}$

$\Large{p\text{ does not exist, and } \tan(x)\text{ is irrational.}}$

Evaluate x :

Radical = 6^1/2 * 6^1/4 * 6^1/8 ...... = 6^S 1/2+1/4+1/8.... to infinite = 6^1 = 6

S 1/n^2 from n=1 to infinite is Problem of Basilea ( https://es.wikipedia.org/wiki/Problema de Basilea) = (Pi^2)/6

X = (1729 * 3* pi^2 ) / (pi^2 ) * (1729 + 0 ) => X=3

Evaluate p :

expresion 1 : p = u^2 + 4

expresion 2 : p = 4 N + 3

for equals : p = u^2 + 4 = 4 N + 3 => (u^2 + 1) / 4 = N => ( u / 2 ) ^2 + 1/4

( u / 2 ) ^2 = integer if u = even or integer + 1/4 if u = odd

expresion : ( u / 2 ) ^2 + 1/4 will be always integer + 1/4 or integer + 1/2 => never N => p not exist

Funtion tg (3) is irrational

First Part of the Solution(computing x ):

$x=\frac{\sqrt{3!\sqrt{3!\sqrt{3!..._\infty}}} \times 1729 \times (\frac{3}{1^2} + \frac{3}{2^2} + \frac{3}{3^2} + \frac{3}{4^2} + ..._\infty)}{\pi^2 \times [(1\times10+9)(9\times10+1) + 1 - 9^3 + 12^3 - 10^3]}$

**

$\sqrt{3!\sqrt{3!\sqrt{3!..._\infty}}}$ = y(say)

Thus $y= \sqrt{3!\times y}$

$y^2 = 6y$

$y=6$

$3! = 3*2*1 = 6$

Click here to know what a Factorial is.

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$(\frac{3}{1^2} + \frac{3}{2^2} + \frac{3}{3^2} + \frac{3}{4^2} + ..._\infty) = 3(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ..._\infty) =3\times \frac{\pi^2}{6} = \frac{\pi ^ 2}{2}$

Click here to find the proof for the above.

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The denominator simplifies to $1729 \times \pi^2.$ using a calculator or just your head, if you can.

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Hence $x = \frac {6 \times 1729 \times \frac{\pi^2}{2}}{\pi^2 \times 1729} = 3$

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Second Part(computing p):

The two equations given to us are:

1. $p\equiv 3(mod 4)$

2. $p = u^2 + \sqrt2^4 = u^2 + 4$

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The first equation tells us that p can be expressed in the form $4n+3$ , where $n \in I$ .

Equating 'p' in both equations gives us: $4n = u^2+1$

This simplifies to $n=\frac{u^2 + 1}{4}$

With our newly acquired knowledge, we can now replace $u^2$ with $4m$ or $4m + 1$ .

Hence: $n=\frac{4m+1}4 or \frac{4m+2}4$ both of which are not integers, but n must be an integer.

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This contradicts our original statement and we can make a conclusion that p does not exist .

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SOLUTION

Since $p$ does not exist, the second statement is followed: "Chetan tells p does not exist and whether $\tan{(x)}$ is irrational or rational, where x is measured in radians."

$\tan(x) = \tan(3)$ (rad) $\approx -0.142$

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Proof that it is irrational:

According to Niven's Theorem , $\tan\theta$ can only be rational if its values are $0$ or $\pm1$

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Hence Chetan says that p does not exist and that $\ tan(x)$ is irrational.