$$ { ${\sqrt a+\sqrt b=\sqrt{2009}}$ } $$
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$$ \sqrt a + \sqrt b = \sqrt{2009}$$

An interesting number theory proof involving concepts of prime numbers and integers.

https://math.stackexchange.com/questions/3871207/an-interesting-number-theory-proof-involving-concepts-of-prime-numbers-and-integers.
...I was trying a British Math Olympiad problem, \( \sqrt a + \sqrt b = \sqrt{2009} \) , find all integers a and b. After solving this problem, https://youtu.be/quECgYPNCXw in a really similar fashion to this solution, i thought to try and come up with a generalisation, if \( \sqrt a + \sqrt b = \sqrt c \) , where \( \sqrt c = m \sqrt n \) , where \( m \) and \( n \) are integers and...
$$ \sqrt{x}+\sqrt{y}=\sqrt{2205}$$
$$ \sqrt{x}+\sqrt{y}=\sqrt{2013}$$

Equation $\sqrt{x}+\sqrt{y}=\sqrt{2013}$ in rationals

https://math.stackexchange.com/questions/564422/equation-%24%5Csqrt%7Bx%7D%2B%5Csqrt%7By%7D%3D%5Csqrt%7B2013%7D%24-in-rationals
...Can we find all rational numbers \( x,y \) such that \( \sqrt{x}+\sqrt{y}=\sqrt{2013} \) ? Certainly possible answers are \( (2013,0) \) and \( (0,2013) \) . If we square the equation, we get \( x+y+2\sqrt{xy}=2013 \) , so \( \sqrt{xy} \) must be ration...
$$ \sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{2013}$$

Equation $\sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{2013}$ in rationals

https://math.stackexchange.com/questions/566141/equation-%24%5Csqrt%7Bx%7D%2B%5Csqrt%7By%7D%2B%5Csqrt%7Bz%7D%3D%5Csqrt%7B2013%7D%24-in-rationals
...Consider the equation \( \sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{2013} \) , where \( x,y,z \) are rational numbers. Are there any solutions other than the trivial ones \( (2013,0,0),(0,2013,0),(0,0,2013) \) ? We can subtract \( \sqrt{z} \) from both sides and square to get...
$$ \sqrt{x}+\sqrt{y}=\sqrt{1376}$$

Find all natural roots of $\sqrt{x}+\sqrt{y}=\sqrt{1376}$ given that $x\leq y$

https://math.stackexchange.com/questions/1830986/find-all-natural-roots-of-%24%5Csqrt%7Bx%7D%2B%5Csqrt%7By%7D%3D%5Csqrt%7B1376%7D%24-given-that-%24x%5Cleq-y%24
...Find all natural roots of \( \sqrt{x}+\sqrt{y}=\sqrt{1376} \) given that \( x\leq y \) I'm confused of this equation because \( 1376 \) is not a square!! So maybe it has no natural root! Am I righ...
$$ 7\sqrt{a} + 17\sqrt{b} + k\sqrt{c} \ge \sqrt{2019}$$

Find the least positive real number $k$ such that $7\sqrt{a} + 17\sqrt{b} + k\sqrt{c} \ge \sqrt{2019}$ over all positive real numbers

https://math.stackexchange.com/questions/3068290/find-the-least-positive-real-number-%24k%24-such-that-%247%5Csqrt%7Ba%7D-%2B-17%5Csqrt%7Bb%7D-%2B-k%5Csqrt%7Bc%7D-%5Cge-%5Csqrt%7B2019%7D%24-over-all-positive-real-numbers
...Working on a problem... Find the least positive real number \( k \) such that \( 7\sqrt{a} + 17\sqrt{b} + k\sqrt{c} \ge \sqrt{2019} \) over all positive real numbers \( a,b,c \) with \( a+b+c=1 \) . Maximizing the "\( a \) " term doesn't seem to work, and expansion through moving a radical to the right side of the equation simply leads to more radicals. Any he...
$$ \sqrt a + \sqrt b = \sqrt n$$

Is there a way to solve $\sqrt a + \sqrt b = \sqrt n$ analytically?

https://math.stackexchange.com/questions/1519204/is-there-a-way-to-solve-%24%5Csqrt-a-%2B-%5Csqrt-b-%3D-%5Csqrt-n%24-analytically%3F#answer-1519253
...A basic fact that comes out of Galois theory is that the set \( \{\sqrt{d_i}\}_{i \in I} \) is linearly independent over \( \mathbb Q \) , where \( \{d_i\}_{i \in I} \) enumerates all square-free integers (including negative integers!). Since \( \sqrt a + \sqrt b = \sqrt n \) expresses a linear dependence among three different square roots, it actually follows that the square free parts of \( a, b, n \) must, in fact, all be the same. So all you need to do is factor the largest square out of \( n \) to get \( n = k^2 m \) with \( m \) square-free - in the case of...
$$ \sqrt a\geq\sqrt{2001}$$

Writing square root of square-free numbers as sum of square roots.

https://math.stackexchange.com/questions/737886/writing-square-root-of-square-free-numbers-as-sum-of-square-roots.
... \( \sqrt a\geq\sqrt{2001} \) and \( \sqrt{b}\leq 0 \) which means \( b=0 \) . With exact method we can prove that \( \sqrt{s}=\sqrt a+\sqrt b \) does not have any natural solutions with \( s \) being a square-free number. Then I tried to generalize the proof for \( 3 \) or more square roots but i failed. The only thing I always get is...
$$ b=\sqrt[2019]{3}$$

degree of a sum of two algebraic numbers

https://math.stackexchange.com/questions/3088909/degree-of-a-sum-of-two-algebraic-numbers
... \( b=\sqrt[2019]{3} \) . How to prove that the degree of \( a+b \) is: greater than 2019? (if it can be done with less advanced methods then the second part) equal \( 2019^2 \) ? (if this value is correct)...
$$ \sqrt{a_1+b_1\sqrt{a_2+b_2\sqrt{a_3+b_3\cdots}}}$$

Evaluating an infinite square root

https://math.stackexchange.com/questions/878363/evaluating-an-infinite-square-root
...How do I evaluate the square root: \( \sqrt{2013+276\sqrt{2027+278\sqrt{2041+280\cdots}}} \) I have tried creating two arithmetic sequences such that \( a_n = 1999+14n \) \( b_n = 274+2n \) so the square root simplifies to \( \sqrt{a_1+b_1\sqrt{a_2+b_2\sqrt{a_3+b_3\cdots}}} \) But I get stuck there. Any help/hints is greatly appreciat...
$$ \sqrt{a+\sqrt{b}}$$

question related to radical sign

https://math.stackexchange.com/questions/171029/question-related-to-radical-sign#answer-171035
...Compute \( p(2012)=\sqrt{2014+3 \sqrt{4019}}-\sqrt{2010+\sqrt{4019}} \) and apply the formula for double square roots, noticing that \( 2010^2-4019 = 2009^2 \) and \( 2014^2-9 \cdot 4019 = 2005^2 \) . Now you can compute the desired result. By the way, the final result is \( 8 \) . Sorry for the italian reference, but you can easily read the formula to simplify an expression like \( \sqrt{a+\sqrt{b}} \) when \( a^2-b \) is a perfect squa...
$$ t=\sqrt{abcd}$$

BMO2 1996 Questio 4 - Algebra Problem

https://math.stackexchange.com/questions/1356680/bmo2-1996-questio-4---algebra-problem#answer-1356691
...I couldn't understand why \( 6\sqrt{abcd}\geq54 \) Note that \( abcd-27\ge 6\sqrt[6]{a^3b^3c^3d^3}=6\sqrt{abcd}. \) Setting \( t=\sqrt{abcd} \) gives you \( t^2-27\ge 6t\iff (t-9)(t+3)\ge 0\iff t\ge 9. \)...
$$ f(\sqrt[3]{2019})=0$$

Prove that $-\sqrt{c}

https://math.stackexchange.com/questions/3783407/prove-that-%24-%5Csqrt%7Bc%7D%3Cab%3C0%24-if-%24a%5E4-2019a%3Db%5E4-2019b%3Dc%24.#comment-7791441
[email protected] von Eitzen \( f \) is a convex function, \( f(0)=0 \) , \( f(\sqrt[3]{2019})=0 \) , \( f \) decreases on \( (-\infty,0] \) and \( f \) increases on \( [\sqrt[3]{2019},+\infty), \) which says the equation \( f(x)=c \) , where \( c >0 \) has two real roots exactly. \( a \) and \( b \) are roots. Thus, \( ab< 0 \) because one of them is negative and other is greater than \( \sqrt[3]{2019}. \) Is it clear n...
$$ \sqrt{2500-a^2}$$

How to count this in a faster way?

https://math.stackexchange.com/questions/514195/how-to-count-this-in-a-faster-way%3F#answer-514206
...For any particular \( a \) , \( b \) can go up to the largest integer below \( \sqrt{2500-a^2} \) . So for any particular \( a \) we have the count equal to \( \lfloor\sqrt{(2500-a^2)}\rfloor \) . The half brackets mean the floor function. Then you would sum this up from \( a=0 \) up to \( a=49 \) . (You could also sum to \( a=50 \) , but the floor of 0 is 0.) Plugging into wolframalpha I get...
$$ 2018 = 2 \cdot 1009$$

Prove that there are no integer solutions to $ c^2 = 2018^a + 2018^b $

https://math.stackexchange.com/questions/3561182/prove-that-there-are-no-integer-solutions-to-%24-c%5E2-%3D-2018%5Ea-%2B-2018%5Eb-%24
...A friend of mine gave me this puzzle and I want to solve it, turns out its harder than I expected. I tried to prove this by contradiction, so let's just assume there is an integer solution. The first thing I noticed was this: \( 2018 = 2 \cdot 1009 \) , 2 and 1009 are both primes, also c must be an even number. My first idea was to rewrite a little: \( c^2 = 2018^a + 2018^b \implies 2018^a = c^2 - 2018^b = (c+\sqrt{2018}^b)(c-\sqrt{2018}^b) \) , but as it turns out, \( (c+\sqrt{2018}^b) \) and \( (c-\sqrt{2018}^b) \) don't necessarily have to be irrational, for example...
$$ \sqrt(1632397825)$$

Square Root Algorithm

https://math.stackexchange.com/questions/439135/square-root-algorithm#answer-439181
..."efficient" rather depends on what you constraints are. For instance, if you have enough memory to store some floats, but little cpu time, then you can store a look up table for all integers until some power of 10. So say you had to find \( \sqrt(1632397825) \) . You could write: \( \sqrt{1732397825} =\sqrt{ 17*10^8 + 32397825}\sim \sqrt{17}\cdot 10^4 + \epsilon \) To calculate \( \epsilon \) , use the fact that: \( \sqrt{a^2+b}\sim a + \frac{b}{2a} - \frac{b^2}{8a^3} + ... \) So in our example, \( \sqrt{1732397825} \sim 41622.06575 \) Quite close to the true value of: \( \sqrt{1732397825}=41622.08338 \)...
$$ a+1 = x, b+1 =y, c+1 = z$$

issues with simple algebraic equations

https://math.stackexchange.com/questions/695496/issues-with-simple-algebraic-equations#answer-695517
...\\ bc+b+c=300 \\ ac+a+c=216 \\ \end{cases} it seems of no difficulty: a system of 3 equations and 3 variables... Let's start by adding 1 to every equations, we obtain: \begin{cases} ab+a+b+1=(a+1)(b+1)=251 \\ bc+b+c+1=(c+1)(b+1)=301 \\ ac+a+c+1=(a+1)(c+1)=217 \\ \end{cases} Now let's substitute \( a+1 = x, b+1 =y, c+1 = z \) , so we can reduce the amount of calculus needed, in fact to solve \begin{cases} xy=251 \\ yz=301 \\ xz=217 \\ \end{cases} you only need to find \( x \) (or...
$$ b=\frac{49}{625}+\frac{\sqrt{(18 808 849)}}{(4375)}$$

Linear estimation of an exponential distribution

https://math.stackexchange.com/questions/1864847/linear-estimation-of-an-exponential-distribution#comment-3819762
...It gives me 2 pairs of coordinates with the same error of 1. \( a=\frac{-4}{625}-\frac{\sqrt{(18 808 849)}}{(26 250)} \) , \( b=\frac{49}{625}+\frac{\sqrt{(18 808 849)}}{(4375)} \) and \( a=\frac{-4}{625}+\frac{\sqrt{(18 808 849)}}{(26 250)} \) , \( b=\frac{49}{625}-\frac{\sqrt{(18 808 849)}}{(4375)} \)...
$$ a_r=\prod_{v=2}^{r} \sqrt[v(v-1)]{v}$$

Infinite product experimental mathematics question.

https://mathoverflow.net/questions/22088/infinite-product-experimental-mathematics-question.#answer-22093
...An experimental observation: if \( a_r=\prod_{v=2}^{r} \sqrt[v(v-1)]{v} \) and \( b_r=\prod_{n=1}^{r} \sqrt[n]{1+\frac{1}{n}} \) , then the numbers \( a_{2r}/b_{2r} \) are, according to Mathematica, \( \frac{1}{\sqrt{3}},\frac{1}{\sqrt[4]{5}},\frac{1}{\sqrt[6]{7}},\frac{1}{\sqrt[4]{3}},\frac{1}{\sqrt[10]{11}},\frac{1}{\sqrt[12]{13}},\frac{1}{\sqrt[14]{15}},\frac{1}{\sqrt[16]{17}},\frac{1}{\sqrt[18]{19}},\frac{1}{\sqrt[20]{21}}, \) \( \frac{1}{\sqrt[22]{23}},\frac{1}{\sqrt[12]{5}},\frac{1}{3^{3/26}},\frac{1}{\sqrt[28]{29}},\frac{1}{\sqrt[30]{31}},\frac{1}{\sqrt[32]{33}},\frac{1}{\sqrt[34]{35}},\frac{1}{\sqrt[36]{37}},\frac{1}{\sqrt[38]{39}},\frac{1}{\sqrt[40]{41}},\dots \) I would imagine the products are the same, then. I don't have time but using this as a hint one should be able to give an actual proof. May you tell us how you ended up with such an identi...
$$ \Re=\frac {24+\sqrt[3]{\frac{12096+\sqrt{134369280}}2}+\frac{144}{\sqrt[3]{\frac{12096+\sqrt{134369280}}2}}}{72}$$

Find $x^n+y^n+z^n$ general solution

https://math.stackexchange.com/questions/3749238/find-%24x%5En%2By%5En%2Bz%5En%24-general-solution#answer-3749266
... \( \Re=\frac {24+\sqrt[3]{\frac{12096+\sqrt{134369280}}2}+\frac{144}{\sqrt[3]{\frac{12096+\sqrt{134369280}}2}}}{72} \) so now we can find \( \Im \) and \( x \) (we know \( 2\Im^2=6\Re^2-4\Re-1 \) and \( x=1-2\Re \) ): \( \Im=\sqrt{\frac {\left(24+\sqrt[3]{\frac{12096+\sqrt{134369280}}2}+\frac{144}{\sqrt[3]{\frac{12096+\sqrt{134369280}}2}}\right)^2}{1728}-\frac {6+\sqrt[3]{\frac{12096+\sqrt{134369280}}2}+\frac{144}{\sqrt[3]{\frac{12096+\sqrt{134369280}}2}}}{36}} \) \( x=\frac {12-\sqrt[3]{\frac{12096+\sqrt{134369280}}2}-\frac{144}{\sqrt[3]{\frac{12096+\sqrt{134369280}}2}}}{36} \) so now using \( (i) \) we get: \( a_n=x^n+y^n+z^n=x^n+(\Re+\Im i)^n+(\Re-\Im i)^n=\left(\frac {12-\sqrt[3]{\frac{12096+\sqrt{134369280}}2}-\frac{144}{\sqrt[3]{\frac{12096+\sqrt{134369280}}2}}}{36}\right)^n+2\sum_{k=0}^{\lfloor\frac n2\rfloor}\binom{n}{2k}\left(\frac {6+\sqrt[3]{\frac{12096+\sqrt{134369280}}2}+\frac{144}{\sqrt[3]{\frac{12096+\sqrt{134369280}}2}}}{36}-\frac {\left(24+\sqrt[3]{\frac{12096+\sqrt{134369280}}2}+\frac{144}{\sqrt[3]{\frac{12096+\sqrt{134369280}}2}}\right)^2}{1728}\right)^k+\left(\frac {24+\sqrt[3]{\frac{12096+\sqrt{134369280}}2}+\frac{144}{\sqrt[3]{\frac{12096+\sqrt{134369280}}2}}}{72}\right)^{n-2k} \) so this is around: \( -0.47533^n+2\sum_{k=0}^{\lfloor\frac n2\rfloor}\binom{n}{2k}(-0.657118)^k(0.737665)^{n-2k} \)...
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