$${ {\sqrt a+\sqrt b=\sqrt{2009}} }$$
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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...
$$\sqrt{x}+\sqrt{y}=\sqrt{2205}$$

How to solve the equation of $\sqrt{x}+\sqrt{y}=\sqrt{2205}$ in integers?

https://math.stackexchange.com/questions/1575392/how-to-solve-the-equation-of-%24%5Csqrt%7Bx%7D%2B%5Csqrt%7By%7D%3D%5Csqrt%7B2205%7D%24-in-integers%3F
...How to solve the equation of $$\sqrt{x}+\sqrt{y}=\sqrt{2205}$$ in integers? How in general to solve the similar equatio...
$$\sqrt{x}+\sqrt{y}=\sqrt{2013}$$

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?

...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...
$$\sqrt{a+\sqrt{b}}$$

...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...
$$t=\sqrt{abcd}$$
...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
... $$\Re=\frac {24+\sqrt[3]{\frac{12096+\sqrt{134369280}}2}+\frac{144}{\sqrt[3]{\frac{12096+\sqrt{134369280}}2}}}{72}$$ 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: $$-0.47533^n+2\sum_{k=0}^{\lfloor\frac n2\rfloor}\binom{n}{2k}(-0.657118)^k(0.737665)^{n-2k}$$...