Me no comprehende.

RaleD wrote: Are there positive integers $a, b, c$ greater than $2011$ such that: $(a+ \sqrt{b})^c=...2010,2011...$? Yes. For example, $a=7010$, $b=25002012$, $c=1$.

nnosipov wrote: RaleD wrote: Are there positive integers $a, b, c$ greater than $2011$ such that: $(a+ \sqrt{b})^c=...2010,2011...$? Yes. For example, $a=7010$, $b=25002012$, $c=1$. OP asked for $c>2011$ too

What does it mean by $\left(a+\sqrt{b}\right)^c = \ldots2010,2011\ldots$? Is it the German decimal notation, and thus, are we to find $a$, $b$, and $c$ for which $\left(a+\sqrt{b}\right)^c = \ldots2010.2011\ldots$?

Batominovski wrote: What does it mean by $\left(a+\sqrt{b}\right)^c = \ldots2010,2011\ldots$? Is it the German decimal notation, and thus, are we to find $a$, $b$, and $c$ for which $\left(a+\sqrt{b}\right)^c = \ldots2010.2011\ldots$? Yes.

Batominovski wrote: What does it mean by $\left(a+\sqrt{b}\right)^c = \ldots2010,2011\ldots$? Is it the German decimal notation, and thus, are we to find $a$, $b$, and $c$ for which $\left(a+\sqrt{b}\right)^c = \ldots2010.2011\ldots$? What is german decimal notation?

RaleD wrote: Are there positive integers $a, b, c$ greater than $2011$ such that: $(a+ \sqrt{b})^c=...2010,2011...$? this is very confusing. What is this? $2010.2011$ or else?If it is German notation as the previous post says,then what is the definition of this notation?

myro111 wrote: Batominovski wrote: What does it mean by $\left(a+\sqrt{b}\right)^c = \ldots2010,2011\ldots$? Is it the German decimal notation, and thus, are we to find $a$, $b$, and $c$ for which $\left(a+\sqrt{b}\right)^c = \ldots2010.2011\ldots$? What is german decimal notation? Simple, "$12,45$" in Germany is $12.45$ in most other countries. I guess the OP wanted $a$, $b$, and $c$ such that $\left(a+\sqrt{b}\right)^c$ contains $2010.2011$ in its decimal presentation. (It may be, for instance, $7642010.201178643513$.)

Problem asks to find $a$, $b$, $c$ (bigger than 2011) such that integer part of $(a+\sqrt{b})^{c}$ is congruent 2010 mod 10000; and it's fractional part is bigger than 0,2011 and smaller than 0,2012. "$,$" is (of course) decimal comma...

Stop arguing about notations guys. It's spam Choose $a = 100000000$, $b = 9999999800000000$, and $c = 1797184159$. Obviosly, all three are more than 2011 Hint: $b = (a-1)^2-1$

mymath7 wrote: Stop arguing about notations guys. It's spam Choose $a = 100000000$, $b = 9999999800000000$, and $c = 1797184159$. Obviosly, all three are more than 2011 Hint: $b = (a-1)^2-1$ How did you calculate it? Did you do it without a calculator?

tom_damrong wrote: How did you calculate it? Did you do it without a calculator? Of course!

mymath7 wrote: Obviosly, all three are more than 2011 Actually, the question wasn't to just find three numbers that are greater than 2011 So, can you show that your numbers satisfy given equation?

Let $S = {\left( {a + \sqrt b } \right)^c} + {\left( {a - \sqrt b } \right)^c}$, for odd $c$ and $a = 10000k$, we have $S \equiv 0(\bmod 10000)$. So if $7989.7989 > {\left( {a - \sqrt b } \right)^c} > 7989.7988$, then we done. Let $b = {(a - 1)^2} - 1 = {a^2} - 2a$, then $\mathop {\lim }\limits_{a \to + \infty } a - \sqrt b = \mathop {\lim }\limits_{a \to + \infty } \frac{{2a}}{{a + \sqrt {a^2-2a} }} = 1$, so for large $k$, that we have $1 < a - \sqrt b < {\left( {\frac{{7989.7989}}{{7989.7988}}} \right)^{1/3000}}$. If $\alpha $ is the biggest integer for $7989.7988 \geqslant {\left( {a - \sqrt b } \right)^\alpha }$, then $\alpha > 3000$, and $7989.7988 < {\left( {a - \sqrt b } \right)^{\alpha + 1}}<{\left( {a - \sqrt b } \right)^{\alpha + 2}}<7989.7988 \times {\left( {\sqrt {\frac{{7989.7989}}{{7989.7988}}} } \right)^2}=7989.7989$. Let $c$ be the odd one of $\alpha+1, \alpha+2$, then $7989.7989 > {\left( {a - \sqrt b } \right)^c} > 7989.7988$. So $a,b,c>3000$, and ${\left( {a + \sqrt b } \right)^c}=S-{\left( {a - \sqrt b } \right)^c}= ...2010.2011...$.