I have only this so far: Let's determine whether $\sqrt{a}+\sqrt{b}$ is rational, where $a$ and $b$ are integers. $\sqrt{a}+\sqrt{b}=\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2}=\sqrt{a+2\sqrt{ab}+b}$ So, it rational only if $ab$ is a perfect square. (If it is a perfect square, we know that it is at least algebraic of degree 2). (I cannot adequetly continue this for more than two terms.)

I think this is helpful .

Zhero's solution is great, but a quicker way to finish Clearly $S > 0$. However: $S = \frac{1}{2} \sum_{i=1001}^{2000} \frac{1}{i} < \frac{1}{2} \cdot \frac{1000}{1001} < 1$ Thus $S$ cannot be an integer, which is a contradiction if the sum if a rational number.

Solution: We call $k$ an algebraic integer as a number for which there exists a monic polynomial $P(X)$ with integer coefficients if $k$ is a root of $P(X)$. It is well known that the set of algebraic integers is a ring, closed under addition and multiplication. We can see that $x=\sqrt{a^2+1}$ is an algebraic integer for any integer $x$ as the monic polynomial $P(x)$ with $x$ as a root is $P(x)=x^2-(a^2+1)=0$. Then $\sqrt{a^2+1}$ is an algebraic integer for $a=1001,...,2000$ and so ${\sum_{a=1001}}^{2000}sqrt{a^2+1}$ is an algebraic integer. Say that $Q(X)$ is a the monic polynomial for which the expression is a root. Then by the rational root theorem, if the expression is rational, then it is an integer. Hence it suffices to show that the expression is not an integer. One can bound the expression by a lower bound and upper bound, which I claim to be $1001+....+2000$ and $1001+...+2000+1$ respectively. It is not difficult to see that the expression is greater than the lower bound as each $\sqrt{a^2+1}>a$. To show that the upper bound is $1001+...+2000+1$ , it suffices to show that $(\sqrt{1001^2+1}-1001)+(\sqrt{1002^2+1}-1002)+..._(\sqrt{200^2+1}-2000)<1$. As the differences between squares increase as $a$ increases, it suffices to show that $\sqrt{1001^2+1}-1001<\frac{1}{1000}$. By a few simple calculations, we are done. As the expression can be bounded between 2 consecutive integers and is not equal to either one, we can see that it is not an integer, therefore not being rational. $\blacksquare$.

"if the expression is rational, then it is an integer" holds only if the first terms's coefficient is 1 (or -1) after eliminated the GCF of the coefficients. For example $4x^2 - 1 = 0$ has rational root 1/2 which is not an integer.

kevin686 wrote: "if the expression is rational, then it is an integer" holds only if the first terms's coefficient is 1 (or -1) after eliminated the GCF of the coefficients. For example $4x^2 - 1 = 0$ has rational root 1/2 which is not an integer. Yes but it is given that a monic polynomial has the expression as a root... Since each $\sqrt{a^2+1}$ is an algebraic integer, it follows that their sum is also an algebraic integer.

Sorry. You are right, I did not read that paragraph carefully.

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