Square Root Sum

The first thing that should be noticed is that $\sqrt{2000}=20\sqrt{5}$ . A possible way to approach this is to take $\sqrt{a}$ , $\sqrt{b}$ , and $\sqrt{c}$ as being equal to $k_1\sqrt{5}$ , $k_2\sqrt{5}$ , and $k_3\sqrt{5}$ and finding the number of ways three nonnegative numbers can add up to $20$ . However, a problem with this is that it ignores the possibility that a the square root of two numbers might simplify to $h_1\sqrt{r}$ and $h_2\sqrt{5}-h_1\sqrt{r}$ , which sum to $h_2\sqrt{5}$ . The third number could be $(20-h_2)\sqrt{5}$ , which would be an acceptable $3$ -tuple.

Therefore, something that needs to be done is to prove that a number's square root cannot factor to $\sqrt{r_1}\pm\sqrt{r_2}$ . This is a relatively straightforward proof by contradiction.

Assume a positive integer $\sqrt{N}$ can factor into $a\sqrt{b}\pm c\sqrt{d}$ where $a$ and $c$ are positive integers and $b$ and $d$ are distinct positive integers not divisible by any square of any prime. Squaring both sides gives $N=a^2b+c^2d\pm2ac\sqrt{bd}$ . Because $N$ is an integer, $b\times d$ must be $0$ or a square. $b$ and $d$ are both positive, so they cannot have a product of $0$ . We now know that $b\times d$ must be a square. In order for two positive integers $b$ and $d$ not divisible by any prime's square to have a product that is a square, the numbers must be equal. Otherwise, when simplified, $\sqrt{bd}$ will still have a number under the radical. Therefore, $b=d$ . However, this contradicts our condition that $b$ and $d$ are distinct. Therefore, $\sqrt{N}≠a\sqrt{b}+c\sqrt{d}$ . $\sqrt{N}$ can only be equal to $a\sqrt{b}$ for $a$ and $b$ being positive integers.

This proves that $20\sqrt{5}=k_1\sqrt{5}+k_2\sqrt{5}+k_3\sqrt{5}$ when $k_1$ , $k_2$ , and $k_3$ are nonnegative integers. This problem can be simplified to a balls-in-urns problem. We are finding the number of ways to put $20$ balls into $3$ urns.

The number of ways to put $n$ balls into $k$ urns is equal to $\dbinom{n+k-1}{n}$

In this problem, $n=20$ and $k=3$ .

$\dbinom{22}{20}=\dfrac{22!}{20!\times2!}=\dfrac{22\times21}{2}=\boxed{231}$

This problem can be generalized using similar logic. $\sqrt{a_1}+\sqrt{a_2}+\ldots+\sqrt{a_k}=n\sqrt{m}$ , then every $a_b$ is of the form $c\sqrt{m}$ for $c$ being a nonnegative integer and $m$ not being divisible by the square of any prime. The number of ways this can be achieved is equivalent to the balls in urns problem, so the number of ways to achieve this is $\dbinom{n+k-1}{n}$

Claim: $a = 5k_1 ^2$ for some non-negative integer $k_1$ .

Proof: We have $20 \sqrt{5} - \sqrt{a} = \sqrt{b} + \sqrt{c}$ . Squaring both sides gives $2000 + a - 40 \sqrt{5a} = b+c - \sqrt{bc}.$

Let $z = b + c -a - 2000$ , which is an integer, so $\sqrt{bc} = 40 \sqrt{5a} +z$

Squaring both sides again, we get $bc = 8000a + 80z \sqrt{5a} + z^2$ , which can be rewritten as $\sqrt{5a} = \frac {bc - 8000a - z^2}{80}$ . Since the RHS is rational, this implies that $5a = K^2$ for some integer $K$ . Hence, $a = 5k_1^2$ for some non-negative integer $k_1$ . $_ \square$

Back to the original problem. We have $a=5k_1 ^2, b = 5k_2 ^ 2, c = 5k_3 ^2$ by symmetry, hence we want to find the number of non-negative solutions to $k_2 + k_2 + k_3 = 20$ . By generating functions, this is the coefficient of $x^{20}$ in the binomial expansion of $(1-x)^{-3}$ , hence is equal to $\frac {21 \times 22} {2} = 231$ .

Claim 1. If √n is rational, then n is a perfect square, where n is a non-negative integer. Proof. Let √n=p/q, where p is a non-negative integer, q is a positive integer and (p,q)=1. n=p²/q², q²n-p²=0. Let P(x)=q²x-p². Observe that P(n)=P(p²/q²)=(q²)(p²/q²)-p²=0. This implies that n is a root of P(x). Since n is a positive integer, by Rational Roots Theorem, q² divides 1. q²=1. Therefore, n=p²/q²=p²/1=p², a perfect square. Claim 2. If √m+√n is rational for non-negative integers m,n, then both m and n are perfect squares. Proof. Let √m+√n=p/q, where p is a non-negative integer, q is a positive integer and (p,q)=1. √m=p/q-√n. Squaring both sides, m=p²/q²+n-2(p/q)(√n), √n is rational, by Claim 1, n is a perfect square. Similarly, m is also a perfect square. Now, we return to the problem. √b+√c=√2000-√a. Squaring both sides, b+c+2√(bc)=2000+a-2√(2000a). Rearranging, √(bc)+√(2000a)=(2000+a-b-c)/2. √(bc)+√(2000a) is rational, by Claim 2, 2000a is a perfect square. 2000a=20²×5a is a perfect square. Therefore, a is divisible by 5. Let a=5x² for some non-negative integer x. Similarly, b=5y² and c=5z² for some non-negative integer y and z. √(5x²)+√(5y²)+√(5z²)=√2000. x+y+z=20. Since a non-negative integer solution of (x,y,z) determines a unique solution for (a,b,c), there is a bijection from (x,y,z) to (a,b,c). So, we only need to determine the number of non-negative solutions for (x,y,z). When z=0, x+y=20, there are 21 solutions. When z=1, x+y=19, there are 20 solutions. … When z=20, x+y=0, there are 1 solution. Therefore, there are a total of 21+20+…+1=231 solutions.

Well, this question's difficulty was pretty out of place I feel.

First, begin by noting that the equation $r\sqrt{x} + s\sqrt{y} = t\sqrt{z}$ where $x, y, z, r, s, t \in \mathbb{Z}$ and $x, y, z$ is not divisible by the square of any prime has no solutions for distinct $x, y, z$ . In addition, $\sqrt{2000} = 20\sqrt{5}$ , so a, b, c all have an odd power of $5$ 's. The equation can be henceforth written as $a + b + c = 20$ , and using stars and stripes, stars and bars, sticks and stones, whatever you want to call it, the answer is clearly ${22 \choose 20} = 231$ .

Note that $20\sqrt{5}=\sqrt{2000}$

So the problem is equivalent to find the solutions of x+y+z=20 where x,y,z are nonnegative integers.

Which can be verified to have 1+2+3+...+21=231 solutions.

Theorem: If $x, y,$ and $z$ are nonnegative integers such that $x+y+z=N$ for some nonnegative integer $N,$ then there are $\binom{N+2}{2}$ ordered triples $(a,b,c)$ that satisfy the equality.

Proof: The number of triples we are looking for is the same as the number of arrangements in a row of $N *$ 's and $2 +$ 's. There are $\binom{N+2}{2}$ such arrangements.

Since $\sqrt{2000}=20\sqrt{5}$ , we are looking for $a,b, c$ that can be written as $x\sqrt{x'}, y\sqrt{y'}, z\sqrt{z'},$ respectively, such that $x,y,z$ are nonnegative integers, $x+y+z=20,$ and $x'=y'=z'=\sqrt{5}.$ The theorem above tells us that there are $\binom{22}{2}=231$ such triples.

Since $\sqrt{2000} = 20\sqrt{5}$ then $a, b$ and $c$ must be multiples of 5, then let $a=5x^2, b=5y^2$ and $c=5z^2$ . Taking note that $a,b,c \geq 0$ . Therefore $x+y+z = 20$ , if $x=0$ , then there are $21$ combiations of $y$ and $z$ , if $x=1$ , then thre are $20$ combinations, if $x=2$ , $19$ combinations, ... if $x=20$ , then only $1$ combination, so the total number: $1+2+3+...+21 = 21(21+1)/2 = 231$ .

We know that $\sqrt{2000}=20\sqrt{5}$ . Also, $\sqrt{a}, \sqrt{b}, \sqrt{c} \geq 0$ .

Square roots can only be added and simplified into one term if the number inside the square root of the simplified radical is the same. Hence, since we are adding non-negative terms, the number inside the square root of $\sqrt{a}, \sqrt{b}$ , and $\sqrt{c}$ should be 5.

Let $\sqrt{a}=x\sqrt{5}$ , $\sqrt{b}=y\sqrt{5}$ , and $\sqrt{c}=z\sqrt{5}$ , where $x$ , $y$ , and $z$ are non-negative integers. Thus,

$\sqrt{5}(x+y+z)=20\sqrt{5}$

$x+y+z=20$

Since $x$ , $y$ and $z$ are non-negative, the number of integer solutions to this equation is $\displaystyle\binom{20+2}{2}=231$ .

We are given that $\sqrt{a} +\sqrt{b} +\sqrt{c} =\sqrt{2000}$ . Simplifying, we get $\sqrt{a} +\sqrt{b} +\sqrt{c} = 20\sqrt{5}$ . Since a,b,c are all nonegative integers, we know that $\sqrt{a} , \sqrt{b} , \sqrt{c}$ can be written in the form $A\sqrt{5} , B\sqrt{5} , C\sqrt{5}$ . We know that A+B+C=20, and that they are nonegative integers. There is a bijection between the triples (a,b,c) and the triples (A,B,C) where A+B+C=20. There are (20+2)C2=231 ordered triples (A,B,C) using the stars and bars method, thus our answer is 231.

First off, we note that $\sqrt{2000} = 20 \sqrt{5}$ . Hence we have that $\sqrt{a} + \sqrt{b} + \sqrt{c} = 20 \sqrt{5}$ , or $\sqrt{\frac{a}{5}} + \sqrt{\frac{b}{5}} + \sqrt{\frac{c}{5}} = 20$ .

The only way for a sum of three square roots to be an integer is if all three square roots are integers themselves, that is, all of $\frac{a}{5}, \frac{b}{5}, \frac{c}{5}$ have to be perfect squares.

Hence, the number of ordered triples of integers $(a,b,c)$ is equal to the number of solutions to $x + y + z = 20$ , where $(x,y,z)$ are nonnegative integers. This is easily solved by the standard choose function, with 20 '1's and 2 'dividers', hence the answer is $22 \choose {2}$ $= 231$ .

first we can see $\sqrt{2000}=20\sqrt{5}$ .

then $a$ is a integer,so $\sqrt{a}$ must be a integer or format like $x\sqrt{y}$ , and $x,y$ is integer.

as $\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{2000}=20\sqrt{5}$ , we can know $\sqrt{a}+\sqrt{b}+\sqrt{c}=(x+y+z)\sqrt{5}=20\sqrt{5}$ , $x,y,z$ are integers and $\geq 0$ .

from combination theory, there are $binomial(20+2,2)=22*21/2=231$ kinds of solution to $x+y+z=20$ , $x,y,z$ are integers and $\geq 0$ .

2000=20^2.5 so (2000)^1/2=20(5)^1/2 it meams that all non zero terms on the LHS must have (5)^1/2 in them. so the problem just boils down to the no of ways to choose 3 numbers such that their sum is 20 which is 22 choose 2

First, we notice that $\sqrt{2000} = 20\sqrt{5}$ . Thus, we have, $\sqrt{a} + \sqrt{b} + \sqrt{c} + 20\sqrt{5}$ Therefore, we can see that $a = 5k^2$ , $b = 5l^2$ , $c = 5m^2$ where $k, l, m \in \mathbb{Z}$ . Now we have, $k\sqrt{5} + l\sqrt{5} + m\sqrt{5} = 20\sqrt{5} \Rightarrow k + l + m = 20$ Thus, we need the number of ways for $3$ non-negative integers to sum to $20$ . For this we will use a technique called stars and bars. Consider the follwowing, $\text{★ ★ ★ ★ ★ ★ |★★ ★ ★ |★ ★ ★ ★★ ★ ★ ★ ★ ★ }$ We have $20$ stars and $2$ bars which split the stars into $3$ groups. The number of ways to place the first bar and second bar is $21$ and $22$ respectively. Therefore, there are, $21 \cdot 22 = 462$ ways to place both bars. However, since it doesn't matter which bar is which, we divide by $2$ to get the final answer of, $\frac{462}{2} = \fbox{231}$

$\sqrt{a}$ + $\sqrt{b}$ + $\sqrt{c}$ = $\sqrt{2000}$

and we know that $\sqrt{2000}$ = 20 $\sqrt{5}$ .

Notice that $\sqrt{2000}$ when simplified ends with a $\sqrt{5}$ .

So now we can rewrite the equation into :

$x_{1}$ $\sqrt{5}$ + $x_{2}$ $\sqrt{5}$ + $x_{3}$ $\sqrt{5}$ = 20 $\sqrt{5}$ . where $x_{1}$ , $x_{2}$ and $x_{3}$ are positive integers.

Now we start adding combinations of $\sqrt{5}$ triple that sum-up to 20 $\sqrt{5}$ .

• 20 $\sqrt{5}$ + 0 + 0 = 20 $\sqrt{5}$
• 0 + 20 $\sqrt{5}$ + 0 = 20 $\sqrt{5}$

• 0 + 0 + 20 $\sqrt{5}$ = 20 $\sqrt{5}$

• 19 $\sqrt{5}$ + $\sqrt{5}$ + 0 = 20 $\sqrt{5}$

• 19 $\sqrt{5}$ + 0 + $\sqrt{5}$ = 20 $\sqrt{5}$
• $\sqrt{5}$ + 19 $\sqrt{5}$ + 0 = 20 $\sqrt{5}$
• $\sqrt{5}$ + 0 + 19 $\sqrt{5}$ = 20 $\sqrt{5}$
• 0 + $\sqrt{5}$ + 19 $\sqrt{5}$ = 20 $\sqrt{5}$
• 0 + 19 $\sqrt{5}$ + $\sqrt{5}$ = 20 $\sqrt{5}$

From the above triples, it is clear that for every distinct triple it gives 6 combinations and for every 2 same variable and 1 distinct it gives 3 combinations.

Counting all the possible combinations starting from 20 $\sqrt{5}$ + 0 + 0 = 20 $\sqrt{5}$ up to 7 $\sqrt{5}$ + 7 $\sqrt{5}$ + 6 $\sqrt{5}$ = 20 $\sqrt{5}$ .

Thus, $33 \times 6$ + $11 \times 3$ = $\boxed{231}$

We can write the solution as follows;

sqrt(2000) = 20(sqrt(5)).

Let sqrt(a),sqrt(b), sqrt(c) be x(sqrt(5)), y(sqrt(5)) and z(sqrt(5)).

By putting these assumptions in the question we get,

x + y +z = 20

Then, for finding no. of triplets we will be using = C((n + r - 1),(r - 1))

Here, n = 3 , r = 20.

Therefore, Total triplets = C((22),(19))

                                          = C((22),(2))

                                          = (22 * 21)/2

                                          = 231 (Ans.)

Firstly, prime factorize $2000 = 2^4$ x $5^3$ .

Hence, $\sqrt{2000} = 20\sqrt{5}$ .

Let $a,b,c = 5A^2,5B^2,5C^2$ respectively, where $A,B,C$ are integers satisfying the condition mentioned.

Thus, $\sqrt{a} + \sqrt{b} + \sqrt{c} = 20\sqrt{5}$ ,

$= \sqrt{5A^2} + \sqrt{5B^2} + \sqrt{5C^2} = 20\sqrt{5}$

$= A\sqrt{5} + B\sqrt{5} + C\sqrt{5} = 20\sqrt{5}$

$= (A+B+C)\sqrt{5} = 20\sqrt{5}$

Hence, $A + B + C = 20$

Thus, when $A = 0, B + C = 20$ , where $B = 20, C = 0 ; B = 19, C = 1; ...B = 0, C = 20$

$A = 0$ yield 21 possible solutions of $B$ and $C$

Similarly, when $A = 1$ , there are 20 solutions of $B$ and $C$ from $B = 19, C = 0$ to $B = 0, C= 19$ for $B+C = 19$

The number of triplets for $(A,B,C) = 21 + 20 + 19 ... + 1 = 231$

As each integer of $A,B,C$ corresponds to a unique value of $a,b,c$ respectively, thus, the number of triplets for $(a,b,c) = \boxed{231}$

$\sqrt {a}+\sqrt{b}+\sqrt{c} = \sqrt{2000} = 20\sqrt{5}.$ Thus, $\sqrt {a}$ , $\sqrt {b}$ , and $\sqrt {c}$ must all be of the form $n \sqrt{5}$ , where $n$ is an integer.

Let $a = x\sqrt{5}$ , $b = y\sqrt{5}$ , and $c = z\sqrt{5}$ . Then, $x + y + z = 20$ and we want to count the number of ordered pairs of non-negative integers $(x,y,z)$ that satisfy this. By Stars and Bars, the answer is $\dbinom {22}{20} = \boxed {231}.$

$\sqrt{2000}=20\sqrt5$

Let

$\sqrt a=x\sqrt5$

$\sqrt b=y\sqrt5$

$\sqrt c=z\sqrt5$

Substitute into the equation we have

$x\sqrt5+y\sqrt5+z\sqrt5=20\sqrt5$

$(x+y+z)\sqrt5=20\sqrt5$

$x+y+z=20$

By Stars and Bars , the ordered triples of $(x,y,z)$ is

$22\choose2$

$=231$

Hence, the ordered triples of integers $(a,b,c)$ is also $231$ .

Simplifying the RHS of the equation, we'll get $\sqrt {a} + \sqrt {b} + \sqrt {c} = 20\sqrt {5}$ , which implies, the LHS can be expressed in such way, so let $a = {d}^{2}q, b = {e}^{2}q, c = {f}^{2}q$ , where $q$ is not a perfect square number, thus, we get, by substitution:

$d\sqrt {q} + e\sqrt {q} + f\sqrt {q} = 20\sqrt {5}$

$(d + e + f)\sqrt {q} = 20\sqrt {5}$ which implies, $d + e + f = 20$ and $q = 5$

Now, since $a = {d}^{2}q , b = {e}^{2}q$ and $c = {f}^{2}q$ and $(a , b, c)$ are all non - negative numbers, otherwise it will yield unreal numbers, the number of ordered triples of integer $(a, b, c)$ that will satisfy the condition is equal to the number of the natural number solutions, $(d, e, f)$ .

Thus, ${22} \choose {20}$ $= 231$ solutions

Simply the problem with:

$\sqrt{a}+\sqrt{b}+\sqrt{c}$ =20.\sqrt{5})

suppose that

$a=x^2, x \geq 0$ ,

$b=y^2 , y \geq 0$ ,

$c=z^2 , z \geq 0$ ,

So,

$x+y+z=20$ , where $x,y,z \geq 0$

Solution for non-negatif integer number are:

$C(20+3-1,3-1)=C(22,2)=231$

So, the answer are $231$