Square Root Sums

Since $\sqrt{432} - \sqrt{a} = \sqrt{b}$ , by squaring both sides we have $432 + a - 2\sqrt{432a} = b$ . As $432 + a - b$ is an integer, $2\sqrt{432a}$ is an integer. Hence $432a = 12^2 \cdot 3a = k^2$ for some integer k. From this, we conclude that a is of the form $3m^2$ , and also $a \leq 432$ . In this case, $m \leq 12$ .

Similarily, we can conclude that $b = 3n^2$ , and thus we have to solve $m\sqrt{3} + n \sqrt{3} = 12 \sqrt{3}$ . This gives 13 integer solutions for (a, b) from m = 0 to 12.

[Slight edits for clarity - Calvin]

Note that $\sqrt{432} = 12\sqrt{3}$

For the square root of two integers, to add up to $12\sqrt{3}$ , the square root of the integers must be in the form $x\sqrt{3}$ and $y\sqrt{3}$ .

Let $\sqrt{a}=x\sqrt{3}$ and $\sqrt{b} = y\sqrt{3}$ , then the original equation becomes $x\sqrt{3}+y\sqrt{3} = 12\sqrt{3}\Longrightarrow x+y =12$ . Since x and y must be nonnegative, there are $\boxed{13}$ solutions. $(x,y) = (0,12),(1,11)(2,10)....(11,1),(12,0)$ For every solution pair $(x,y)$ , the pair of integers $(3x^2, 3y^2)$ satisfies the original equation.

Note that the prime factorization of 432 is 2^4*3^3 and that a, b \geq 0

We start by moving \sqrt{b} to the RHS and squaring both sides of the equation. This gives us a=432+b-2*\sqrt{432b}

Since a, b, 432 are all integers, we find that \sqrt{432b} must be a perfect square; hence b must be some multiple of 3. Note that the same process can be applied to a as well, so a, b are both multiples of 3. Call a and b 3x and 3y respectively, so that the new equation becomes \sqrt{3x} + \sqrt{3y} = \sqrt{432}. Dividing by \sqrt{3}, we find that \sqrt{x} + \sqrt{y} = 12. Since every ordered pair (x, y) maps to a unique (a, b), it suffices to find the solutions to our new equation.

Since sqrt{x} + sqrt{y} sum to an integer, it's clear that x and y must individually be perfect squares as well. And, because a, b \geq 0, so must x and y. The possible ordered pairs, then are (0, 12), (1, 12), \ldots , (12, 0). Thirteen solutions.

Claim: If $\sqrt{N}$ is rational then $N$ must be a perfect square.

Proof: Let $\frac{a}{b} = \sqrt{N}$ for co-prime integers $a$ and $b$ . So we have $b^2N = a^2$ . Suppose a prime $p$ divides $b$ then it must divide $b^2N$ thus it must divide $a$ . Since $a$ and $b$ are co-prime, there is no prime $p$ that divides both $a$ and $b$ . Hence no prime $p$ divides $b$ which implies that $b=1$ is the only possibility. Therefore $N = a^2$ must be a perfect square.

Squaring both sides gives $a + b + 2 \sqrt{ab} = 432$ . Since $\sqrt{ab} = \frac {432-a-b}{2}$ is rational, therefore $ab$ must be a perfect square by the above claim. Let $a = A^2 x$ and $b = B^2 y$ , where $x$ and $y$ are the product of distinct prime factors (not divisible by the square of any prime). Since $ab$ is a perfect square, then $ab = A^2B^2xy$ must be a perfect square. As such, any prime $p$ that divides x must divide $y$ and vice versa. This shows that $x = y$ . Hence, $(A+B) \sqrt{ x } = \sqrt{432} = 12\sqrt{3}$ , which shows that $x = 3$ (since $x$ is not divisible by a square) and $A+B =12$ . Thus, $A+B = 12$ has unique integer solutions, for all integers from $A=0$ to $A=12$ , giving $12-0+1 = 13$ total solutions.

Let's assume we know a value of $a$ that is part of a solution. Then, we can conclude that $b = 432 + a - \sqrt{1728a}$ . The solution will be valid for all integers $a$ such that $0 \le a \le 432$ and $1728a = k^2$ for some $k \in N$ .

The prime factorisation of the coefficient is $1728 = 2^6 * 3^3$ . Therefore, the integers we take for $a$ must be of the form $a = 3 * x$ , where $x$ is a complete square.

Since $0 \leq a \leq 432$ , it is implied that $0 \leq x \leq 432 / 3 = 144$ . As $144 = 12^2$ , we can conclude that there are exactly $\boxed{13}$ possible values for $a$ , and hence for $(a,b)$ as well.

Firstly, if we square the equation we will get $a+b+2\sqrt{ab} = 432$

Let $d = gcd(a,b), a = dp, b = dq$ where $gcd(p,q) = 1$ . Substituting this to above implies $d(p+q+2\sqrt{pq}) = 432$ so $pq$ must be a perfect square and since they are relatively prime, each of them is a perfect square, i.e. $p = k^2,q=l^2$

Firstly separate the case for $k=0$ and $l=0$ , we have 2 solutions. Now we can consider $k,l$ as positive integers

We have $d(k+l)^2 =432 = 2^4*3^3$ , so we consider all possible choices of $(k+l)^2$ which is $1,3^2,2^2,2^2*3^2,2^4,2^4*3^2$ (value of $d$ will automatically follow), and for each of them count all the possible $k,l)$ relatively prime to each other.

For example if $k+l = 2^2*3^2$ then $k+l = 2*3 = 6$ , so we have $(k,l) = \{(1,5),(5,1)\}$ and so there are 2 solutions to this case.

In total we have 13 solutions.

Simplify sqrt(432). This is 12sqrt(3). Since a and b are non negative integers, we know that sqrt(3) has to be part of the terms sqrt(a) and sqrt(b). Thus, we have sqrt(3a) + sqrt(3b) = 12sqrt(3). Dividing sqrt(3) we get a +b = 12. Therefore, we just have to find all the integer pairs which make this statement true and that is 13.

I started by simplifying sqrt(432) to 12sqrt(3). then because a and b are both integers, the only way for sqrt(a) and sqrt(b) to add up to 12 sqrt(3), they must both have a constant times sqrt(3). This gives us 13 possible solutions because sqrt(a) can equal to any number from 0 to 12 times sqrt(3) and sqrt(b) would just have to match sqrt(a) so that they added up to twelve.

Note that $\sqrt{432}=12\sqrt{3}$ . Hence we must find integers $a$ and $b$ such that $\sqrt{a}+\sqrt{b}=12\sqrt{3}$ . Next, we realize that $\sqrt{a}+\sqrt{b}$ can only be combined into a single radical (in this case $\sqrt{432}$ ) if $a$ and $b$ can be represented in the forms $3m^2$ and $3n^2$ respectively, where $m$ and $n$ are not necessarily distinct positive integers. This is due to the fact that square roots are irrational, and as a result only a few pairs of radicals can be combined to form a single radical. As a result, we will have $\sqrt{a}+\sqrt{b}=\sqrt{3m^2}+\sqrt{3n^2}=\sqrt{3}(m+n)$ , which clearly has the same form as the RHS of the equation. In fact, since this expression is equal to $12\sqrt{3}$ , we have $m+n=12$ . Finally, we note that $m$ and $n$ have to both be positive integers. To prove this, suppose WLOG that $m$ is negative. Then as a result we must have $n>12$ , so $\sqrt{3m^2}+\sqrt{3n^2}>\sqrt{3m^2}+\sqrt{432}>\sqrt{432}$ , contradiction. Therefore, since $m$ and $n$ have to be positive, the equation has $13$ solutions in integers: $(m,n)=(0,12),(1,11),(2,10),\cdots,(12,0)$ .

$\sqrt{a} + \sqrt{b} = \sqrt{432}$ and $\sqrt{432} = 12\sqrt{3}$

We can say a = 3 $A^2$ and b = 3 $B^2$ a and b are integers, the sum of their roots is irrational so: A $\sqrt{3}$ + B $\sqrt{3}$ = 12 $\sqrt{3}$

A + B = 12 The non negative solutions are: 6,6; 5,7; 7,5; 4,8; 8,4; 3,9; 9,3; 2,10; 10,2; 1,11; 11,1; 0,12; 12,0.

We have 13 Solutions.

Find the largest non-negative 'a' and 'b' For above question!!

sqrt(432) = 12sqrt(3) sqrt(a)+sqrt(b) = sqrt(c) implies a,b must have common irrational term implies 12sqrt(3) = 0+12 = 1+11 = .... 12+0

√a+√b=√432 √a+√b=12√3 from 0+12√3=12√3 √3+11√3=12√3 2√3+10√3=12√3 . . . 12√3+√3=12√3 so all together there is 13 pairs of integers (a,b)

(a^0.5)+(b^0.5)=(432)^0.5 (432)^0.5=12 (3)^0.5 therefore the no.of ordered pairs should sum up to 12 (3)^0.5 by this,from (0,12 (3)^0.5).................................to (12 (3)^0.5,0),there are 13 ordered pairs.

Note that $\sqrt{432} = 12 \sqrt{3}.$

So we have $\sqrt{a} + \sqrt{b} = 12 \sqrt{3}.$

For the above expression to happen, $\sqrt{a} = n \sqrt{3}$ and $\sqrt{b} = m \sqrt{3},$ for some $n,m \in \mathbb{Z}^{+} \cup \{0\}.$ (Also, $n,m$ can never be both zero otherwise the equality won't hold.)

Hence, there are only 13 possible ordered pairs $(n,m)$ such that $n \sqrt{3} + m \sqrt{3} = 12 \sqrt{3}.$

Since $\sqrt{a} = n \sqrt{3}$ and $\sqrt{b} = m \sqrt{3},$ it follows that there are 13 possible ordered pairs $(a,b)$ such that $\sqrt{a} + \sqrt{b} = 12 \sqrt{3} = \sqrt{432}.$

First, we simplify $\sqrt{432}$ to $\sqrt{144 \cdot 3} = 12\sqrt{3}$ . The fact that $a$ and $b$ are integers limits our possibilities greatly. What we essentially need to do is choose $a$ and $b$ so that they simplify to $m\sqrt{3}$ and $n\sqrt{3}$ where $m + n = 12$ . We start with $(m, n) = (0, 12)$ , and then continue by increasing $m$ by $1$ and decreasing $n$ by $1$ each time up to $(m, n) = (12, 0)$ . That gives us a total of $12 - 0 + 1 = \boxed{13}$ possible ordered pairs of $(a, b)$ .

The \sqrt{432} can be simplified to 12\sqrt{3}, and as such, you know you will be working with multiples of \sqrt{3}. You can replace \sqrt{ a } with x \sqrt{3} and \sqrt{ b } with y \sqrt{3}. Dividing both sides by \sqrt{3} yields the equation x + y =12. Since you are not solving for the actual values of a and b , you can simply find the number of ordered pairs of positive (as the original requires square roots without complex numbers) integers that proves the equation x + y =12. This should result in 13 ordered pairs.

Firstly, √ 432 is simplified till it is irreducible.

So, √ 432 = 12√ 3

It given that √ a & √ b are integers, then inevitably (a) & (b) are non-negative integers.(i.e. from within a square root negative outcomes are not possible unless the case of imaginary number comes) (reason 1).

Now, 12√ 3 can be expressed as: x√3 + y√3 =√3(x + y)

Matching, 12√ 3 with (1), we can say that (x + y) is equal to 12, specifying that (x) and (y) are non-negative integers. (again using reason 1).

Now, we need to find the number of such pairs which sum up to 12.

Conditions applied while choosing:

(x) and (y) are non-negative integers

From the above , both of them can have maximum value of 12, exceeding 12 violates the above .

Repetition of pairs are not allowed .

Using formula, C(N+R-1, R-1) (where N is number of objects or specimens , R is the number of specimens taken at a time, and it is valid for specimens <= 0)

Here, N = 13, R = 2.

Solving , we get: [(N+R-1)!]÷[(R-1)!(N+R-1-R+1)! ]

                                 = 13

Now, 13 pairs are obtained which are: (0,12);(1,11);(2,10);(3,9);(4,8);(5,7);(6,6);(7,5);(8,4);(9,3);(10,2);(11,1);(12,0)

Putting x√3=a & y√3=b, desired values are obtained if required.