Square Root Sum

$\sqrt{x} +\sqrt{y} = \sqrt{z}$

square both sides

$x+y+2 \sqrt{xy} = z$

since $1 \leq z \leq 25$ ,

therefore $1 \leq x+y+2 \sqrt{xy} \leq 25$ where $2 \sqrt{xy} >0$

We also know that $2 \sqrt{xy}$ must be integer because on the Right Hand Side only $z$ exists, which is an integer.

Without Loss of Generality, $x \leq y$ , so the list of possible values of $(x,y)$ are $(1,1), (1,4), (1,9), (1,16), (2,2), (2,8), (2,18), (3,3), (3,12), (4,4), (4,9), (4,16), (5,5), (6,6), ... , (k,k), ... , (12,12)$ without considering the value of $z$ .

By substituting each possible $(x,y)$ values we get that only $(1,1), (1,4), (1,9), (1,16), (2,2), (2,8), (3,3), (4,4), (4,9), (5,5), (6,6)$ fulfill the conditions, counted with multiplicity. So, there are $16$ possible triples $(x,y,z)$ .

Ít's easy to prove that zy must be a perfect square and z>x,y. Then z cannot be a prime number or equal pq ( p,q are distinct prime numbers) or else y >= z. Therefore z can only be {25,24,20,28,16,12,9,8,4} z=25 gives (x,y)=(16,1),(9,4),(4,9) or (1,16) z=24 gives (x,y)=(6,6) z=20 gives (x,y)=(5,5) z=18 gives (x,y)=(2,8),(8,2) z=16 gives (x,y)=(9,1),(4,4),(1,9) z=12 gives (x,y)=(3,3) z=9 gives (x,y)=(4,1),(1,4) z=8 gives (x,y)=(2,2) z=4 gives (x,y)=(1,1) Hence the answer is 16

$1 \leq x, y, z \leq 25 \Rightarrow z \geq 2$

Let z = $k \times m^2$ , with k and m are 2 positive integers, k is not divisble by any square number greater than 1

$\sqrt{x} + \sqrt{y} = \sqrt{z}$

$\Rightarrow \sqrt{x} = m\sqrt{k} - \sqrt{y}$

$\Rightarrow x = km^2- 2m\sqrt{k*y} + y$

So, $ky$ must be a square, so y can be written as $y = k \times n^2$ , n is a positive integer. Similarly, we can prove that $x=k \times p^2$ , p is a positive integer

Now we only have to find the number of positive integer triples $m,n,p$ such that $\sqrt{m^2} + \sqrt{p^2} = \sqrt{n^2}$ or $m + p =n$ with $1 \leq m, n ,p \leq 5$

• If n = 5, there are 4 solutions
• If n = 4, there are 3 solutions
• If n = 3, there are 2 solutions
• If n = 2, there is 1 solution
• n can't be 1

We have:

• If n = 5 or n =4, k is always 1
• If n = 3, k can be 1 or 2
• If n = 2, k can be 1 or 3 or 4 or 5 or 6 ( k can't be 2 because the solution has been counted in the case n =4

Then, the answer is: $1*4 + 1*3 + 2*2 + 5*1 = 16$

There can be 2 cases:

1. √z is rational

2. √z is irrational

Case 1)

We can prove that if √z is an integer then (√x,√y) must also be integers.

Suppose , k is an integer and z = k^2

Squaring gives:

x^2 + y^2 + 2 √xy = k^2 .

This means √xy = p^2 , where p is an integer

so, y = p^2 / x

thus, √x + p/√x = k

So x and hence y must be perfect squares.

So for case 1 , we can easily find the triples of solutions: the possible values of z are (25, 16 , 9 , 4 ) as these are perfect squares .

[Note : I have left √y to meet up the difference.]

when , z = 25 , √x +√y = 5 , √x = {1,2,3,4}, 4 solutions.

when , z = 16 , √x +√y = 4 , √x = {1,2,3} , 3 solutions .

when , z = 9 , √x +√y = 3 , √x = {1,2} , 2 solutions .

when , z = 4 , √x +√y = 2 , √x = {1} , 1 solution .

so,10 solutions in case 1

Case 2

If √z is irrational. Then none of √x or √y is rational. We can show this by an argument similar from case 1. So x and y are not squares.

F(√z) has degree of extension 2 over F . In this case, as √z = √x + √y ,√x + √y must also have degree of extension 2 over F .

Suppose that x = (A^2)m and y = (B^2)n where m , n are square free and m,n >1 .

assume ,√z = r = √x + √y = A√m + B√n . m and n cannot be equal unless . A = B = 0 . so, assume m and n are unequal . so,F(r) must contain both √m and √n . But then the degree of extension of F(r) over is at least 3 , when m and n are unequal. But since the degree is known to be 2 so m and n must be equal.

value of z has to be such that √z = C√q , C,q > 0

so , the possible value of √z = { √24=2√6 ,√20=2√5,√18=3√2,√12=2√3 , √8=2√2 }

[Note : I have again left √y to meet up the difference.]

when , √z = 2√6 = √x +√y ,√x = { 1√6 }

when , √z = 2√5 = √x +√y ,√x = { 1√5 }

when , √z = 3√2 = √x +√y ,√x = { 1√2 ,2√2 }

when , √z = 2√3 = √x +√y ,√x = { 1√3 }

when , √z = 2√2 = √x +√y ,√x = { 1√2 }

so, 6 solutions in case 2

so , case 1 + case 2 = 10 + 6 = 16

root 24 = 2 root 6 . root 20 = 2 root 5 . root 18 = 3 root 2 . root 16 = 2 root 4 . root 12 = 2 root 3 . root 8 = 2 root 2 . As root 24 = root 6 + root 6 . Similarly no.s 20 , 12 , 8 , 24 will have one - one solution each . But 18 will have 2 solutions as root 18 = root 2 + 2 root 2 . and also 18 = 2 root 2 +root 2 . ..Now counting for perfect squares 1,4,9,16,25 . 1 will have no solution . 4 will have 1 solution . As root 4 = root 1 +root 1 . 9 will have two solutions . As root 9 = root 1 +root 4 or root 4 + root 1 . 16 will have 3 solutions . As root 16 = root 9 + root 1 or root 1 + root 9 or root 4 + root 4 . 25 will have 4 solutions . As root 25 = root 1 + root 16 or root 16 + root 1 or root 4 + root 9 . or root 9 + root 4 . Hence in total we have . 1+1+1+1+2+1+2+3+4 = 16 solutions.

First, I assume $x \leq y \leq z$ .

My attempt here is to find the upper bound of $x$ . If $x \geq 7$ , $LHS \geq 2 \sqrt{7} = \sqrt{28} > \sqrt{25}$ , since $z \leq 25$ , we have have no solution with $x \geq 7$ .

So values of $x$ must be in the range of $[1, 6]$

Now there's two cases. Either $x=y$ or $x \neq y$

If $x = y$ .

Suppose $x=1 \rightarrow LHS = 2 = \sqrt{4} \rightarrow (1,1,4)$ is a solution

Suppose $x=2 \rightarrow LHS = 2 \sqrt{2} = \sqrt{8} = RHS \rightarrow (2,2,8)$ is a solution

Suppose $x=3 \rightarrow LHS = 2 \sqrt{3} = \sqrt{12} = RHS \rightarrow (3,3,12)$ is a solution

Suppose $x=4 \rightarrow LHS = 2 \sqrt{4} = \sqrt{16} = RHS \rightarrow (4,4,16)$ is a solution

Suppose $x=5 \rightarrow LHS = 2 \sqrt{5} = \sqrt{20} = RHS \rightarrow (5,5,20)$ is a solution

Suppose $x=6 \rightarrow LHS = 2 \sqrt{6} = \sqrt{24} = RHS \rightarrow (6,6,24)$ is a solution

Which is a total of $6$ solution

If $x \neq y$ , then all $x,y,z$ are perfect squares, Knowing that $1+2=3,1+3=4,1+4=5,2+3=5$ , note that we only consider $RHS \leq \sqrt{25} =5$ , we have

$\sqrt{1} + \sqrt{4} = \sqrt{9}, \sqrt{1} + \sqrt{9} = \sqrt{16}, \sqrt{1} + \sqrt{16} = \sqrt{25}, \sqrt{4} + \sqrt{9} = \sqrt{25}$

Which is a total of $4$ solution. However, when we relax the constraint, $x \leq y$ , values of $x$ and $y$ are interexchangeable. So the solution here is doubled: $8$ solution.

Lastly, note that if $(x,y,z)$ is a solution, then $(ax, ay, az)$ is a solution as well, where $a$ is a positive integer. Thus since $(1,4,9)$ and $(4,1,9)$ are solutions, then $(2,8,18)$ and $(8,2,18)$ are solutions as well. We need not consider other triples because any multiple of their value of $z$ is greater than $25$ .

Hence, this gives a total of $6 + 8 + 2 = 16$ solution.

Let \­(\sqrt{x}=a\sqrt{n}\­) \­( \sqrt{y}=b\sqrt{n} \­) \­( \sqrt{z}=c\sqrt{n} \­)

We now have reduced the problem to finding possible values of \­( a,b,c \­) that work for each \­( n \­).

Case \­( n=1 \­) This implies that \­( \sqrt{1} \geq a,b,c \geq \sqrt{25} \­), resulting in \­( 10 \­) possible outcomes of \­( a+b=c \­).

Case \­( n=2 \­) \­( \sqrt{\frac{1}{2}} \geq a,b,c \geq \sqrt{\frac{25}{2}} \implies 1 \geq a,b,c \geq 3 \­), which results in \­( 3 \­) outcomes. (1+1=2, 1+2=3, 2+1=3)

We continue doing this for higher values of \­( n \­) with no perfect square factor (because if it had such a factor, it would have already been counted by a previous case) and find that when \­( n=3,5,6 \­), there are \­( 1,1,1 \­) outcomes, respectively.

The sum of the outcomes gives the answer, namely \­( 10+3+1+1+1=\boxed{16} \­).

We only consider values for z where $\sqrt{z}$ is an integer or if it is of the form $a\sqrt{b}$ , where a is prime and b is not the square of a prime. When $\sqrt{z}$ is not of one of these forms, it is impossible for it to be of the form $\sqrt{x}+\sqrt{y}$ .

We first consider the case when $\sqrt{z}$ is of the form $2\sqrt{n}$ , where n is not the square of a prime. The values for z when this is true are 8, 12, 20, and 24. In this case, the values for x and y must both be n, so there are a total of 4 ordered triples for this case.

Next, we consider when $\sqrt{z}$ is of the form $a\sqrt{n}$ , where a > 2 and n is not the square of a prime. There is only one such number z, 18, and x and y must equal 2 and 8. Since we can order the 2 and 8 two ways, there are 2 ordered triples for this case.

Lastly, we consider the case where z is a perfect square. The values for z can be 4, 9, 16, and 25. 1 is not included since it is impossible for there to be x and y which match the conditions. x and y must be perfect squares for their square roots to add up to z, which is an integer. We find pairs of values which add up to $\sqrt{z}$ . There is 1 pair for z=4, 1 pair for z=9, 2 pairs for z=16, and 2 pairs for z=25. Since we can order each pair two ways when the x and y values are unique, there are a total of 10 ordered triples for this case (the pair (x, y) for z=4 is (1,1) and one of the pairs for z=16 is (4,4), and since the x and y values are the same, we can order them only one way).

We add up the number of pairs from each of the cases, getting a total of 4+2+10= 16 ordered triples.

Squaring both sides, we have $2 \sqrt{xy} = z-x-y$ . Rearranging and squaring, we have $2\sqrt{yz}=z+y-x$ and $2\sqrt{xz}=x+y-z$ . Since $xy, yz$ and $xz$ are integers and $\sqrt{xy}, \sqrt{yz}$ and $\sqrt{yz}$ are rational numbers, this implies that $\sqrt{xz}, \sqrt{yz}$ and $\sqrt{xz}$ must be integers (proof in given here ).

Let $xy = A^2$ , $yz = B^2$ and $xz=C^2$ for some positive integers $A, B$ and $C$ . Let $x=x_1 ^2 \times x_2$ , where $x_2$ is not divisible by the square of any prime. Similarly, we let $y = y_1^2 \times y_2$ and $z = z_1^2\times z_2$ . Then $xy = x_1^2y_1^2x_2y_2 = A^2$ , $yz = y_1 ^2 z_1 ^2 y_2z_2$ and $xz = x_1^2z_1^2x_2z_2 = C^2$ . Since $x_2$ , $y_2$ and $z_2$ are not divisible by the square of any prime they must all be equal. Thus $(x, y, z) =\left( x_1 ^2 d, y_1^2 d, (x_1+y_1)^2 d \right)$ , where $d$ is not divisible by the square of any prime.

For $d=1$ , $5 \geq \sqrt{z} = x_1+y_1$ . This yields $(x,y,z) = (1,1,4)$ , $(1,4,9)$ , $(1,9,16)$ , $(1,16,25)$ , $(4,1,9)$ , $(4,4,16)$ , $(4,9,25)$ , $(9,1,16)$ , $(9,4,25)$ , $(16,1,25)$ for a total of 10 solutions.

For $d=2$ , $5 \geq \sqrt{z} = \sqrt{2} x_1+ \sqrt{2} y_1 \Rightarrow 4 > x_1 + y_1$ . This yields $(2, 2, 8), (2, 8, 18), (8, 2, 18)$ for a total of 3 solutions.

For $d=3$ , $5 \geq \sqrt{z} = \sqrt{3} x_1 + \sqrt{3} y_1 \Rightarrow 3 > x_1 + y_1$ . This yields $(3, 3, 12)$ as the only solution.

( $d=4$ is ignored since $d$ is not divisible by the square of any prime.)

For $d=5$ , $5 \geq \sqrt{z} = \sqrt{5} x_1 + \sqrt{5} y_1 \Rightarrow 3 > x_1 + y_1$ . This yields $(5, 5, 20)$ as the only solution.

For $d =6$ , $5 \geq \sqrt{z} = \sqrt{6} x_1 + \sqrt{6} y_1 \Rightarrow 3 > x_1 + y_1$ . This yields $(6, 6, 24)$ as the only solution.

For $d\geq 7$ , $5 \geq \sqrt{z} = \sqrt{7} x_1 + \sqrt{7} y_1$ . Since $\sqrt{6.25}=2.5$ , we know that $\frac {5}{\sqrt{7}} < 2$ . Thus $2 > x_1 + y_1$ . Since $x_1 \geq 1$ and $y_1\geq 1$ , we have no further solutions.

Thus, the total number of solutions is $10+3+1+1+1 = 16$ .

We know that $\sqrt{x} + \sqrt{y} = \sqrt{z}$ with $(x, y, z) \in [1, 25]$ .

Therefore, $z - (x+y) = 2\sqrt{xy}$ .

It is obvious that z - (x+y) is a positive integer. Therefore $xy \in N^{2}$ with $(x+y) < 25$ .

Let $S$ = { $1, 4, 9, 16, 25, 36$ }, $xy \in S$ .

By looking at the 6 cases we got

$(x,y)$ = { $(1, 1), (1, 4), (2, 2), (1, 9), (3, 3), (1, 16), (2, 8), (4, 4), (5, 5), (4, 9), (6, 6)$ } and its permutations.

Therefore we got 16 positive integers $(x, y, z)$ satisfying the above equality.

i get the answer by record it one by one

1,1,4 1,4,9 4,1,9 2,2,8 1,9,16 9,1,16 3,3,12 1,16,25 16,1,25 2,8,18 8,2,18 4,4,16 5,5,20 4,9,25 9,4,25 6,6,24