How many ordered positive integer triples (x,y,z), such that 1 ≤ x,y,z ≤ 25, are there to √x+√y=√z?
16
To find the number of ordered positive integer triples (x, y, z) that satisfy the equation √x + √y = √z, we can use a systematic approach.
Step 1: Start by listing all the possible values of x, y, and z. Since 1 ≤ x, y, z ≤ 25, we can create a table to keep track of the values.
x y z
--------------
1 ? ?
2 ? ?
3 ? ?
... ... ...
25 ? ?
Step 2: Now, let's fill in the table by solving the equation √x + √y = √z. From the given equation, we can square both sides to eliminate the square roots. So, we have x + 2√xy + y = z.
For each value of x from 1 to 25, we can substitute it into the equation and solve for y and z.
Let's take x=1 as an example:
1 + 2√1y + y = z
1 + 2√y + y = z
3√y + y = z - 1
Now, let's perform the same process for all values of x from 1 to 25 and fill in the table.
x y z
----------------------------------
1 y z (from 1st equation)
2 y z (from 2nd equation)
3 y z (from 3rd equation)
... ... ...
25 y z (from 25th equation)
Step 3: Count the number of potential solutions that meet the given constraints 1 ≤ x, y, z ≤ 25.
In this case, we don't need to solve every equation since the possible values of the square roots (√x, √y, √z) are integers only when x, y, and z are perfect squares. We can see that there are 5 perfect squares between 1 and 25: 1, 4, 9, 16, and 25.
Now, let's consider each case:
Case 1: x = 1
We have 3√y + y = z - 1. Since y and z can take integer values from 1 to 25, we'll compute the number of solutions.
Case 2: x = 4
We have 2√y + y = z - 4. Similar to Case 1, we'll compute the number of solutions.
Case 3: x = 9
We have √y + y = z - 9. Compute the number of solutions.
Case 4: x = 16
√y + 4√y + y = z - 16. Compute the number of solutions.
Case 5: x = 25
√y + 5√y + y = z - 25. Compute the number of solutions.
Finally, you need to sum up the number of solutions from all the cases to get the total number of ordered positive integer triples (x, y, z) that satisfy the equation √x + √y = √z.