what is the number of all positive integer solutions [x,y] of the equation x^2+y^2=2010 [] is not G.I.F

To find the number of all positive integer solutions \((x, y)\) for the equation \(x^2 + y^2 = 2010\), we can approach the problem step by step. The first thing to note is that both \(x\) and \(y\) must be positive integers, meaning they are whole numbers greater than zero.

Understanding the Equation

The equation \(x^2 + y^2 = 2010\) represents a circle in the coordinate plane with a radius of \(\sqrt{2010}\). However, since we are only interested in positive integer solutions, we need to find pairs \((x, y)\) such that both values are integers and satisfy the equation.

Finding the Range of Possible Values

First, let's determine the maximum possible value for \(x\) and \(y\). Since both \(x^2\) and \(y^2\) are non-negative, we can deduce that:

• Both \(x\) and \(y\) must be less than or equal to \(\sqrt{2010}\).

Calculating \(\sqrt{2010}\) gives approximately \(44.83\). Therefore, the maximum integer values for \(x\) and \(y\) are \(44\).

Checking Possible Values

Next, we will check each integer value of \(x\) from \(1\) to \(44\) and see if \(y^2 = 2010 - x^2\) results in a perfect square. If it does, then \(y\) will also be a positive integer.

Iterating Through Values

We can iterate through values of \(x\) and calculate \(y^2\) as follows:

• For \(x = 1\): \(y^2 = 2010 - 1^2 = 2009\) (not a perfect square)
• For \(x = 2\): \(y^2 = 2010 - 2^2 = 2006\) (not a perfect square)
• For \(x = 3\): \(y^2 = 2010 - 3^2 = 2001\) (not a perfect square)
• For \(x = 4\): \(y^2 = 2010 - 4^2 = 1994\) (not a perfect square)
• Continue this process up to \(x = 44\).

Identifying Perfect Squares

After checking each integer value, we find that:

• For \(x = 9\): \(y^2 = 2010 - 9^2 = 2010 - 81 = 1929\) (not a perfect square)
• For \(x = 27\): \(y^2 = 2010 - 27^2 = 2010 - 729 = 1281\) (not a perfect square)
• For \(x = 36\): \(y^2 = 2010 - 36^2 = 2010 - 1296 = 714\) (not a perfect square)
• For \(x = 39\): \(y^2 = 2010 - 39^2 = 2010 - 1521 = 489\) (not a perfect square)
• For \(x = 44\): \(y^2 = 2010 - 44^2 = 2010 - 1936 = 74\) (not a perfect square)

Final Count of Solutions

After checking all values from \(1\) to \(44\), we find that none of the calculations yield a perfect square for \(y^2\). Therefore, there are no positive integer solutions \((x, y)\) that satisfy the equation \(x^2 + y^2 = 2010\).

In summary, the number of all positive integer solutions \((x, y)\) for the equation \(x^2 + y^2 = 2010\) is 0.