How many pairs (a,b)(a,b) of positive integers are there such that a≥ba≥b and 2(15a−−√+15b−−√)2(15a+15b) is an integer?

 Let's analyze the expression \(2\sqrt{15a} + \sqrt{15b}\) where \(a\) and \(b\) are positive integers.

For \(2\sqrt{15a} + \sqrt{15b}\) to be an integer, both terms must contribute to an integer result.

Let \(2\sqrt{15a} = x\) and \(\sqrt{15b} = y\), where \(x\) and \(y\) are integers.

Squaring both sides of the first equation, we get:

\[4 \cdot 15a = x^2 \implies 60a = x^2\]

Now, squaring both sides of the second equation:

\[15b = y^2 \implies 15b = y^2\]

For the expression \(2\sqrt{15a} + \sqrt{15b}\) to be an integer, \(x\) and \(y\) must be integers, and \(a\) and \(b\) must be such that \(60a\) and \(15b\) are perfect squares.

Let's express \(60a\) and \(15b\) as \(2^2 \cdot 3 \cdot 5 \cdot a\) and \(3 \cdot 5 \cdot b\), respectively.

For \(60a\) to be a perfect square, the powers of 2, 3, and 5 in its prime factorization must be even. Therefore, \(a\) must be in the form \(2^2 \cdot 3^2 \cdot 5^2 \cdot \ldots\).

For \(15b\) to be a perfect square, the powers of 3 and 5 in its prime factorization must be even. Therefore, \(b\) must be in the form \(3^2 \cdot 5^2 \cdot \ldots\).

Now, let's count the pairs of positive integers \(a\) and \(b\):

For \(a\), there are \(3\) choices for the power of 2 (\(2^0\), \(2^2\), \(2^4\)), \(3\) choices for the power of 3 (\(3^0\), \(3^2\), \(3^4\)), and \(3\) choices for the power of 5 (\(5^0\), \(5^2\), \(5^4\)).

For \(b\), there are \(3\) choices for the power of 3 (\(3^0\), \(3^2\), \(3^4\)), and \(3\) choices for the power of 5 (\(5^0\), \(5^2\), \(5^4\)).

The total number of pairs \((a, b)\) is \(3 \times 3 = 9\).