How many pairs $(a, b)$ of positive integers are there such that $a \le b$ and $ab =$ $4^{2017}$?

We need to find pairs $(a,b)$ of positive integers where $a \le b$ and $ab = 4^{2017}$.

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Since $ab = 4^{2017}$, we can rewrite this using prime factorization:

$4^{2017} = (2^2)^{2017} = 2^{4034}$

So our constraint becomes: $ab = 2^{4034}$

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Since $a$ and $b$ are positive integers whose product is $2^{4034}$, both $a$ and $b$ must be powers of 2.

Because $2^{4034}$ has only one prime factor: the number 2. If $a$ had any other prime factor, then $b$ would need to "cancel it out" to make $ab = 2^{4034}$, but that's impossible since $b$ is also a positive integer.

So we can write: $a = 2^i$ and $b = 2^j$ where $i,j \geq 0$ are non-negative integers.

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From $ab = 2^{4034}$:

$2^i \times 2^j = 2^{4034}$

$2^{i+j} = 2^{4034}$

Therefore: $i + j = 4034$

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Since $a \le b$, we have:

$2^i \le 2^j$

This means: $i \le j$

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From $i + j = 4034$, we can express $j = 4034 - i$.

Substituting into $i \le j$:

$i \le 4034 - i$

$2i \le 4034$

$i \le 2017$

Since $i \geq 0$ (as $a$ is a positive integer), we have: $0 \le i \le 2017$

This gives us exactly $2017 - 0 + 1 = 2018$ possible values for $i$.

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Therefore, there are $2018$ pairs $(a,b)$ of positive integers such that $a \le b$ and $ab = 4^{2017}$.