How many different pairs $(a, b)$ of positive integers are there such that $a ≤ b$ and $\frac{1}{a} + \frac{1}{b} = \frac{1}{9}$

We need to find all pairs $(a, b)$ of positive integers where $a ≤ b$ that satisfy the equation $\frac{1}{a} + \frac{1}{b} = \frac{1}{9}$.

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Working with fractions is messy, so let's clear them by finding a common denominator.

$9ab \cdot \frac{1}{a} + 9ab \cdot \frac{1}{b} = 9ab \cdot \frac{1}{9}$

This gives us:

$9b + 9a = ab$

$ab - 9a - 9b = 0$

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Here's where we use a clever algebraic technique called Simon's Favorite Factoring Trick.

When we have $ab - 9a - 9b$, we can force it to factor by adding and subtracting the same constant.

$ab - 9a - 9b + 81 = 81$

We need $(9)(9) = 81$ to make the left side factorable.

The left side factors as:

$a(b - 9) - 9(b - 9) = (a - 9)(b - 9)$

So we get: $(a - 9)(b - 9) = 81$

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Now we need to find all ways to write 81 as a product of two integers.

First, let's find the prime factorization: $81 = 3^4$

The positive divisors of 81 are: $1, 3, 9, 27, 81$

So the factor pairs $(x, y)$ where $xy = 81$ are:

$(1, 81)$

$(3, 27)$

$(9, 9)$

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Since $(a - 9)(b - 9) = 81$, we have:

$a - 9 = x$ and $b - 9 = y$ for each factor pair $(x, y)$

Therefore: $a = x + 9$ and $b = y + 9$

Factor Pair (x, y)	x + 9 =a	y + 9=b	Solution (a, b)
$(1, 81)$	$1 + 9 = 10$	$81 + 9 = 90$	$(10, 90)$
$(3, 27)$	$3 + 9 = 12$	$27 + 9 = 36$	$(12, 36)$
$(9, 9)$	$9 + 9 = 18$	$9 + 9 = 18$	$(18, 18)$

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All our solutions satisfy $a ≤ b$:

$(10, 90)$: $10 ≤ 90$

$(12, 36)$: $12 ≤ 36$

$(18, 18)$: $18 ≤ 18$

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There are 3 different pairs $(a, b)$ of positive integers that satisfy the given conditions:

$(10, 90), (12, 36), (18, 18)$

When solving equations like $\frac{1}{a} + \frac{1}{b} = \frac{1}{n}$, the factoring trick $(a-n)(b-n) = n^2$ is incredibly powerful. This transforms a fraction problem into a much simpler factorization problem!