To show that a² + b² / ab + 1 is the square of an integer, we need to find an integer k such that:

a² + b² / ab + 1 = k²

To start, we can simplify the expression on the left-hand side by multiplying both the numerator and denominator by a - b:

(a² + b²)(a - b) / (ab + 1)(a - b) = k²

Expanding the numerator on the left-hand side and canceling out the common factor of a - b in both the numerator and denominator, we get:

(a³ - b³) + ab(a - b) / (a²b - ab²) + 1 = k²

Since ab + 1 divides a² + b², we can write:

a² + b² = n(ab + 1)

where n is some positive integer. Substituting this expression into the numerator of the left-hand side, we get:

(a³ - b³) + ab(a - b) = a³ - b³ + ab(a - b) = (a - b)(a² - ab + b²) = (a - b)n

Substituting this expression back into our original equation and simplifying, we get:

(a - b)n / ab - 1 = k²

Multiplying both sides by ab - 1, we get:

(a - b)n = k²(ab - 1)

Since a, b, and n are positive integers, we have:

a - b > 0 and ab - 1 > 0

Therefore, k² must also be a positive integer, and we can write:

k² = (a - b)n / (ab - 1)

Now we need to show that k is an integer. To do this, we will use the fact that ab + 1 divides a² + b². This can be written as:

a² + b² = m(ab + 1)

where m is some positive integer. Substituting this expression into the original equation, we get:

m(ab + 1) / ab + 1 = k²

Multiplying both sides by ab + 1, we get:

m(ab + 1) = k²(ab + 1)

Expanding the left-hand side and simplifying, we get:

mab + m = k²ab + k²

Dividing both sides by ab, we get:

m/a + m/b = k² + m/(ab)

Since m/(ab) is a positive integer and m/a and m/b are both positive integers, k² must also be a positive integer, and k is an integer.

Therefore, we have shown that a² + b² / ab + 1 is the square of an integer.
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