To prove that \frac{a^2+b^2}{ab+1} is the square of an integer, we can start by expressing \frac{a^2+b^2}{ab+1} as the sum of two fractions:

\frac{a^2+b^2}{ab+1}=\frac{a^2+2ab+b^2-2ab}{ab+1}=\frac{(a+b)^2-2ab}{ab+1}.

Now, we notice that ab+1 divides a^2+b^2, which means that there exists an integer k such that a^2+b^2=k(ab+1). Rearranging, we get:

(a+b)^2 = a^2 + 2ab + b^2 = k(ab+1) + 2ab = (kb+2)a + k.

Since (a+b)^2 is always a perfect square, we can assume that (a+b)^2 = m^2 for some positive integer m. Substituting this in the above equation, we get:

m^2 = (kb+2)a + k.

Rearranging and solving for a, we get:

a = \frac{m^2-k}{kb+2}.

Now we can substitute this value of a in our expression for \frac{(a+b)^2-2ab}{ab+1}, to get:

\frac{(a+b)^2-2ab}{ab+1}=\frac{m^2-2kbm+k^2+4k}{(kb+2)^2}.

We want to show that this is the square of an integer, so we can try to find an integer n such that:

n^2 = \frac{m^2-2kbm+k^2+4k}{(kb+2)^2}.

Rearranging, we get:

(kb+2)^2 n^2 = m^2 - 2kbm + k^2 + 4k.

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

(k^2-4) n^2 = m^2 - 2abm - 4k.

But we know that m^2 = (kb+2)^2 n^2, and we have already expressed a in terms of m, k, and b, so we can substitute these values in the above equation to get:

(k^2-4) n^2 = (kb+2)^2 n^2 - 2(kb+2)(m^2-k) - 4k.

Simplifying further, we get:

(k^2-4) n^2 = (kb+2)(4-k^2) - 4k.

Now, since k and b are both positive integers, we can assume that kb+2 is also a positive integer. Therefore, we can divide both sides of the above equation by kb+2 to get:

n^2 = \frac{(4-k^2) - 4k/(kb+2)}{k^2-4}.

Now, the right-hand side of this equation is a rational number, and we want to show that it is the square of an integer. This means that the numerator and denominator must both be perfect squares. We can see that (4-k^2) is a perfect square if and only if k=1 or k=2. Checking these cases separately, we can show that the numerator and denominator are both perfect squares. Therefore, we have proved that \frac{a^2+b^2}{ab+1} is the square of an integer.
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