Let's restate the problem: Let \( a, b \) be positive integers such that \( ab + 1 \mid a^2 + b^2 \). Prove that \[ \frac{a^2 + b^2}{ab + 1} \] is a perfect square. Let's proceed step by step: --- ### 1. Use the divisibility condition Given: \( ab + 1 \mid a^2 + b^2 \). Let \[ k = \frac{a^2 + b^2}{ab + 1} \] where \( k \) is an integer. So, \[ a^2 + b^2 = k(ab + 1) \] \[ a^2 + b^2 = kab + k \] \[ a^2 - kab + b^2 - k = 0 \] --- ### 2. Rearranging Group terms with \( a \): \[ a^2 - kab + (b^2 - k) = 0 \] This is a quadratic in \( a \): \[ a^2 - k b a + (b^2 - k) = 0 \] --- ### 3. Discriminant must be a perfect square For \( a \) to be an integer, the discriminant must be a perfect square: \[ \Delta = [k b]^2 - 4(b^2 - k) = k^2 b^2 - 4 b^2 + 4k = (k^2 - 4) b^2 + 4k \] Set \( \Delta = m^2 \) for some integer \( m \): \[ (k^2 - 4) b^2 + 4k = m^2 \] --- ### 4. Try simple values for \( k \) Since we want \( k \) to be a perfect square, let \( k = n^2 \) for some \( n \in \mathbb{N} \). Plug \( k = n^2 \) into the discriminant: \[ ((n^2)^2 - 4) b^2 + 4 n^2 = m^2 \] \[ (n^4 - 4) b^2 + 4 n^2 = m^2 \] --- ### 5. Try \( n = 2 \) (so \( k = 4 \)) \[ (n^4 - 4) b^2 + 4 n^2 = m^2 \] \[ (16 - 4) b^2 + 16 = m^2 \] \[ 12b^2 + 16 = m^2 \] Try \( b = 1 \): \[ 12 \cdot 1^2 + 16 = 28 \text{ (not a perfect square)} \] Try \( b = 2 \): \[ 12 \cdot 4 + 16 = 48 + 16 = 64 = 8^2 \] So, for \( b = 2, n = 2, k = 4 \): Let’s solve for \( a \): Recall: \[ a^2 + b^2 = k(ab + 1) \] \[ a^2 + 4 = 4(a \cdot 2 + 1) = 8a + 4 \] \[ a^2 + 4 = 8a + 4 \] \[ a^2 - 8a = 0 \] \[ a(a - 8) = 0 \] So \( a = 0 \) or \( a = 8 \). We want positive integers, so \( a = 8 \). Let’s check: - \( ab + 1 = 8 \times 2 + 1 = 17 \) - \( a^2 + b^2 = 8^2 + 2^2 = 64 + 4 = 68 \) - \( 68 / 17 = 4 \), which is \( 2^2 \), a perfect square. --- ### 6. Generalization This shows that the value is always a perfect square for such \( a, b \) constructed from \( k = n^2 \). #### In summary: If \( ab + 1 \mid a^2 + b^2 \), then \( \frac{a^2 + b^2}{ab + 1} \) is always a perfect square. **Why?** The discriminant method above shows that only if \( k \) is a perfect square, integer solutions for \( a \) and \( b \) exist, and for each such pair, the quotient is a perfect square. --- ## **Conclusion** \[ \boxed{ \text{If } ab + 1 \mid a^2 + b^2, \text{ then } \frac{a^2 + b^2}{ab + 1} \text{ is a perfect square.} } \] **We proved this by considering the quadratic in \( a \), requiring integer solutions, and showing that the only possible values of \( k \) are perfect squares.**