Let P1(x) = x2 + a1x + b1 and P2(x) = x2 + a2x + b2 be two quadratic p

1.	Let P1(x) = x2 + a1x + b1 and P2(x) = x2 + a2x + b2 be two quadratic polynomials with integer coeffcients. Suppose a1 ≠ a2 and there exist integers m ≠ n such that P1(m) = P2(n), P2(m) = P1(n). Prove that a1 - a2 is even.
Answer» We have m² + a₁m + b₁ = n² + a₂n + b₂ n² + a₁n + b₁ = m²+ a₂m + b_2. Hence (a₁ - a₂)(m + n) = 2(b₂ - b₁); (a₁ + a₂)(m - n) = 2(n² - m²). This shows that a₁+ a₂ = -2(n + m). Hence 4(b₂ - b₁) = a²₂ - a²₁. Since a₁ + a₂ and a₁ - a₂ have same parity, it follows that a₁ - a₂ is even.

Let P1(x) = x2 + a1x + b1 and P2(x) = x2 + a2x + b2 be two quadratic polynomials with integer coeffcients. Suppose a1 ≠ a2 and there exist integers m ≠ n such that P1(m) = P2(n), P2(m) = P1(n). Prove that a1 - a2 is even.

Answer»

We have

m² + a₁m + b₁ = n² + a₂n + b₂

n² + a₁n + b₁ = m²+ a₂m + b_2.

Hence

(a₁ - a₂)(m + n) = 2(b₂ - b₁); (a₁ + a₂)(m - n) = 2(n² - m²).

This shows that a₁+ a₂ = -2(n + m).

Hence

4(b₂ - b₁) = a²₂ - a²₁.

Since a₁ + a₂ and a₁ - a₂ have same parity, it follows that a₁ - a₂ is even.

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Let P1(x) = x2 + a1x + b1 and P2(x) = x2 + a2x + b2 be two quadratic polynomials with integer coeffcients. Suppose a1 ≠ a2 and there exist integers m ≠ n such that P1(m) = P2(n), P2(m) = P1(n). Prove that a1 - a2 is even.

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