Cho mình xin lời giải bài 5 với
Show that non-negative integers a ≤ b satisfy (a2 + b2) = n2(ab + 1), where n is a positive integer, iff they are consecutive terms in the sequence ak defined by a0 = 0, a1 = n, ak+1 = n2ak - ak-1.
If n = 1, then the sequence is 0, 1, 1, 0, -1, -1, 0, 1, 1, ... . Thus the only consecutive non-negative terms a, b with a ≤ b are 0, 1 and 1, 1 both of which satisfy the equation.
Conversely suppose that a ≤ b is a solution for n = 1. Then a2 + b2 = ab + 1. If 1 < a, then 1 < a2, ab ≤ b2, so 1 + ab < a2 + b2. Contradiction. So a = 0 or 1. If a = 0, then b2 = 1, so b = 1. If a = 1, then 1 + b2 = b + 1, so b = 0 or 1, but b ≥ a = 1, so b = 1. Thus the only solutions are a = 0, b = 1, or a = 1, b = 1.
So assume n > 1. It is a trivial induction to show that ak < ak+1. Now it is an easy induction on k to show that consecutive terms ak-1, ak satisfy the equation. It is true for k = 1: (02 + n2) = n2(0.n + 1). Suppose it is true for k. Then we have ak+1 + ak-1 = n2ak. Hence ak+12 - ak-12 = n2ak(ak+1 - ak-1). Adding to ak2 + ak-12 =n2(akak-1 + 1), we get ak+12 + ak2 =n2(ak+1ak + 1), which completes the induction.
Now suppose that a ≤ b is any solution in non-negative integers of a2 + b2 = n2(ab + 1). The idea is to show that n2a - b, a is a smaller solution.
If a = b, then 2a2 = n2(a2 + 1) ≥ 4(a2 + 1) > 2a2. Contradiction. So a < b. If a = 0, then b2 = n2, so b = n. This solution belongs to the sequence. So assume a > 0.
If b > n2a, then b >= n2a + 1, so b2 ≥ n2ab + b > n2ab + n2a ≥ n2(ab + 1), so a2 + b2 > n2(ab + 1). Contradiction. So n2a - b ≥ 0. If n2a ≥ a + b, then n2ab ≥ ab + b2 > a2 + b2. Contradiction. So n2a - b < a. Finally, (n2a - b)2 + a2 = n4a2 - 2n2ab + a2 + b2 = n2( a(n2a - b) + 1) + (a2 + b2 - n2(ab + 1) ), so if a, b is a solution, then so is n2a - b, a.
Thus if we start with any solution 0 < a ≤ b, we can derive a solution a' < a, where the relationship between a', a and b is the same as that between ak-1, ak and ak+1. This process must terminate, so eventually we get a solution 0, c. But we have shown that this must be 0, n. So we must have been moving down the sequence ak. Hence a, b must be consecutive terms in that sequence.
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