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Algebra Difficulty 6.5 National olympiad Prove it Ukraine

nn natural numbers are written on the board. You can add only natural numbers in the form a+bab\frac{a+b}{a-b} where aa and bb are the numbers already written on the board. It appears that by doing so you can make any natural number appear on the board. Calculate the least value of nn and find the numbers initially written (consider all the cases).

Solution

As (a+b)>(ab)(a + b) > (a - b), you can not obtain 11 performing the operations allowed. Therefore it should be written on the board, but one number is not enough. Let's show that two numbers will be enough. Let another number of the two be xx. x+1x1\frac{x+1}{x-1} is the only number which can be obtained in the first step. Since it is a natural number, x+1x12(x+1)2x2\frac{x+1}{x-1} \ge 2 \Rightarrow (x+1) \ge 2x-2 or x3x \le 3. Thus the second number should be 22 or 33. We obtain the two possible sets: {1,2}\{1,2\} and {1,3}\{1,3\}.

Let's prove that they both satisfy the condition. As 2+121=3\frac{2+1}{2-1} = 3 and 3+131=2\frac{3+1}{3-1} = 2, in the first step we obtain the set {1,2,3}\{1,2,3\} in both cases. Now we have to prove that any natural number greater than 33 can be obtained from these three numbers.

Let's assume that we've already obtained the set {1,2,3,...,(2k+1)}\{1,2,3,...,(2k+1)\}. Let's show how we can obtain the next two numbers. We obtain number (k+2)+(k+1)(k+2)(k+1)=2k+3\frac{(k+2)+(k+1)}{(k+2)-(k+1)} = 2k+3 from numbers (k+1),(k+2)(k+1), (k+2). Next we obtain (2k+3)+(2k+1)(2k+3)(2k+1)=2k+2\frac{(2k+3)+(2k+1)}{(2k+3)-(2k+1)} = 2k+2 from numbers (2k+3),(2k+1)(2k+3), (2k+1). This implies the desired result.

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