Problem 1.   Let n ≥ 100 be an integer. Ivan writes the numbers n, n + 1, . . . , 2n each on different cards. He then shuffles these n + 1 cards, and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.
proposed by Australia

Solution

The problem requirement is equivalent to the proof: When dividing the set  {n, n + 1,...,2n) into two subsets, there exists a two-element subset whose sum is a perfect square.

Let x be the largest positive integer such that  x2 ≤ n. Then we have  n < (x + l)2. ie n ≤ x2 + 2x. Since  n ≥ 100 so (x + l)2 > 100, ie  x > 9. From here, we have

2n - (2x2 - 2) ≥ 2x2 - (2x2 - 2) > 0,

(2x2 - 2) - (2x2 - 4x + 3) = 4x - 5 > 0,

(2x2 -4x + 3)- (2x2 - 8x + 6) = 4x - 3 > 0.

(2x2 - 8x + 6)-n ≥ (2x2 - 8x + 6)- (x2 + 2x) = x(x - 10) + 6 > 0.

I guess

                     n < 2x2 - 8x + 6 < 2x2 - 4x + 3 < 2x2 - 2 <2 n.

Thus, three numbers 2x2-8x+6, 2x2-4x+3 and 2x2-2 are three numbers in the set  A ={n, n + 1,...,2n).

When the set A is divided into two subsets, two of these three numbers will belong to the same subset. Which (2x2 - 8x + 6) + (2x2 - 4x + 3) = (2x - 3)2, (2x2 - 8x + 6) + (2x2 - 2) = (2x - 2)2 and (2x2 - 4x + 3) + (2x2 - 2) = (2x -1)2 are all perfect squares, so we have something to prove.

 
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