Find the 1973rd non-perfect square.
Clarification: Non-perfect squares are positive integers which cannot be expressed as , where is an integer. For example: 2, 3, 5, 7, 8,... and 2 is the first non-perfect square.
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Let q ( n ) be the n th non-perfect integer. Then the sequence of non-perfect integers up to q ( n ) = k 2 + 1 , where k 2 is the k th perfect square is as follows:
{ q ( n ) } ⟹ n q ( k 2 − k + 1 ) = k 2 + 1 − k terms 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 1 0 , ⋯ k 2 , k 2 + 1 = k 2 − k + 1 = k 2 + 1
Now the largest perfect square smaller than 1973 is ⌊ 1 9 7 3 ⌋ = 4 4 . Then, we have:
q ( k 2 − k + 1 ) q ( 4 4 2 − 4 4 + 1 ) q ( 1 8 9 3 ) q ( 1 8 9 3 + 8 0 ) q ( 1 9 7 3 ) = k 2 + 1 = 4 4 2 + 1 = 1 9 3 7 = 1 9 3 7 + 8 0 = 2 0 1 7 Putting k = 4 4