If there is a natural number whose sum with 160 is equal to the square of a certain number, and whose sum with 84 is equal to the square of another number, then the natural number is______ ．

Let this natural number be x, the sum of it and 160 is the square of M, and the sum of it and 84 is the square of N, so there are: x + 160 = m2, x + 84 = N2, m2-n2 = (m-n) × (M + n) = 76, because M-N and M + n have the same parity, 76 is even, so M-N and M + n are both even, and 76 is decomposed into two even multiplicative products, only 76 = 2 × 38, so: (m-n) (M + n) = 2 × 38 = (20-18) × (20 + 18) m = 20, n = 18 So x + 160 = 202, x = 240

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