A positive integer, if 100 is a perfect square, if 168 is another perfect square, then the positive integer is______ ．

Let the number be n. from the meaning of the question, we can get: n + 168 = A2 （1）n+100=b2… (2) (1) - (2), 68 = A2-B2 = (a + b) (a-b), because 68 = 1 × 68 = 2 × 34 = 4 × 17, there are only three cases, namely: ① a + B = 68, A-B = 1; ② a + B = 34, A-B = 2; ③ a + B = 17, A-B = 4; because ① A and B have no integer solution, exclude; ② calculate a = 18, B = 16, so: n = 182-168 = 162-10 = 156; ③ A and B have no integer solution, exclude To sum up, only n = 156, that is, the number sought. So the answer is: 156

If 100 is added to a positive integer, it is a complete square number; if 168 is added, it is another complete square number
100 + M = 10 ^ 2 + 20n + n ^ 2, that is: M = 20n + n ^ 2. How is this m = 20n + n ^ 2 obtained? Is it arbitrary or what
Thank you
Are all unknowns set at will???

Let this positive integer be a. according to the meaning of the question 100 + a = B ^ 2168 + a = C ^ 2, then C ^ 2-B ^ 2 = 68 (c + b) (C-B) = 2 × 2 × 17, there are three possibilities: C-B = 1, C + B = 68c-b = 2, C + B = 34c-b = 4, C + B = 17

A positive integer, if 100 is a perfect square, if 168 is another perfect square, then the positive integer is______ ．