If you add 100 to a positive integer, it's a perfect square number. If you add 168, it's another perfect square number?

If you add 100 to a positive integer, it's a perfect square number. If you add 168, it's another perfect square number?

In the process of solution, the unknown number is added continuously. Let this number be m, then 100 + M = 10 ^ 2 + 20n + n ^ 2, that is: M = 20n + n ^ 2; 168 + M = 10 ^ 2 + 68 + M = 10 ^ 2 + 68 + n ^ 2 + 20n = 10 ^ 2 + 20x + x ^ 2 68 + n ^ 2 + 20n = 20x + x ^ 2 x ^ 2-N ^ 2 + 20 (x-n) = 68 (x-n) * (x + n) + 20 (x-n) = 68 (x-n) (x + N + 20) = 68

A positive integer, plus 100 or plus 168, is a complete square. This positive integer is______ ．

Let this positive number be a, then a + 100 = X2, a + 168 = Y2, y2-x2 = 68, (Y-X) (y + x) = 68, because if x, y are odd and even, then y + X, Y-X are odd, and 68 is even, so y + X and Y-X should be the same odd or even. For Factorization of 68, because they are the same odd and even, only 2 and 34 can be decomposed, so y + x = 34, Y-X = 2 (y + X is greater than Y-X), y = 18 From a + 168 = Y2: a = 156, so the answer is 156

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