The congruence equation solution of n ^ 2 + N + 24 divisible by 2010

The congruence equation solution of n ^ 2 + N + 24 divisible by 2010

nn+n+24==0 mod 2010
Its solution is
4 (NN + N + 24) = = 0 mod 2010 part
First solve the latter, and then replace it with the test
(2n+1)^2+95==0 mod 2010
Let y = 2n + 1, that is, YY = - 95 mod 2010
It is easy to see that y = = 0 Mod 5, let y = = 5Z, that is 25zz = - 95 mod 2010, that is 5zz = - 19 mod 402
402 = 2 * 3 * 67, so ZZ = 1 mod 2,5zz = - 1 mod 3,5zz = - 19 mod 67
That is Z = = 1 mod 2, z = = 1 or - 1 mod 3, ZZ = (- 19-67 * 3) / 5 = - 44 = = 23 mod 67
So 4zz = = 92 = = 25 mod 67, so 2Z = = 5 or - 5 mod 67
Z = 36 or 31 mod 67
from
2n + 1 = 5Z, z = = 1 mod 2, z = = 1 or - 1 mod 3, z = 36 or 31 mod 67. Using the Chinese remainder theorem inversely, four kinds of solutions are obtained and then replaced

Can 2009 square + 2009 be divided by 2010

2009²+2009
=2009(2009+1）
=2009×2010
It's divisible by 2010

Given that a and B are positive integers and satisfy a + Ba2 + AB + B2 = 449, the value of a + B is obtained

From 49 (a + b) = 4 (A2 + AB + B2) and a, B are all positive integers, so there exists a positive integer k, so that a + B = 4K ① and A2 + AB + B2 = 49K, that is, (a + b) 2-AB = 49K, so ab = 16k2-49k. ② so a, B are the equations about X, x2-4kx + (16k2-49k) = 0 ③ (this can also be regarded as substituting ① into ② and arranging into a similar equation about a, ③). Two positive integer roots are obtained. From △ = 16k2-4 (16k2-49k) ≥ 0, 0 is obtained When k = 1, 2 and 3, equation 3 has no positive integer root; when k = 4, equation 3 is x2-16x + 60 = 0, and the solution is X1 = 10, X2 = 6. So a + B = 4K = 16