If in the polynomial a ^ m * B ^ 2 + A ^ 3 * B ^ n + 3A ^ m + 3B ^ n + 2 (where m and N are positive integers), there are exactly two terms of the same kind, find the value of M and n

If in the polynomial a ^ m * B ^ 2 + A ^ 3 * B ^ n + 3A ^ m + 3B ^ n + 2 (where m and N are positive integers), there are exactly two terms of the same kind, find the value of M and n

There are five items in a ^ m * B ^ 2 + A ^ 3 * B ^ n + 3A ^ m + 3B ^ n + 2. The only items with the same base are a ^ m * B ^ 2 and a ^ 3 * B ^ n. according to the definition of similar items, the base must be the same, and the index of unknowns must be the same, then M = 3, n = 2

If m n is an integer and 7m-n is a multiple of 6, it is proved that 28m ^ 2 + 31mn-5n ^ 2 can be divided by 18

28m^2+31mn-5n^2=(7m-n)(4m+5n)
7m-n is a multiple of 6 and of course a multiple of 3
4m + 5N = 7m-n-3m + 6N, obviously, - 3M + 6N is a multiple of 3, then 7m-n-3m + 6N is a multiple of 3
That is to say, 4m + 5N is a multiple of 3
7m-n is a multiple of 6
So (7m-n) (4m + 5N) is a multiple of 3 * 6 = 18
That is 28m ^ 2 + 31mn-5n ^ 2 can be divided by 18

For integer n, (n + 2) (n + 7) - (n-1) (n + 4), is it a multiple of 6

(n+2)(n+7)-(n-1)(n+4)
=(n²+9n+14)-(n²+3n-4)
=6n+18
=6(n+3)
For the integer n, it must be a multiple of 6