In known sequence {an}, A1 = 1, A2 = 0, for any positive integer n, m (n > m) satisfies (an) ^ 2 - (AM) ^ 2 = an man + m, then a119

In known sequence {an}, A1 = 1, A2 = 0, for any positive integer n, m (n > m) satisfies (an) ^ 2 - (AM) ^ 2 = an man + m, then a119

(an) ^ 2 - (AM) ^ 2 = an man + m holds for any positive integer n, m (n > m),
Let n = 119, M = 1 or 2
(a119)^2-(a1)^2=a119-a119+1=1
(a119)^2-(a2)^2=a119-2a119+2=-a119+2
(a119)^2=2
(a119)^2+a119=2
There is no solution

The sequence {an} satisfies A1 = 1, an (n = 2,3,.) is a non-zero integer, and for any positive integer m and natural number k, there is - 1 ≤ am + am
The sequence {an} satisfies A1 = 1, an (n = 2,3,.) is a non-zero integer, and for any positive integer m and natural number k, - 1 ≤ am + a (M + 1) +. + a (M + k) ≤ 1
(1) Find A2, a3.a4 and write the general term formula of sequence {an}

Let k = 0, then - 1

Given that an ≠ 0 in the arithmetic sequence {an}, if M ＞ 1, m ∈ n *, and am-1 + am + 1-am2 = 0, s2m-1 = 38, then the value of M is?
M-1 and M + 1 are subscripts, AM2 should be a, M is subscript, 2 is square

m=10
If the tolerance is D, the original formula is (am-d) + (am + D) - AM2 = 0
The solution is am = 2
S2m-1=[(2m-1)(a1+a2m-1)]\2=2(2m-1)am\2=am(2m-1)=2(2m-1)=38
The solution is m = 10