Part 1 (Why GCD?):

Firstly, let's find the answer for a fixed array (not subarray). How can we do that? Let's use the fact that $$$(x + m * y) \mod m = x \mod m$$$ for every $$$x$$$, $$$y$$$ and $$$m$$$. Every difference of two numbers must be divisible by $$$m$$$ because $$$y$$$ must be an integer. What's the biggest number that divides every $$$a_i$$$? It's just their GCD! $$$m$$$ can't be equals to $$$1$$$ according to the statement, so now the problem is to find the longest subarray with GCD > 1.

Part 2 (Solving for every subarray efficiently):

Let's build GCD sparse table on differences of $$$a$$$. If you're familiar with two pointers technique (if not, there's a very nice lesson in EDU!), you will easily come up with such solution:

1. Let's maintain two pointers — $$$L = 0$$$ and $$$R = 0$$$.

2. While $$$R < n$$$:

   2.1. While $$$GCDQuery(L, R) = 1$$$ increase L by 1. Corner case: $$$a_i = 1$$$. Stop the movement of $$$L$$$ pointer when $$$L = R$$$.

   2.2. If $$$GCDQuery(L, R) > 1$$$, then do $$$ans = max(ans, R - L + 2)$$$. Why am I using $$$R - L + 2$$$ instead of $$$R - L + 1$$$? That's because the number of differences in $$$a$$$ equals to $$$n - 1$$$.

1. Print $$$m$$$.

Implementation:

int my_gcd(int a, int b) {
    return (a == 0ll ? b : gcd(b % a, a));        
}
 
int st[200000][20];
 
void build(vector<int> b) {
    for(int i = 0; i < len(b); i++)
        st[i][0] = b[i];
 
    for(int j = 1; j < 20; j++) {
        for(int i = 0; i + (1 << j) <= len(b); i++) {
            st[i][j] = my_gcd(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]); 
        }
    }
}
 
int query(int l, int r) {
    return my_gcd(st[l][log(r - l + 1)], st[r - (1 << log(r - l + 1)) + 1][log(r - l + 1)]);
}
 
void test_case() {
    int n;
    cin >> n;
    vector<int> a(n);
    cin >> a;
    vector<int> b(n - 1);
    for(int i = 1; i < n; i++)
        b[i - 1] = abs(a[i] - a[i - 1]);
 
    debug(b);
    build(b);
    int ans = 1; 
    for(int L = 0, R = 0; R < n - 1; R++) {
        while(L < R && query(L, R) == 1) L++;
        if(query(L, R) != 1) ans = max(ans, R - L + 2);
    }   
 
    cout << ans << "\n";
}