Square-free integer Is it true that for any positive integer, the prime decomposition of N = 10^2n + 10^n + 1 is of the form N = pqrs... with distinct numbers p < q < r < s <...?
For the first values of n,hereafter the prime decompositions:
n = 1, N = 111 = 3*37
n = 2, N = 10101 = 3*7*13*37
n = 3, N = 1001001 = 3*333667
n = 4, N = 100010001 = 3*7*13*37*9901
n = 5, N = 10000100001 = 3*31*37*2906161
n = 6, N = 1000001000001 = 3*19*52579*333667
n = 7, N = 100000010000001 = 3*37*43*1933*10838689
n = 8, N = 10000000100000001 = 3*7*13*37*9901*99990001
etc....

No, for $n=14$ we have
$$
10^{2n}+10^n+1=3^1\cdot 7^2\cdot 13\cdot 37\cdot 43\cdot 127\cdot 1933\cdots .
$$
So this number is not squarefree. In principle, we have seen this question here, a bit differently:
Prove that all numbers $10^n + 1$ are square free
This claim also was not true.
