We need to prove that $z = \cfrac {5^{125 }- 1}{5^{25} - 1} $ isn't a prime number . This form comes out to be the sum of a geometric progression with the first term $1$ and ratio $5^{25}$, with five terms. So it reduces to the form:
$$z=1 + 5^{25} + 5^{50} + 5^{75} + 5^{100} $$
Now you could present it as a difference of two squares, but that turns out to be a tedious task.
Any help after this? Thanks a lot!
It is enough to exploit an Aurifuillean factorization, in particular the one relative to $b=5$ in the table on the linked Wikipedia page, or just the identity
$$ \frac{z^5-1}{z-1}=1+z+z^2+z^3+z^4=(1+3z+z^2)^2 - 5z(1+z)^2 \tag{1}$$ If $z$ is a power of $5$ with and odd exponent, the RHS of $(1)$ is the difference of two squares as wanted. As pointed by wendy.krieger, $(1)$ is a consequence of the theory of Gauss sums applied to cyclotomic polynomials.
The number $x^4+x^3+x^2+x+1$ is the product of two values of $x^2+3x+1 \pm (x+1) \sqrt{5x}$
Since $x = 5^{25}$ makes $\sqrt{5x}=5^{13}$, it has a solution.
This has been known since the time of K.F.Gauss.
Cunningham Project
The number $b^n-1$ represents a number of $n$ digits, each $b-1$, in base b, and thus the divisors correspond to numbers with period of $n$. For example, with $b=10, n=7$, we get the number $9999999$. A good deal of number theory has gone into this.
One can find that there is a unique factor $^bA_n$ for every value of b and n, and that any pair are coprime, unless $p \mid n$ and $m/n$ is a power of $p$. Apart from these and sevenites, the list of factors of $^bA_n$ is just the list of primes with this as a period.
Primes that have a period of n, by Fermat's little theory, are of the form $p=mx+1$. While this greatly reduces the number of primes to check, it really is not enough.
Gauss's quadratic law tells us if the period n divides p an even or odd number of times, and for most values of n, this halves the number of primes to search. But where the base is of the form $b=xy^2$, then the values of either x (for 1 mod 4), or 2x (for 2, 3 mod 4), have an algebraic factorisation of the kind given above.
So, for example, where b=120, x=30, and we would find that ^{120}A_{120a+60} has an algebraic factorisation, that is, odd numbers times 2*30. For b=10, x=10, and we find that ^{10}A_{40a+20} has an algebraic factorisation.
The Cunningham project is to find the factors of these A values, as far as technology permits, for b=2-10, 12. I have done similar runs for base 120. The tables contain some unimaginately large numbers, and more than half of them contain a value larger than $\sqrt{A}$.
The number in this example is $^5A_{125}$, which passes the second test (ie algebraic and gaussian factorisation), but the rest is an uninspiringly large number best left to utilities like factor.
For $b^n-a^n$, the same general rules hold, except that the divisors of the base are $ab$, and $abx$ as square defines the value of $x$. So $3^n-2^n$ is a sixth-square, and like all sixth-squares (6, 24, 54, ...), it follows that $A_n$ has a factorisation if $n$ is an twelve times an odd number.
