Denote $X_{n}=5^{5^{n}}$, we have to prove that $X_{2}^{4}+X_{2}^{3}+X_{2}^{2}+X_{2}+1$ isn't prime. Since the polynomial is irreducible we will look for a factorisation of the form $(X_{2}^{2}+aX_{2}+b)^{2}-5X_{2}(X_{2}+c)^{2}$, since $5X_{n}$ is a perfect square. Filling in coefficients gives us \[\begin{array}{rcl}\frac{X_{3}-1}{X_{2}-1}&=& (X_{2}^{2}+3X_{2}+1)^{2}-5X_{2}(X_{2}+1)^{2}\\ &=& (5^{50}+5^{38}+3\cdot 5^{25}+5^{13}+1)(5^{50}-5^{38}+3\cdot 5^{25}-5^{13}+1).\end{array}\]

Peter wrote: Prove that $ \frac {5^{125} - 1}{5^{25} - 1}$ is a composite number. An epsilon remark. This was proposed by Korea

Let $ x=5^{25}$ , then $ 5^{125}-1$ $ =x^{5}-1$ $ =(x-1)(x^{4}+x^{3}+x^{2}+x+1)$ $ =(x^{4}+9x^{2}+1+6x^{3}+6x+2x^{2}-5x^{3}-10x^{2}-5x)(x-1)$ $ =((x^{2}+3x+1)^{2}-5x(x+1)^{2})(x-1)$ $ =((x^{2}+3x+1)^{2}-(5^{13}(x+1)^{2})(x-1)$ $ =(x^{2}+3x+1+5^{13}(x+1))(x^{2}+3x+1-5^{13}(x+1))(x-1)$ Which implies that $ \frac {5^{125} - 1}{5^{25} - 1}$ is a composite number!

Yes... that's exactly what I wrote, no?

With the similar method,one can prove that when p=4k+1 is prime,then $ \frac {p^{p^{3}} - 1}{p^{p^{2}} - 1}$is a composite number.

http://www.mathlinks.ro/viewtopic.php?t=283377

Can I just ask, how does one come up with these factorisations? The solutions described above are very beautiful, I just cannot see how to come up with them!

Yes! What are the motivations of these factorizations ?

lightstar_01 wrote: Can I just ask, how does one come up with these factorisations? The solutions described above are very beautiful, I just cannot see how to come up with them! Had the same thought. Someone told me that it's a whole class of factorisation known as aurifeuillean factorization